Chọn A

PT: \(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}=\sqrt{2x-1}-10\) (1) (ĐK: \(x\ge\dfrac{1}{2}\)) 

Đặt: \(y=\sqrt{2x-1}\) (ĐK: \(y\ge0\)) 

\(\Leftrightarrow x=\dfrac{y^2+1}{2}\)  

Thay vào (1) ta có: 

\(\sqrt{2\cdot\dfrac{y^2+1}{2}+2y}-\sqrt{2\cdot\dfrac{y^2+1}{2}-2y}=y-10\)

\(\Leftrightarrow\sqrt{y^2+1+2y}-\sqrt{y^2+1-2y}=y-10\)

\(\Leftrightarrow\sqrt{\text{ }y^2+2y+1}-\sqrt{y^2-2y+1}=y-10\)

\(\Leftrightarrow\sqrt{\left(y+1\right)^2}-\sqrt{\left(y-1\right)^2}=y-10\)   

\(\Leftrightarrow\left|y+1\right|-\left|y-1\right|=y-10\)

TH1: Với: \(0\le y< 1\) 

\(\Leftrightarrow y+1-1+y=y-10\)

\(\Leftrightarrow2y-y=-10\)

\(\Leftrightarrow y=-10\left(ktm\right)\)

TH2: \(y\ge1\)

\(\Leftrightarrow y+1-y+1=y-10\)

\(\Leftrightarrow2=y-10\)

\(\Leftrightarrow y=10+2\)

\(\Leftrightarrow y=12\left(tm\right)\)

Mà: y=12

\(\Rightarrow x=\dfrac{12^2+1}{2}=\dfrac{145}{2}\left(tm\right)\) 

Vậy: ...

x = 145/2

a: =>14x+20+5=6x-9-9x

=>14x+25=-3x-9

=>17x=-34

=>x=-2

b: =>\(2x^2-30x+2x-30=2x^2+10x-10x-50\)

=>-28x-30=-50

=>-28x=-20

=>x=20/28=5/7

c: =>2x+x^3-x=x^3+1

=>x=1

d: =>x^3-3x^2+3x-1-x(x^2+2x+1)=10x-2x^2-11x-22

=>x^3-3x^2+3x-1-x^3-2x^2-x=-2x^2-x-22

=>-5x^2+2x-1+2x^2+x+22=0

=>-3x^2+3x+21=0

=>x^2-x-7=0

=>\(x=\dfrac{1\pm\sqrt{29}}{2}\)

a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1

\(2x+10⋮x+1\)

\(\Rightarrow2x+2+8⋮x+1\)

\(\Rightarrow2\left(x+1\right)+8⋮x+1\)

\(\Rightarrow x+1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(x\in\left\{0;1;-2;\pm3;-5;7;-9\right\}\)

\(2x+10⋮x-1\)

\(\Rightarrow2x-2+12⋮x-1\)

\(\Rightarrow2\left(x-1\right)+12⋮x-1\)

\(\Rightarrow x-1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

x - 1 = 1 => x = 2

x -1 = -1 => x = 0 

... tg tự 

\(3x+10⋮x+1\)

\(\Rightarrow3x+3+7⋮x+1\)

\(\Rightarrow3\left(x+1\right)+7⋮x+1\)

\(\Rightarrow x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x\in\left\{0;-2;6;-8\right\}\)

(2x + 1) + (2x + 2) +...+ (2x + 10) = 650

=> 2x+1+2x+2+...+2x+10=650

=> (2x+2x+...+2x)+(1+2+3+...+10)=650

=> 10.2x+55=650

=> 20x=650-55

=> 20x=595

=> x=595:20

=> x=29,75

(x-3)² - (2x-1)(2x+1) = 10

<=> x² - 6x + 9 - 4x² + 1 = 10

<=> -3x² - 6x = 0

<=> x² + 2x = 0

<=> x(x + 2) = 0

<=> \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

vậy ....

\(4(x-3)\)\(^2\) \(-(2x-1)(2x+1)=10\)

\(=> 4(x\)\(^2\) \(-6x+9)-4x\)\(^2\)\(-1=10\)

\(=>\)\(4x\)\(^2\)\(-24x+36-4x\)\(^2\) \(+1=10\)

            \(=>-24x+27=10\)

            \(=>-24x=-27\)

             \(=>x=\)\(\dfrac{-27}{-24}\)\(=\) \(\dfrac{9}{8}\)

                             \(Vậy \)\(X=\)\(\dfrac{9}{8}\)

         Nhớ vote cho mik hen

3-10=-3x+2x

-7=x

=>x=7

1+1=3x-2x

2=x

-1+5=2x-x

4=x

1)

<=> 3 - 10 = -3x + 2x

<=>   -7     = -x

<=>    7     = x

Các câu sau tương tự :)

3-2x=10-3x

<=>3x-2x=10-3

<=> x=7

2x+1=3x-1

<=>2x-3x= -1-1

<=>-x= -2

<=> x=2

x-1=2x-5

<=> x-2x = 1-5

<=> -x= -4

<=> x=4