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

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

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

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

=>\(\dfrac{3x^3-9x^2+9x-2x^3+2x^2-6x}{\left(x^2-3x+3\right)\left(x^2-x+3\right)}=-1\)

=>x^3-7x^2+3x=-[(x^2+3)^2-4x(x^2+3)+3x^2]

=>x^3-7x^2+3x+(x^2+3)^2-4x(x^2+3)+3x^2=0

=>x^3-4x^2+3x+x^4+6x^2+9-4x^3-12x=0

=>x^4-3x^3+2x^2-9x+9=0

=>(x-3)(x-1)(x^2+x+3)=0

=>x=3;x=1

Bạn gõ Latex để mọi người hiểu đề dễ hơn nhà =))

đk : x khác -1 ; -2 

sửa đề \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x+5}{\left(x+1\right)\left(x+2\right)}\)

\(\Rightarrow2x-4-x-1=3x+5\Leftrightarrow x-5=3x+5\Leftrightarrow2x+10=0\Leftrightarrow x=-5\left(tm\right)\)

1/ ( x-3) ²=16

\(\Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

2/ (3x-1)³=8

\(\Rightarrow3x-1=2\\ \Rightarrow3x=3\\ \Rightarrow x=1\)

3/ (x-11)³=-27

\(\Rightarrow x-11=-3\\ \Rightarrow x=8\)

phần 4 mình ko rõ đề

4) \(x^3-3x^2+3x-1=-64\)

\(\Rightarrow x^3-3x^2+3x+63=0\\ \Rightarrow\left(x^3+3x^2\right)-\left(6x^2+18x\right)+\left(21x+63\right)=0\\ \Rightarrow x^2\left(x+3\right)+6x\left(x+3\right)+21\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x^2+6x+21\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x^2+6x+21=0\end{matrix}\right.\)

\(x+3=0\\ \Rightarrow x=-3\)

\(x^2+6x+21=0\\ \Rightarrow\left(x^2+6x+9\right)+12=0\\ \Rightarrow\left(x+3\right)^2+12=0\)

Vì \(\left(x+3\right)^2\ge0;12>0\Rightarrow\left(x+3\right)^2+12>0\Rightarrow x^2+6x+21vônghiệm\)

Vậy \(x=-3\)

Nếu \(x< 1\)thì phương trình trở thành : \(-x+1-x+2=3x+1\)

\(\Leftrightarrow-2x+3=3x+1\Leftrightarrow-5x+2=0\Leftrightarrow x=\frac{2}{5}\)

Nếu : \(x>2\)thì phương trình trở thành :\(x-1+x-2=3x+1\Leftrightarrow2x-3-3x-1=0\)

\(\Leftrightarrow-x-4=0\Leftrightarrow x=-4\)

Nếu : \(1\le x\le2\)thì phương trình trở thành : \(x-1-x+2=3x+1\Leftrightarrow x=0\)

\(ĐK:-1\le x< 0;x\ge1\\ PT\Leftrightarrow x+2\sqrt{x-\dfrac{1}{x}}=3+\dfrac{1}{x}\\ \Leftrightarrow x-\dfrac{1}{x}+2\sqrt{x-\dfrac{1}{x}}-3=0\)

Đặt \(\sqrt{x-\dfrac{1}{x}}=a\ge0\)

\(PT\Leftrightarrow a^2+2a-3=0\\ \Leftrightarrow\left(a-1\right)\left(a+3\right)=0\\ \Leftrightarrow a=1\left(a\ge0\right)\\ \Leftrightarrow x-\dfrac{1}{x}=1\\ \Leftrightarrow x^2-x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\end{matrix}\right.\)

1) Ta có: \(x^3-3x^2+2x=0\)

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

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

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

Vậy: S={0;1;2}

2) Ta có: \(\dfrac{x^2-x-1}{x+1}=2x-1\)

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

\(\Leftrightarrow x^2-x-1=2x^2+2x-x-1\)

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

\(\Leftrightarrow-x^2-2x=0\)

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

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

Vậy: S={0;-2}

       3x²+2x=0

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

<=>x=0 hoặc 3x+2=0

từ đó bạn giải ra x thuộc{0;-2/3}

chúc bạn học tốt và nhớ tích đúng cho mình