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Bài 1:
3: ĐKXĐ: x>=1
\(x-\sqrt{x+3+4\sqrt{x-1}}=1\)
=>\(x-\sqrt{x-1+2\cdot\sqrt{x-1}\cdot2+4}=1\)
=>\(x-\sqrt{\left(\sqrt{x-1}+2\right)^2}=1\)
=>\(x-\left|\sqrt{x-1}+2\right|=1\)
=>\(x-\left(\sqrt{x-1}+2\right)=1\)
=>\(x-\sqrt{x-1}-2-1=0\)
=>\(x-1-\sqrt{x-1}-2=0\)
=>\(\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}+\sqrt{x-1}-2=0\)
=>\(\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}+1\right)=0\)
=>\(\sqrt{x-1}-2=0\)
=>\(\sqrt{x-1}=2\)
=>x-1=4
=>x=5(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
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3:
ĐKXĐ: x>=0; x<>1
a: \(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)
b: \(x+\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}+1\right)+1>=0+1=1\)
=>\(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
mà 2>0
nên \(P=\dfrac{2}{x+\sqrt{x}+1}>0\forall x\) thỏa mãn ĐKXĐ
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\dfrac{6+2\sqrt{5}}{2\sqrt{2}+\sqrt{2}\cdot\left(\sqrt{5}+1\right)}-\dfrac{6-2\sqrt{5}}{2\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)}\)
\(=\dfrac{6+2\sqrt{5}}{2\sqrt{2}+\sqrt{10}+\sqrt{2}}-\dfrac{6-2\sqrt{5}}{2\sqrt{2}-\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{6+2\sqrt{5}}{3\sqrt{2}+\sqrt{10}}-\dfrac{6-2\sqrt{5}}{3\sqrt{2}-\sqrt{10}}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\left(3\sqrt{2}-\sqrt{10}\right)-\left(6-2\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)}{8}\)
\(=\dfrac{18\sqrt{2}-6\sqrt{10}+6\sqrt{10}-10\sqrt{2}-18\sqrt{2}-6\sqrt{10}+6\sqrt{10}+10\sqrt{2}}{8}\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)m=2`
`pt<=>x^2-4x+4-1=0`
`<=>x^2-4x+3=0`
`Delta'=4-3=1`
`<=>x_1=1.x_2=3`
Vậy `m=2` thì `S={1,3}`
`b)Delta'=m^2-(m^2-1)`
`=m^2-m^2+1=1>0`
`=>` pt có 2 nghiệm pb `AAm`
`c)Delta'=1`
`<=>x_1=(-b'+sqrtDelta')/a=m+1,x_2=(-b'-sqrtDelta')/a=m-1`
`=>x_1-x_2`
`=m+1-m+1=2`
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Xét tứ giác BCEF có
\(\widehat{BFC}=\widehat{BEC}=90^0\)
Do đó: BCEF là tứ giác nội tiếp
1. was made
2. have been taken
3. is visited
4. is being painted
5. will be given
6. must be kept
7. was being moved
8. had cancelled
9. were blown
10. was published
11. has been translated
12. to have escaped
13. taken
14. repaired
15. to wait
16. keeping
17. to be
18. will be shown
19. washed
20. being told