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)    \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

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

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

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

\(\Leftrightarrow\)\(\left(3x+1\right)\left(5x+4\right)=0\)

đến đây tự lm nha

b)   \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)    (1)

ĐKXĐ:    \(x\ne\pm\frac{1}{3}\)

\(\left(1\right)\)\(\Leftrightarrow\)\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

\(\Rightarrow\)\(\left(1-3x\right)^2-\left(1+3x\right)^2=12\)

\(\Leftrightarrow\)\(\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)

\(\Leftrightarrow\)\(-12x=12\)

\(\Leftrightarrow\)\(x=-1\)   (t/m ĐKXĐ)

Vậy....

a) \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

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

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

\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-\frac{4}{5}\end{cases}}}\)

b) ĐKXĐ: \(x\ne\pm\frac{1}{3}\)

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

\(\Leftrightarrow\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)

\(\Leftrightarrow\left(1-3x\right)^2-\left(1+3x\right)^2=12\)

\(\Leftrightarrow\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)

\(\Leftrightarrow-12x=12\)

\(\Leftrightarrow x=-1\) (thỏa mãn)

Vậy x = -1

(3x – 1)(x2 + 2) = (3x – 1)(7x – 10)

⇔ (3x – 1)(x2 + 2) – (3x – 1)(7x – 10) = 0

⇔ (3x – 1)(x2 + 2 – 7x + 10) = 0

⇔ (3x – 1)(x2 – 7x + 12) = 0

⇔ (3x – 1)(x2 – 4x – 3x + 12) = 0

⇔ (3x – 1)[(x2 – 4x) – (3x - 12)] = 0

⇔ (3x – 1)[x(x – 4) – 3(x – 4)] = 0

⇔ (3x – 1)(x – 3)(x – 4) = 0

⇔ 3x – 1 = 0 hoặc x – 3 = 0 hoặc x – 4 = 0

+ 3x – 1 = 0 ⇔ 3x = 1 ⇔ x = 1/3.

+ x – 3 = 0 ⇔ x = 3.

+ x – 4 = 0 ⇔ x = 4.

Vậy phương trình có tập nghiệm là 

a) \(6x^2-5x+3=2x-3x\left(3-2x\right)\)

\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)

\(\Leftrightarrow2x=-3\Leftrightarrow x=-\dfrac{3}{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

=>(9x^2+24x-6x-16)(x^2+2x+1)=-16

=>(9x^2+18x-16)(x^2+2x+1)=-16

=>(9x^2+18x+9-25)(x^2+2x+1)=-16

=>[9(x+1)^2-25](x+1)^2=-16

=>9(x+1)^4-25(x+1)^2+16=0

Đặt (x+1)^2=a

=>9a^2-25a+16=0

=>a=1 hoặc a=16/9

=>(x+1)^2=1 hoặc (x+1)^2=16/9

=>\(x\in\left\{0;-2;\dfrac{1}{3};-\dfrac{7}{3}\right\}\)

CẢM ƠN NHÌU NHA

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)[x\left(x-4\right)-3\left(x-4\right)]=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\)

Tương đương với 1 trong 3 biểu thức trên bằng 0.

Giải ra 3 nghiệm là \(x=\frac{1}{3};x=4;x=3\)

Đặt\(3x^2+5x-8=a\)\(\Rightarrow3x^2+5x+1=a+9\), từ đó phương trình có dạng \(\sqrt{a}-\sqrt{a+9}=1\), bn giải phương trình bình thường tìm ra a, rồi suy ra x nha!. Học tốt

PT \(\Leftrightarrow9x^2-6x+1-9x+6=9x^2-18x-27\)

\(\Leftrightarrow9x^2-6x+1-9x+6-9x^2+18x+27=0\)

\(\Leftrightarrow3x+34=0\)

\(\Leftrightarrow x=-\dfrac{34}{3}\)

Vậy ...

Ta có: \(\left(3x-1\right)^2-3\left(3x-2\right)=9\left(x+1\right)\left(x-3\right)\)

\(\Leftrightarrow9x^2-6x+1-9x+6=9\left(x^2-3x+x-3\right)\)

\(\Leftrightarrow9x^2-15x+7=9x^2-18x-27\)

\(\Leftrightarrow9x^2-15x+7-9x^2+18x+27=0\)

\(\Leftrightarrow3x+34=0\)

\(\Leftrightarrow3x=-34\)

\(\Leftrightarrow x=-\dfrac{34}{3}\)

Vậy: \(S=\left\{-\dfrac{34}{3}\right\}\)