a. \(2.\left(5x-8\right)-3.\left(4x-5\right)=4.\left(3x-4\right)+11\Leftrightarrow10x-16-12x+15=12x-16+11\\ \)

\(\Leftrightarrow-2x-1=12x-5\Leftrightarrow14x-4=0\Leftrightarrow x=\frac{2}{7}\)

\(a,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow10x-12x-12x=-16+11+16-15\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\frac{-4}{-14}=\frac{2}{7}\)

giải hết đống này chắc @@ quá,để tối đi,giờ t đi làm mấy bài ngắn ngắn

tuỳ bạn, qua Tết mik đăng lại

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Bài 3:

a) Ta có: \(x^3-16x=0\)

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

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

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

Vậy: \(x\in\left\{0;4;-4\right\}\)

b) Ta có: \(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)

\(\Leftrightarrow x\left[x^2\left(x-2\right)+10\left(x-2\right)\right]=0\)

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

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

Vậy: \(x\in\left\{0;2\right\}\)

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

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

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

\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{2}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-\frac{2}{3}\right\}\)

d) Ta có: \(x^2\left(x-1\right)-4x^2+8x-4=0\)

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

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

\(\Leftrightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\)

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

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

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

Vậy: \(x\in\left\{1;2\right\}\)

\(a,x^4+2x^3+x^2=\left(x^2+x\right)^2\)

\(b,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)\)\(=\left(x-1\right)\left(x+6\right)\)

\(c,5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\) \(e,x^2+x-6=x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)\(f,x^2-2x-3=x^2-3x+x-3=x\left(x-3\right)+\left(x-3\right)=\left(x+1\right)\left(x-3\right)\)\(h,2x^2+5x-3=0\Leftrightarrow2x^2-6x+x-3=0\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)

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

\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b,\)\(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

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

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

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

\(c,\)\(9x-6x^2-3=0\)

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

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)

\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)

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

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

\(d,\)\(2x^2+5x+2=0\)

\(\Leftrightarrow2x^2+4x+x+2=0\)

\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)

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

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

Có làm theo hàng đẳng thức ko bạn?

có

a.Ta có : \(\dfrac{x^2-4x+4}{x^3-2x^2-4x+8}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\)

Để \(\dfrac{1}{x+2}>0\) thì 1 và x+2 cùng dấu

mà 1>0

=>x + 2 > 0 <=> x > 2

\(\Rightarrow S=\left\{x|x>2\right\}\)

b, Ta có : \(x^2\ge0\Rightarrow x^2+1>0\)

Để \(\dfrac{7-8x}{x^2+1}>0\) thì 7 - 8x và \(x^2+1\) cùng dấu

mà \(x^2+1>0\Rightarrow7-8x>0\Leftrightarrow x< \dfrac{7}{8}\)

\(\Rightarrow S=\left\{x|x< \dfrac{7}{8}\right\}\)

c. Ta có bảng xét dấu:

Bổ xung câu c:

Vậy : \(-1< x\le\dfrac{-1}{2}\)