\(\left(x-2\right)^2=\left(2x+3\right)^2\Leftrightarrow x-2=2x+3\text{ hoặc:}x-2=-3-2x\)

\(\Leftrightarrow x+5=0\text{ hoặc: }-1-3x=0\Leftrightarrow x=-5\text{ hoặc: }x=\frac{1}{3}\)

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

 \(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)

\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\) 

 \(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

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

\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)

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

\(\Rightarrow\hept{\begin{cases}2x-2=0\\6+3x=0\\2x+2=0\end{cases}}\)(hoặc chứ ko pk và nha)

\(x\in\left\{1;-2;-1\right\}\)

Nhóm này bao gồm các nguyên tố hóa học là flo, clo, brôm, iốt, astatin và tennessine

ghi rõ lại đề bạn mk không hiểu là 1/2 * (x+1) hay 1/2(x+1) 

là 1/2*(x+1)

a) Vì n thuộc AC nên \(AN+NC=AC\)

Thay số: AN + 8 = 12

\(\Rightarrow AN=12-8=4\left(cm\right)\)

Ta có: \(\frac{AM}{AB}=\frac{5}{15}=\frac{1}{3}\)và \(\frac{AN}{AC}=\frac{4}{12}=\frac{1}{3}\)

\(\Rightarrow\frac{AM}{AB}=\frac{AN}{AC}\)

Áp dụng định lý Thales đảo suy ra MN // BC (đpcm)

b) Vì MN //BC (cmt) nên áp dụng định lý Thales, ta có:

\(\frac{AM}{AB}=\frac{MN}{BC}\Rightarrow\frac{MN}{20}=\frac{1}{3}\Rightarrow MN=\frac{20}{3}\)

Vậy MN = \(\frac{20}{3}\)