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

a/. x³ - 9x² +27x - 19 = 0

<=> (x³ - 3.x² .3 + 3.3² .x - 3³) + 8 = 0

<=> (x - 3)³ + 8 = 0

<=> (x - 3 + 2) [(x - 3)² - 2(x-3) +4] = 0

<=> (x -1)(x² - 6x+ 9 -2x +6 +4) =0

<=> (x - 1)(x²  - 8x + 19) = 0

<=> x - 1 = 0 => x = 1

Vậy S = {1}

Xem lại đề câu b nha bạn?

c/. x³ + 1 -7x -7 =0 

<=> (x³ + 1) -7(x+1)=0

<=> (x+1)(x²-x+1) -7(x+1)=0

<=> (x+1)(x²-x+1-7)=0

<=> x + 1 = 0 hay x² -x - 6 = 0

<=> x = -1 hay (x² - 3x) + (2x - 6) = 0 

<=>                   x(x - 3) +2(x-3) = 0

<=>                 (x - 3)(x+2) = 0

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

Vậy S = {-1;3;-2}

X^{3 -} X²-8X²+8X+19X-19=0

<=>X²(X-1)-8X(X-1)+19(X-1)=0

<=>(X-1)(X²-8X+19)=0

vi X^2-8X+19=(X-4)²+3>3

a, x³-x²- 8x² +8x+19x-19=0
<=> (x-1)(x²- 8x+19)=0
<=> x-1=0 <=> x=1
 

Đặt t =x^2 (t>=0)

Pt trở thành t^2-7t-18=0

Giải pt bậc 2 được t =9 ( nhận) và t=-2(loại)

--> x^2=9--> x= +-3

\(x^4-7x^2-18=0\)

\(=>x^4-9x^2+2x^2-18=0\)

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

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

\(=>x^2-9=0\) (vì \(x^2+2\ge0\forall x\))

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

\(x^4-7x^2-18=0\)

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

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

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

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)

=>\(-2\sqrt{3x}=-4\)

=>\(\sqrt{3x}=2\)

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

=>\(\sqrt{1-3x}=2\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

1/ \(x^3-7x+6=0\)

\(\Leftrightarrow x^3+3x^2-3x^2-9x+2x+6=0\)

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

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

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

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

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

\(\Leftrightarrow\)\(x+3=0\)

hoặc   \(x-1=0\)

hoặc   \(x+2=0\)

\(\Leftrightarrow\)\(x=-3\)

hoặc   \(x=1\)

hoặc   \(x=-2\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-3;1;-2\right\}\)

2/ \(x^3-6x^2-x+30\)

\(\Leftrightarrow x^3+2x^2-8x^2-16x+15x+30=0\)

\(\Leftrightarrow x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+15\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-3x-5x+15\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[x\left(x-3\right)-5\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\)\(x+2=0\)

hoặc   \(x-3=0\)

hoặc   \(x-5=0\)

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

hoặc   \(x=3\)

hoặc   \(x=5\)

Vậy tập nghiệm của phương trình là :\(S=\left\{-2;3;5\right\}\)

3/ \(x^3-9x^2+6x+16=0\)

\(\Leftrightarrow x^3+x^2-10x^2-10x+16x+16=0\)

\(\Leftrightarrow x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-10x+16\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-8x-2x+16\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x\left(x-8\right)-2\left(x-8\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-8\right)\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x+1=0\)

hoặc  \(x-8=0\)

hoặc  \(x-2=0\)

\(\Leftrightarrow\)\(x=-1\)

hoặc   \(x=8\)

hoặc   \(x=2\)

Vậy tập nghiệm của phương trình là :\(S=\left\{-1;8;2\right\}\)

4/ Đề bài sai ! Sửa lại nhé :

 \(2x^3-x^2+5x+3=0\)

\(\Leftrightarrow2x^3+x^2-2x^2-x+6x+3=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2-x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(tm\right)\\\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-\frac{1}{2}\right\}\)

1.   3x( x - 2 ) - ( x - 2 ) = 0

<=> ( x-2).(3x-1)  = 0 => x = 2 hoặc x = \(\dfrac{1}{3}\)

2.    x( x-1 ) ( x² + x + 1 ) - 4( x - 1 )

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

(phần này tui giải được x = 1 thôi còn bên kia giải ko ra nha )

3 \(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)

\(1. 3x^2 - 7x +2=0\)

=>\(Δ=(-7)^2 - 4.3.2\)

        \(= 49-24 = 25\)

Vì 25>0 suy ra phương trình có 2 nghiệm phân biệt:

\(x_1\)=\(\dfrac{-\left(-7\right)+\sqrt{25}}{2.3}=\dfrac{7+5}{6}=2\)

\(x_2\)=\(\dfrac{-\left(-7\right)-\sqrt{25}}{2.3}=\dfrac{7-5}{6}=\dfrac{1}{3}\)

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):}`