`(x - 2)/3 = (x + 1)/4`

`(x - 2) . 4 = (x + 1) . 3`

`<=> 4x - 8 = 3x + 3`

`<=> 4x - 3x = 3 + 8`

`<=> (4 - 3)x = 11`

`=> x = 11`

`=>` `x = 11`

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\(\frac{8}{5}=\frac{-12}{x}\left(x\ne0\right)\)\(\Leftrightarrow8x=-60\)\(\Leftrightarrow x=\frac{-60}{8}=\frac{-15}{2}\)(tmđk)

\(\frac{x-1}{-4}=\frac{-4}{x-1}\left(x\ne1\right)\)\(\Leftrightarrow\left(x-1\right)^2=16\)\(\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=-3\left(tm\right)\end{cases}}}\)

\(\frac{8}{5}=\frac{-12}{x}\)

\(\Rightarrow8x=-60\)

        \(x=-60:8\)

        \(x=-7,5\)

Vậy x=-7,5

\(\frac{x-1}{-4}=\frac{-4}{x-1}\)

\(\Rightarrow\left(x-1\right)^2=16\)

     \(\left(x-1\right)^2=4^2\)

 \(\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=4+1=5\\x=-4+1=-3\end{cases}}\)

vậy x=5 hoặc x=-3

Bài làm

a) 8/5 = -12/x

=> x = -12.5/8 = -7,5

Vậy x = -7,5

b) x-1/-4 = -4/x-1

=> ( x - 1 )( x - 1 ) = -4 . (-4)

=> x² - 2x + 1 = 16

=> x² - 2x + 1 - 16 = 0

=> x² - 2x - 15 = 0

=> x² - 5x + 3x - 15 = 0

=> x( x - 5 ) + 3( x - 5 ) = 0

=> ( x - 5 )( x + 3 ) = 0

=> x - 5 = 0 hoặc x + 3 = 0

=> x = 5 hoặc x = -3

Vậy x = 5 hoặc x = -3

c) -2/x = y/4

=> xy = -8

d) 4/y+2 = 7/3y+1 

=> 4( 3y + 1 ) = 7( y + 2 )

=> 12y + 4 = 7y + 14

=> 5y = 10

=> y = 2

Vậy y = 2

e) 2x-1/3 = 3x+1/4

=> 4( 2x-1) = 3( 3x + 1 )

=> 8x - 4 = 9x + 3

=> -x = 7

=> x = -7

Vậy x = -7

a) pt

<=> (x - 5)(x + 5) - (x - 5) = 0

<=> (x - 5)(x + 4) = 0

<=> x - 5 = 0 hoặc x + 4 = 0

<=> x = 5 hoặc x = -4

b) pt

<=> (2x - 1)(2x - 1 - 2x - 1) = 0

<=> (2x - 1).(-2)=0

<=> 2x - 1 = 0

<=> x = 1/2

c) pt

<=> (x - 1)(x + 1)(x^2 + 4) = 0

<=> x - 1 = 0 hoặc x + 1 = 0 hoặc x^2 + 4 = 0

<=> x = 1 hoặc x = -1

a,x2−52−(x−5)=0<=>(x−5)(x+5)−(x−5)=0<=>(x−5)(x+4)=0=>x=5;x=−4.b,x2−x−6=0<=>x2−3x+2x−6=0<=>x(x−3)+2(x−3)=0<=>(x+2)(x−3)=0=>x=3;x=−2

a. x² - 25 - (x - 5) = 0

<=> x² - 5² - (x - 5) = 0

<=> (x - 5)(x + 5) - (x - 5) = 0

<=> (x + 5 - 1)(x - 5) = 0

<=> (x + 4)(x - 5) = 0

<=> \(\left[{}\begin{matrix}x+4=0\\x-5=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-4\\x=5\end{matrix}\right.\)

b. (2x - 1)² - (4x² - 1) = 0

<=> (2x - 1)² - (2x - 1)(2x + 1) = 0

<=> (2x - 1)(1 - 2x + 1) = 0

<=> (2x - 1)(2 - 2x) = 0

<=> \(\left[{}\begin{matrix}2x-1=0\\2-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

c. x²(x² + 4) - x² - 4 = 0

<=> x²(x² + 4) - (x² + 4) = 0

<=> (x² - 1)(x² + 4) = 0

<=> (x - 1)(x + 1)(x² + 4) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x+1=0\\x^2+4=0\left(VLí\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Trước khi xem lời giải bài toán này bạn nên xem qua video để hiểu cách biến đổi biểu thức 1 cách nhanh,gọn:Khai triển, rút gọn đa thức bằng máy tính casio . Bài này nhìn rồi mắt chứ rút gọn thì easy

a) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=0\Leftrightarrow-\left(2x-8\right)\) ( Dùng máy tính casio để biến đổi cho nhanh nha =))

\(\Leftrightarrow-2x+8=0\Leftrightarrow8-2x=0\Leftrightarrow2x=8\Leftrightarrow x=4\)

b) \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)

\(\Leftrightarrow\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)-2=0\)

\(\Leftrightarrow3x-42=0\Leftrightarrow3x=42\Leftrightarrow x=14\)

1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)

Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

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