计算：3次根号3又8分之3+3次根号0.001 ； 3次根号125分之61-1-3次根号-2又27分之10_百度知道
计算：3次根号3又8分之3+3次根号0.001 ； 3次根号125分之61-1-3次根号-2又27分之10
算； ②3次根号125分之61-1-3次根号-2又27分之10化简：①3次根号3又8分之3+3次根号0.001
提问者采纳
1=1.5+01,3次根号3又8分之3=3次根号8分之27=2分之3=1，3次根号0.5.001 =3次根号1000分之一=0，3次根号-2又27分之10=-3分之4
-5分之4-（-3分之4）=15分之8化简过程打着太麻烦.62,3次根号125分之61-1=-5分之4.1
3,4-2*根号24,3*根号2-6*根号5-3*根号2=-6*根号5
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太感谢了，真心有用
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出门在外也不愁已知直线x-ay-4=0与直线y=-2x+4的夹角为arccos5分之2倍根号5，求a_百度知道
已知直线x-ay-4=0与直线y=-2x+4的夹角为arccos5分之2倍根号5，求a
数学好的大哥帮下忙
) 再算出两个式子的斜率 用夹角公式|(1/根号5 sin^2+cos^2 =1 算得sinA=1/(1-2/根号5 (夹角小于等于90'2化简得3a^2+4a=0解得a=0 a=-4&#47首先得cosA=2/a)|=1/a+2)&#47
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a)|=1/a+2)/(1-2&#47，利用夹角公式|(1&#47由夹角为arccos5分之2倍根号5那么可得这个角的余弦是5分之2倍根号5根据sin^2+cos^2 =1
且夹角小于90度可得到该角正弦为5分之根号5然后算出两条直线的斜率;2解得a=0 或者a=a=-4&#47
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出门在外也不愁问几道题 1：已知a减a分之一为2，求a方减a方分之一 2：已知a方减3a加1为零，求a的4次方加1的和分之一_百度知道
问几道题 1：已知a减a分之一为2，求a方减a方分之一 2：已知a方减3a加1为零，求a的4次方加1的和分之一
要过程，详细一点
用完全平方公式，我看不懂根号
2a^2= { (-3±根号5)/(a^4+1)=1/a^2+2=8(a+1/a=2(a-1&#47：a^2-3a+1=0a^2=3a-1X^2-3X+1=0a={-3±根号(3^2-4)}/a^2=4+2=6a^2+1/a)=±2根号2 * 2 = ±4根号22;21/{7*[7^2-(3根号5)^2}=(7±3根号5)/2]}=2(7±3根号5)/a)^2=8a+1/(9a^2-6a+2)=1&#471a-1/(7a^2)=1/a=±2根号2a^2-1/2={-3±根号5}/a^2-2=4a^2+1/{9a^2-2a^2}=1/a)^2=4a^2+1/a^2=(a+1/{7*[7±3根号5)/2}^2=（7±3根号5)/a)(a-1/{9a^2-2(3a-1)}=1/{(3a-1)^2+1}=1&#47
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a^2=62？解;a^2-2=4
a^2+1&#47：已知a方减3a加1为零1，求a的4次方加1的和分之一;(7a^2) ;a=2
(a-1&#47：a^2-3a+1=0
(a^2+1)^2=9a^2
a^4+2a^2+1=9a^2
a^4+1=7a^2 即，求a方加a方分之一 ？解;(a^4+1)=1&#47：a-1&#47：已知a减a分之一为2：1/a)^2=4
1、因a-1/a=2则(a-1/a)^2=4a^2+1/a^2=6(a+1/a)^2=8a+1/a=±2√2∴a^2-1/a^2=(a+1/a)(a-1/a)=2*(±2√2)=±4√22、a^2-3a+1=0
a=(3±√5)/2
a^2=(7±3√5)/2又a^2+1=3aa^4+1=7a^2∴1/(a^4+1)=1/(7a^2)=2/[7(7±3√5)]=(7+3√5)/14或(7-3√5)/14
1,平方得a^2-2+1/a^2=4,a^+2+1/a^2=8,a+1/a=+-2*根号2，则所求式=(a-1/a)(a+1/a)=+-8根号22，a^4+1=(a^2+1)^2-2a^2=(3a)^2-2a^2=7a^2=21a-7
由原式解得a=(3+-根号5)/2,21a-7=49+-21*根号5/2
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出门在外也不愁已知a=根号2+ 1，求代数式a?-2a-3分之a?-a÷a-3分之a的值（要具体计算过程）
已知a=根号2+ 1，求代数式a?-2a-3分之a?-a÷a-3分之a的值（要具体计算过程）
不区分大小写匿名
等于根号2减1
我要具体过程
1.计算：1÷（2x+x分之1-x?）
a分之a-b÷（a-a分之2ab-b?）（要具体计算过程）
2.解下列方程：（1）x?+x分之5x+2=x+1分之3
（2）2x-5分之2x-2x+5分之2=1
（要具体解方程过程和检验过程）
能给分么？
解下列方程:(1)x?+x分之5x+2=x+1分之3
（2）2x-5分之2x-2x+5分之2=1 （要具体解方程过程和检验过程）
计算：（1）|-4|-根号9+3的-2次方-（2011-2分之π）的0次方；
（2）（b分之-a）?÷（5b分之2a?）?×5b分之a；
（3）x?-4分之3x-6÷x?+4x+4分之x+2-2；（要具体计算过程）
解分式方程：（1·）x?-1分之2=负x-1分之1；
（2）x-2分之x-x?-4x+4分之2=1.（要具体解方程过程和具体验算过程）
没必要回答你了 自己去找别人
[a^2-a]/(a^2-2a-3) ×（a-3）/a
=a(a-1)/[(a-3)(a+1)]*(a-3)/a
=(a-1)/(a+1)
将a=根号2+1代入可得：
原式=根号2/（2+根号2）=根号2*（2-根号2）/2=根号2-1
1.计算：1÷（2x+x分之1-x?） a分之a-b÷（a-a分之2ab-b?）（要具体计算过程） 2.解下列方程：（1）x?+x分之5x+2=x+1分之3 （2）2x-5分之2x-2x+5分之2=1 （要具体解方程过程和检验过程）
1.计算：1÷（2x+x分之1-x?） a分之a-b÷（a-a分之2ab-b?）（要具体计算过程） 2.解下列方程：（1）x?+x分之5x+2=x+1分之3 （2）2x-5分之2x-2x+5分之2=1 （要具体解方程过程和检验过程）
计算：（1）|-4|-根号9+3的-2次方-（2011-2分之π）的0次方； （2）（b分之-a）?÷（5b分之2a?）?×5b分之a； （3）x?-4分之3x-6÷x?+4x+4分之x+2-2；（要具体计算过程） 解分式方程：（1·）x?-1分之2=负x-1分之1； （2）x-2分之x-x?-4x+4分之2=1.（要具体解方程过程和详细验算过程）
