\(y^2-7y-8=0\Rightarrow\orbr{\begin{cases}y=-1\\y=8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\sqrt[3]{-1}=-1\\x=\sqrt[3]{8}=2\end{cases}}\)

tại sao lại làm đc như vậy

\(x^3-7x+6=0\)

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

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

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

\(\Leftrightarrow\left(x^2-2x+3x-6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[x\left(x-2\right)+3\left(x-2\right)\right]\left(x-1\right)\)

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

\(\Rightarrow x=\left\{-3;1;2\right\}\)

\(\dfrac{7x+11}{2}-\dfrac{5x-3}{4}=\dfrac{x-6}{8}+\dfrac{3+x}{16} \)

⇔\(8\left(7x+11\right)-4\left(5x-3\right)=2\left(x-6\right)+\left(x+3\right)\)

⇔\(56x+88-20x+12=2x-12+x+3\)

⇔\(56x-20x-2x-x=-12+3-88-12\)

⇔\(33x=-109\)

⇔\(x=\dfrac{-109}{33}\)

a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu

Bài `1:`

`h)(3/4x-1)(5/3x+2)=0`

`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`

______________

Bài `2:`

`b)3x-15=2x(x-5)`

`<=>3(x-5)-2x(x-5)=0`

`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`

`d)x(x+6)-7x-42=0`

`<=>x(x+6)-7(x+6)=0`

`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`

`f)x^3-2x^2-(x-2)=0`

`<=>x^2(x-2)-(x-2)=0`

`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`

`h)(3x-1)(6x+1)=(x+7)(3x-1)`

`<=>18x^2+3x-6x-1=3x^2-x+21x-7`

`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`

`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`

`j)(2x-5)^2-(x+2)^2=0`

`<=>(2x-5-x-2)(2x-5+x+2)=0`

`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`

`w)x^2-x-12=0`

`<=>x^2-4x+3x-12=0`

`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`

`m)(1-x)(5x+3)=(3x-7)(x-1)`

`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`

`<=>(1-x)(5x+3+3x-7)=0`

`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`

`p)(2x-1)^2-4=0`

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

`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`

`r)(2x-1)^2=49`

`<=>(2x-1-7)(2x-1+7)=0`

`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`

`t)(5x-3)^2-(4x-7)^2=0`

`<=>(5x-3-4x+7)(5x-3+4x-7)=0`

`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`

`u)x^2-10x+16=0`

`<=>x^2-8x-2x+16=0`

`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`

Ta có:

2(a − 1)x − a(x − 1) = 2a + 3

⇔(a − 2)x = a + 3       (3)

Do đó, khi a = 2, phương trình (2) tương đương với phương trình 0x = 5.

Phương trình này vô nghiệm nên phương trình (2) vô nghiệm.

\(x^3-2x^2-2x^2+4x+3x-6=0\)

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

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

\(\Leftrightarrow x=2\)