1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

a: \(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{x}{x^2-2x+1}-\dfrac{1}{x^2-1}\right)\)

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

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

\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{x^2+x-x+1}{x-1}\)

\(=\dfrac{1-x}{x-1}=-1\)

b: \(\dfrac{x}{6-x}+\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right):\dfrac{2x-6}{x^2+6x}\)

\(=\dfrac{x}{6-x}+\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}\)

\(=\dfrac{x}{6-x}+\dfrac{x^2-x^2+12x-36}{x-6}\cdot\dfrac{1}{2\left(x-3\right)}\)

\(=\dfrac{x}{6-x}+\dfrac{12\left(x-3\right)}{2\left(x-3\right)\left(x-6\right)}\)

\(=\dfrac{x}{6-x}+\dfrac{6}{x-6}=\dfrac{-x+6}{x-6}=-1\)

Bài 1.

a) ( x³ - 8) : ( x² + 2x + 4 )

= ( x - 2)( x² + 2x + 4 ) : ( x² + 2x + 4 )

= x - 2

b) ( 3x² - 6x ) : ( 2 - x)

= 3x( x - 2) : ( 2 - x)

= -3x( 2 - x ) : ( 2 - x)

= - 3x

Bài 2 .

\(\dfrac{2x-1}{x^2-x}\)

a) Để A có nghĩa tức là A xác định :

ĐKXĐ : x( x - 1) # 0

=> x # 0 ; x # 1

Vậy,...

b) Vì : x = 0 không thỏa mãn ĐKXĐ nên tại x = 0 giá trị của A không xác định

Vì : x = 3 thỏa mãn ĐKXĐ nên ta thay x = 3 vào A , ta có :

\(A=\dfrac{2.3-1}{3^2-3}=\dfrac{5}{6}\)

Vậy , tại : x = 3 thì A = \(\dfrac{5}{6}\)

Bài 3 .

a) ( 6x + 1)² + ( 6x - 1)² - 2( 1 + 6x )( 6x - 1)

= ( 6x + 1)² - 2( 1 + 6x )( 6x - 1) + ( 6x - 1)²

= ( 6x + 1 - 6x + 1)²

= 1

b) 3( 2² + 1)( 2⁴ + 1)( 2⁸ + 1)( 2¹⁶ + 1)

= ( 2² - 1)( 2² + 1)( 2⁴ + 1)( 2⁸ + 1)( 2¹⁶ + 1)

= ( 2⁴ - 1)( 2⁴ + 1)( 2⁸ + 1)( 2¹⁶ + 1)

= ( 2⁸ - 1)( 2⁸ + 1)( 2¹⁶ + 1)

= ( 2¹⁶ - 1)( 2¹⁶ + 1)

= 2³² - 1

c) x( 2x² - 3) - x²( 5x + 3 ) + 3x²

= 2x³ - 3x - 5x³ - 3x² + 3x²

= - 3x³ - 3x

d) 3x( x - 2) - 5x( 1 - x) - 8( x² - 3)

= 3x² - 6x - 5x + 5x² - 8x² + 24

= -11x + 24

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

bn ... ơi...mik ...bỏ...cuộc ...hu...hu

. Huhu T^T mong sẽ có ai đó giúp mình "((

\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)

\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)

\(\Leftrightarrow12x-9=29x-145\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x+136=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\left(tm\right)\)

Vậy \(S=\left\{8\right\}\)

\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)

\(\Rightarrow2x-1=2\left(5-3x\right)\)

\(\Leftrightarrow2x-1=10-6x\)

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x-11=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{11}{8}\right\}\)

\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)

\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)

\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)

\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)

\(\Rightarrow4x-5=3x-2\)

\(\Leftrightarrow4x-5-3x+2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{3\right\}\)

\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)

\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)

\(\Rightarrow15x+25=0\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)

\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\)

\(2,\dfrac{2x-1}{5-3x}=2\)

\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\)

\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)

\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)

\(\Leftrightarrow4x-5-2x+2+2x=0\)

\(\Leftrightarrow4x=3\)

\(\Leftrightarrow x=\dfrac{3}{4}\)

\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)

\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=-\dfrac{5}{3}\)

a) Ta có: \(5x\left(x+1\right)-5\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[5x-5\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(5x-5x+10\right)=0\)

\(\Leftrightarrow10\left(x+1\right)=0\)

mà \(10\ne0\)

nên x+1=0

hay x=-1

Vậy: x=-1

b) Ta có: \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)=4\)

\(\Leftrightarrow4x^2-8x+x-2-2x+3-4=0\)

\(\Leftrightarrow4x^2-9x-3=0\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot\frac{9}{4}+\frac{81}{16}-\frac{129}{16}=0\)

\(\Leftrightarrow\left(2x-\frac{9}{4}\right)^2=\frac{129}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{9}{4}=\frac{\sqrt{129}}{4}\\2x-\frac{9}{4}=-\frac{\sqrt{129}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{9+\sqrt{129}}{4}\\2x=\frac{9-\sqrt{129}}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9+\sqrt{129}}{8}\\x=\frac{9-\sqrt{129}}{8}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{9+\sqrt{129}}{8};\frac{9-\sqrt{129}}{8}\right\}\)

c) Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x^2-9\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

mà \(2\ne0\)

nên \(\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;3\right\}\)

d) Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)

\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)

\(\Leftrightarrow-13x-2=11\)

\(\Leftrightarrow-13x=13\)

hay x=-1

Vậy: x=-1

e) Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+8\right)=3-3x^2\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8-3+3x^2=0\)

\(\Leftrightarrow3x-12=0\)

\(\Leftrightarrow3x=12\)

hay x=4

Vậy: x=4

f) Ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-1\)

\(\Leftrightarrow6x^2-\left(6x^2-4x+15x-10\right)+1=0\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10+1=0\)

\(\Leftrightarrow-11x+11=0\)

\(\Leftrightarrow-11x=-11\)

hay x=1

Vậy: x=1

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)