Nhân cả 2 vế vs 7-xta dc

1=(8-x)(7-x)-8(7-x)=(x-7)x

còn lại tự làm

8,

b, (-x²+12x+4)/(x²+3x-4) = 12/(x+4) + 12/(3x-3)

(=) (-x²+12x+4)/(x-1)(x+4) -12(x-1)/(x-1)(x+4) - 4(x+4)/(x-1)(x+4) = 0

(=) -x² +12x +4 -12x +12 -4x -16 = 0

(=) -x² -4x = 0

(=) -x(x+4) = 0

(=) -x = 0 hoặc x +4 = 0

(=) x=0 hoặc x=-4

Vậy S={0;4}

Chúc bạn học tốt.

a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)

=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)

=>\(x^2+2x+1-x^2+2x-1=16\)

=>4x=16=>x=4

b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)

=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)

=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)

=>\(12-x^2-3x-2+x^2+5x-14=0\)

=>2x-4=0=>2x=4=>x=2

c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)

=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)

=>\(12=x^3+8+x^2-2x+4\)

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

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

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

=>\(x\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)

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

=>x=?

\(\Leftrightarrow\hept{\begin{cases}x\left(x+2\right)-\left(x-2\right)=2\\x\left(x-2\right)\ne0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0,1\\x\ne0,2\end{cases}}\Rightarrow x=1\)

\(\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{8}\) (ĐKXĐ: x \(\ne\) -2; x \(\ne\) -3; x \(\ne\) -4; x \(\ne\) -5; x \(\ne\) -6)

\(\Leftrightarrow\) \(\frac{1}{x^2+2x+3x+6}+\frac{1}{x^2+3x+4x+12}+\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{1}{x\left(x+2\right)+3\left(x+2\right)}+\frac{1}{x\left(x+3\right)+4\left(x+3\right)}+\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\) \(\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\)

\(\Rightarrow\) (x + 2)(x + 6) = 32

\(\Leftrightarrow\) (x + 2)(x + 6) - 32 = 0

\(\Leftrightarrow\) x² + 6x + 2x + 12 - 32 = 0

\(\Leftrightarrow\) x² + 8x - 20 = 0

\(\Leftrightarrow\) x² + 8x + 16 - 36 = 0

\(\Leftrightarrow\) (x + 4)² - 36 = 0

\(\Leftrightarrow\) (x + 4 - 6)(x + 4 + 6) = 0

\(\Leftrightarrow\) (x - 2)(x + 10) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TMĐK\right)\\x=-10\left(TMĐK\right)\end{matrix}\right.\)

Vậy S = {2; -10}

Chúc bn học tốt!!

thank you very very..... much!

OMG!