2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

a: \(B=\dfrac{x-2\sqrt{x}}{\sqrt{x}-2}-\dfrac{2x+12\sqrt{x}+18}{\sqrt{x}+3}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}-\dfrac{2\left(x+6\sqrt{x}+9\right)}{\sqrt{x}+3}\)

=căn x-2(căn x+3)

=-căn x-6

b: B+8>0

=>-căn x-6+8>0

=>-căn x+2>0

=>-căn x>-2

=>căn x<2

=>0<=x<4

a) \(\sqrt{\left(2x-3\right)^2}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)

\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)

\(\Leftrightarrow5\sqrt{x+2}=20\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

a. \(\sqrt{\left(2x-3\right)^2}=7\)

<=> \(\left|2x-3\right|=7\)

<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)

b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\)  ĐK: \(x\ge-2\)

<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)

<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)

<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)

<=> \(5\sqrt{x+2}=20\)

<=> \(\sqrt{x+2}=4\)

<=> \(\left(\sqrt{x+2}\right)^2=4^2\)

<=> \(\left|x+2\right|=16\)

<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)

c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)             ĐK: \(x\ge3\)

<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)

<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)

<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)

a) I2x-3I=7

        2x-3=7     =>x=5

        2x-3=-7     =>x=-2

b) \(8\sqrt{3x}-5\sqrt{3x}+2\sqrt{3x}=20\)

   5\(\sqrt{3x}=20\)

   3x=16  =>x=16/3

c)  vì câu c dài nên mình chỉ cho đáp án thôi là 0,3,6

vì \(\sqrt{ }\) của 1 số luôn dương nên 3,6 thỏa mãn 

a/ x <hoac= -23/4

b/ x=2

a/ có 2xcăn6 > 2x2=4

=> 2 căn 6 > 3+1

<=> 2 căn 6 - 3 >1

b/ có 3 căn 2 > 3 

=> 3 căn 2 - 9 > -6 

=> 6 > 9- 3 căn 2

a, đổi dấu ở phân số cuối để mẫu thành x-4

rồi sau quy đồng mẫu chung là x-4

bn sẽ rút gọn được

b, theo câu a ta có P = \(\frac{3x-3\sqrt{x}-3}{\left(\sqrt{x-2}\right)\left(\sqrt{x+2}\right)}\)

2 trường hợp

th1 tử và mẫu cùng dương

th2 

tử và mẫu cùng âm

c, thay x= 4 vào biểu thức đã rút gọn ở câu a

`#3107.101107`

`1/2x + 4/5 = 2x - 8/5`

`=> 1/2x - 2x = -4/5 - 8/5`

`=> -3/2x = -12/5`

`=> x = -12/5 \div (-3/2)`

`=> x = 8/5`

Vậy, `x = 8/5`

_____

`\sqrt{x} = 5`

`=> x = 5^2`

`=> x = 25`

Vậy, `x = 25`

___

`x^2 = 3`

`=> x^2 =  (+-\sqrt{3})^2`

`=> x = +- \sqrt{3}`

Vậy, `x \in {-\sqrt{3}; \sqrt{3}}.`

1: \(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

=>căn x-3=0

=>x-3=0

=>x=3

2: =>\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+2\cdot\sqrt{2x-3}\cdot4+16}=5\)

=>\(\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)
=>2*căn 2x-3+5=5

=>2x-3=0

=>x=3/2

a/ x²-2x-8=0

x²-2x+4x-8=0

x(x-2)+4(x-2)=0

(x-2)(x+4)=0

Th1: x-2=0 <=>x=2

Th2: x+4=0 <=> x=-4

b/ sorry bạn mình ko viết được căn bậc 

x căn bậc x -8=0

x căn bậc x =8

căn bậc x bình nhân căn bậc x=8

căn bậc x³=8

căn bậc x³=2³

căn bậc x=2

<=> x=2²=4

c/ áp dụng hằng đẳng thức số 3:

(x-2)²-(x+3)²=8

(x-2-x-3)(x-2+x+3)=8

-5(2x+1)=8

2x+1=8/-5

2x=-13/5

x=-13/10

nhớ t ick cho mình nha

giúp tớ vớiiiiiiiiiiiiiiiiiiiiii

Ta có : \(\sqrt{3}.x-\sqrt{75}=0\)

\(\Leftrightarrow\sqrt{3}.x-5\sqrt{3}=0\)

\(\Leftrightarrow\sqrt{3}\left(x-5\right)=0\)

Vì \(\sqrt{3}\ne0\)

Nên : x - 5 = 0

Vậy x = 5. 

b) Ta có : \(\sqrt{2}.x+\sqrt{2}=\sqrt{8}+\sqrt{32}\)

\(\Leftrightarrow\sqrt{2}\left(x+1\right)=6\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}\left(x+1\right)-6\sqrt{2}=0\)

\(\Leftrightarrow\sqrt{2}.\left(x+1-6\right)=0\)

\(\Leftrightarrow\sqrt{2}.\left(x-5\right)=0\)

Vì \(\sqrt{2}\ne0\)

Nên x - 5 = 0

Suy ra : x = 5

b)đk:\(x\ge\dfrac{1}{2}\)

Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)

\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)

=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\) 

Dấu = xảy ra\(\Leftrightarrow x=1\)

Vậy....

c) đk: \(x\ge0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)

\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)

pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)

\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...

 

a)ĐKXĐ: x≥-1/3; x≤6

<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)

(vì x≥-1/3 nên3x+1≥0 )