如右图所示，若物重50N，拉力为20N。（1）动滑轮的重力式多少？(2)若物体上升2m，求F移动的距离？（3）若F移动距离为15M，求物体上升的高度是多少？
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数学领域专家已知实数a，b，有下列结论： （1）a=1，b=1，则根号ab≤1； （2）a=2分之1，b=5分之2，则根号ab≤9分之20_百度知道
已知实数a，b，有下列结论： （1）a=1，b=1，则根号ab≤1； （2）a=2分之1，b=5分之2，则根号ab≤9分之20
b与根号ab的关系，并判断当a=9.根据以上计算提供的信息，则根号ab≤3(3)a=2，b=5,b=3，请猜想a,则根号ab≤2分之5，根号ab的范围，b=10时；（4）a=1
提问者采纳
=2ab a=9，b=10时;=2根号下ab因为a^2+b^2&gta+b&gt，根号下ab&=（a+b）/2=19&#47
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系为，根号ab≤2分之19想了很久：根号ab≤2分之（a+b)当a=9，b=10时
猜想 (a+b )/2 &= 根号(ab)根号(ab) &= (9+10)/2 = 19/2
The treatment of gastroesophageal reflux disease overviewNow that is mainly due to the esophagus and gastrointestinal motility disorder, because the esophagus to the stomach and duodenal contents reflux and reflux down the defense mechanisms of the esophageal mucosa matter the result of the attacks increased, which is the next most important esophageal sphincter (LES) pressure drop. Therefore, its treatment principles include ① reduce o ② prev ③ prevention and treatment of serious complications. I general treatment, including lifestyle and eating habits change, raise the bed to reduce nighttime reflux, low fat, low sugar diet, quit smoking wine, to avoid satiation, limit spicy food and so on. 2 acid-suppressing drugs, including H, receptor antagonists and proton pump inhibitors, the former through a competitive H2 receptor on the parietal cells to inhibit acid secretion, reducing gastric acid secretion, thereby reducing the acid on the esophageal mucosa injury, drugs, including West for D, ranitidine and so on. Proton pump inhibitors for the wall membrane of a proton pump H a K ATP inhibitor, can effectively inhibit gastric acid secretion lasting. H2 blocker maintenance therapy as effective as proton pump inhibitors. Author: 054001 Hebei Province, soldiers honor 3 prokinetic drug rehabilitation hospital of these drugs increase the esophageal motility, esophageal mucus secretion. ① metoclopramide include: dopamine receptor antagonists, the LES pressure increases, preventing the stomach, duodenum fluid reflux, but it is through blood-brain barrier, causing the central nervous system side effects. Such as insomnia, tremor, movement disorders. Not long- ② domperidone: Peripheral dopamine receptor antagonist, increased LES can promote esophageal motility,Links of London Charms, nausea and vomiting symptoms are particularly effective, side effects can cause high lactation hyperlipidemia cable, such as galactorrhea, breast enlargement, etc. ; ③ 5 a HT4 receptor agonists, cisapride, mosapride, acting on the myenteric plexus to ask the release of acetylcholine, the increase in lower esophageal sphincter pressure, esophageal peristalsis strengthened, accelerated gastric emptying, attention is not with large ring lactone antibiotics, antifungal agents combined, and the other can cause QT interval prolongation, heart disease patients with caution. 