1) \(\frac{4x-8}{2x^2+1}=0\)

<=> \(\frac{4\left(x-2\right)}{2x^2+1}=0\)

<=> 4(x - 2) = 0

<=> x - 2 = 0

<=> x = 2

2) \(\frac{x^2-x-6}{x-3}=0\)

<=> \(\frac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)

<=> x + 2 = 0

<=> x = -2

3) xem ở đây Câu hỏi của Vương Thanh Thanh

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

<=> \(\frac{12}{\left(1+3x\right)\left(1-3x\right)}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

<=> 12 = (1 - 3x)² - (1 + 3x²)

<=> 12 = 1 - 6x + 9x² - 1 - 6x - 9x²

<=> 12 = -12x

<=> x = -1

5) ĐKXĐ: \(x\ne1,x\ne3\)

\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)

<=> \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)

<=> (x + 5)(x - 3) = (x + 1)(x - 1) - 8

<=> x² - 3x + 5x - 15 = x² - x + x - 1 - 8

<=> x² + 2x - 15 = x² - 9

<=> x² + 2x - 15 - x² = -9

<=> 2x - 15 = -9

<=> 2x = -9 + 15

<=> 2x = 6

<=> x = 3 (ktm)

=> pt vô nghiệm

6) ĐKXĐ: \(x\ne\pm2\)

\(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\)

<=> \(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{\left(x-2\right)\left(x+2\right)}+1\)

<=> (x + 1)(x + 2) - 5(x - 2) = 12 + (x - 2)(x + 2)

<=> x² + 2x + x + 2 - 5x + 10 = 12 + x² + 2x - 2x - 4

<=> x² - 2x + 12 = x² + 8

<=> x² - 2x + 12 - x² = 8

<=> -2x + 12 = 8

<=> -2x = 8 - 12

<=> -2x = -4

<=> x = 2 (ktm)

=> pt vô nghiệm

\(\frac{a}{x-2}+\frac{b}{\left(x+1\right)^2}=\frac{a\left(x+1\right)^2+b\left(x-2\right)}{\left(x-2\right)\left(x+1\right)^2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)

\(\Rightarrow\frac{x^2+5}{x^3-3x-2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)

Đồng nhất hệ số, ta có :

\(\hept{\begin{cases}a=1\\2a+b=0\\a-2b=5\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=-2\end{cases}}}\)

cái thứ 2 tương tự

\(\frac{1}{x+3}+\frac{8}{\left(x-1\right)\left(x-3\right)}=\frac{x+3}{x^2-2x=3}\)

\(\Leftrightarrow\frac{1}{x+3}+\frac{8}{\left(x-1\right)\left(x-3\right)}=\frac{x+3}{\left(x-3\right)\left(x+1\right)}\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+1\right)+8\left(x+3\right)\left(x+1\right)=\left(x+3\right)^2\left(x-1\right)\)

\(\Leftrightarrow x^3+5x^2+31x+27=x^3+5x^2+3x-9\)

\(\Leftrightarrow5x^2+31x+27=5x^2+3x-9\)

\(\Leftrightarrow31x+27=3x-9\)

\(\Leftrightarrow31x+27-3x=-9\)

\(\Leftrightarrow28x+27=-9\)

\(\Leftrightarrow28x=-36\)

\(\Leftrightarrow x=\frac{-36}{28}=-\frac{9}{7}\)

Vậy \(x=-\frac{9}{7}\)

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

\(ĐKXĐ:x\ne\pm2\)

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

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

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

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

\(ĐKXĐ:x\ne\pm2;x\ne0\)

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

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

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

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

\(A=\frac{4-2x}{x+2}\)

Bài làm

a) Vì \(\widehat{AMN}=\widehat{ABC}\)( gt )

Mà hai góc này đồng vị

=> MN // BC

Xét tam giác ABC có:

Theo hệ quả của định lí Thales cóL

\(\frac{MN}{BC}=\frac{AN}{AC}\)

hoặc\(\frac{6}{15}=\frac{AN}{NC+NA}\)

hay \(\frac{6}{15}=\frac{AN}{6+AN}\)

=> 36 + 6AN = 15AN

=> 6AN - 15AN = -36

=> AN( 6 - 15 ) = -36

=> AN . ( -9 ) = - 36

=> AN = 4

Vậy AN = 4 cm