Rối mắt , loạn thần kinh toàn là x không

Nhiều quá bạn ơi 

Bài 2: 

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

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

         \(x^2-5x-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

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

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

\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

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

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

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

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

Vậy...

\(\left(x+\frac{2}{3}\right)\left(\frac{5}{4}-2x\right)>0\)

th1 : 

\(\hept{\begin{cases}x+\frac{2}{3}>0\\\frac{5}{4}-2x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-\frac{2}{3}\\-2x>-\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x>-\frac{2}{3}\\x>\frac{5}{8}\end{cases}\Rightarrow}x>\frac{5}{8}}\)

th2 : 

\(\hept{\begin{cases}x+\frac{2}{3}< 0\\\frac{5}{4}-2x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -\frac{2}{3}\\-2x< -\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x< -\frac{2}{3}\\x< \frac{5}{8}\end{cases}\Rightarrow}x< -\frac{2}{3}}\)

vậy_

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=4\\3x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=4\\x=-\frac{2}{3}\end{cases}}\)

(2x – 1)2 – (x + 3)2 = 0

[(2x - 1) - (x + 3)][(2x - 1) + (x + 3)] = 0 (phân tích ra hằng đẳng thức số 3)

(2x - 1 - x - 3)(2x - 1 + x + 3) = 0 (bỏ ngoặc)

(x - 4)(3x + 2) = 0 (rút gọn)

Hoặc x - 4 = 0 => x = 4

=> A=0 hoặc B=0 Hoặc 3x + 2 = 0 => 3x = 2 => x = -2/3

 Vậy x = { 4 ; -2/3} 

(2x-1)^2=0 hoặc (x+3)^2=0

2x-1=0             x+3=0

2x=0+1           x=-3

x=1/2

a) |2x-1|+|3y-5|=0

\(\Rightarrow\left|2x-1\right|=0\) và  \(\left|3y-5\right|=0\) 

\(\Rightarrow2x-1=0\)   và   \(3y-5=0\) 

\(\Rightarrow2x=1\)   và    \(3y=5\)

\(\Rightarrow x=\frac{1}{2}\) Và  \(y=\frac{5}{3}\) 

Vậy.........

b,   /2x-3/^2 + /y-1/ = 0

\(\Rightarrow\)/2x-3 /^2=0         \(\Rightarrow\)x= 3/2^6    \(\Rightarrow\)x= 9/4

\(\Rightarrow\)/y-1/ =0            \(\Rightarrow\)y=1