4 mucosal protective agent mucosal protective agent for patients with esophagitis, may consider the application. Including sucralfate, cover the stomach flat, well close up, bismuth preparations, sucralfate which can reduce the symptoms and treatment of reflux esophagitis may be sucralfate and ulcer, erosive surface positively charged protein binding, Live + charge in forming a barrier to prevent mucous membrane is digested, can cover the stomach and saliva levels, gastric juice works, forming a layer floating in the viscous glue tour. 5 Surgical treatment Surgical treatment of reflux esophagitis on esophageal stenosis when combined,timberland shoes, can be considered for endoscopic dilation. Tips for endoscopy in patients with Barrett's esophagus, endoscopic suturing can be used or added anti-reflux fundoplication surgery. Currently,vibram five fingers singapore, the majority of gastroesophageal reflux disease need medical treatment. Application mainly mucosal protective agent, acid-suppressing agents and prokinetic drugs, of which the best efficacy of proton pump. Elbow joint nerve disease in 1 case Liubao Xun Xie Hongfeng Han Fusheng Yang Qijun Xiegang Long patient,GHD gl?tteisen, female, 55 years old. Due cervical syringomyelia 1O years, a fall two weeks left elbow swelling with fever 1 day hospitalization. Physical examination: T39 ℃, no abnormal cardiopulmonary abdomen, right convex thoracic spine, thoracic asymmetry. His hands were claw deformity, left elbow was swollen, red skin, skin temperature high, no tenderness, palpable sense of volatility,belstaff outlet, the left elbow against Zhang, both upper limbs was significantly decreased pain and temperature sensation of skin to the ulnar to the double upper limbs Ⅳ, increased muscle tone, brachial two triceps hyperreflexia, Hoffman's sign (+), trunk from the nipple pain and temperature sensation below the level of decline. Lower limb muscle strength grade Ⅳ, increased muscle tone, knee tendon reflex and Achilles tendon hyperreflexia, ankle clonus patellar (+), bilateral Babinski'S sign (+). X ray showed dislocation of the elbow joint, bone-side frosted glass-like changes in intra-articular loose bodies, soft tissue swelling. Diagnosis: cervical syringomyelia, left Author: 065,300 Dachang Hui Autonomous County, Hebei Province People's Hospital, Case Report elbow joint nerve disease complicated by infection. Line elbow puncture, pus out about 400ml of liquid, bad taste. The E. coli bacterial culture on Cyprus oxime cephalosporin sodium sensitive, the anti-infection and improved and discharged after the joint incision and drainage. Discussion of joint disease, also known as Chareot nerve or neuropathic joint arthritis, 25% of the relevant section of the spinal cord lesions of patients with syringomyelia 【l0J. The disease is characterized by pain, joint dysfunction and damage is not proportional. Treatment to remove the cause of treatment mainly due to disease in patients with syringomyelia to conservative therapy.
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