\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\left(đkxđ:x\ne-2;-8;-4;-6\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)

\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)

\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)

\(\Leftrightarrow\dfrac{2}{x+2}-1+\dfrac{8}{x+8}-1=\dfrac{4}{x+4}-1+\dfrac{6}{x+6}-1\)

\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)

\(\Leftrightarrow\left(-x\right)\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

ĐKXĐ: x\(\ne\) -2; x\(\ne\) -4; x\(\ne\) -6; x\(\ne\) -8;

\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\) \(\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)

\(\Leftrightarrow\left(x+2+\dfrac{2}{x+2}\right)+\left(x+8+\dfrac{8}{x+8}\right)=\)

\(\left(x+4+\dfrac{4}{x+4}\right)+\left(x+6+\dfrac{6}{x+6}\right)\)

\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)

=> 2.(x+4)(x+8)(x+6) + 8(x+2)(x+4)(x+6)=4(x+2)(x+6)(x+8)

+ 6(x+2)(x+4)(x+8)

<=>(2x+8)(x² + 14x+64) + (8x+48)(x²+6x+8) - (4x+8)(x² + 14x+64)

-(6x+48)(x²+6x+8)

<=> (x² + 14x+64)(2x+8 -4x -8) + (x²+6x+8)(8x+48+6x-48)=0

<=> -2x(x² + 14x+64)+ 2x(x²+6x+8) = 0

<=> -2x³ -28x² -128x+ 2x³ +12x² +16x = 0

<=> -16x² - 112x = 0

<=> -x(16x+112) = 0

<=>\(\left[{}\begin{matrix}x=0\\16x+112=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=7\left(tmđk\right)\end{matrix}\right.\)

vậy S={0;7}

sửa bài:

<=>﴾2x+8﴿﴾x² + 14x+48﴿ + ﴾8x+48﴿﴾x² +6x+8﴿ ‐ ﴾4x+8﴿﴾x² + 14x+48﴿

‐﴾6x+48﴿﴾x² +6x+8﴿
<=> ﴾x² + 14x+48﴿﴾2x+8 ‐4x ‐8﴿ + ﴾x² +6x+8﴿﴾8x+48+6x‐48﴿=0
<=> ‐2x﴾x² + 14x+48﴿+ 2x﴾x² +6x+8﴿ = 0
<=> ‐2x³ ‐28x² ‐96x+ 2x³ +12x² +16x = 0
<=> ‐16x² ‐ 80x = 0
<=> ‐x﴾16x+80﴿ = 0

<=>\(\left[{}\begin{matrix}x=0\\16x+80=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

vậy : S={0;-5}

\(pt\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)

\(\Rightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)

\(\Rightarrow2x+10+\dfrac{2}{x+2}+\dfrac{8}{x+8}=2x+10+\dfrac{4}{x+4}+\dfrac{6}{x+6}\)

\(\Rightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)

\(\Rightarrow\dfrac{2x+16+8x+16}{\left(x+2\right)\left(x+8\right)}=\dfrac{4x+24+6x+24}{\left(x+4\right)\left(x+6\right)}\)

\(\Rightarrow\dfrac{\left(x+2\right)\left(x+8\right)}{2\left(5x+16\right)}=\dfrac{\left(x+4\right)\left(x+6\right)}{2\left(5x+24\right)}\)

\(\Rightarrow\dfrac{x^2+8x+2x+16}{5x+16}=\dfrac{x^2+6x+4x+24}{5x+24}\)

\(\Rightarrow\dfrac{x^2+10x+16}{5x+16}=\dfrac{x^2+10x+24}{5x+24}=\dfrac{x^2+10x+24-x^2-10x-16}{5x+24-5x-16}=\dfrac{8}{8}=1\)(dãy ts bằng nhau)

Dễ dàng tìm được x

ảnh k đc rõ mấy, mong bạn thông cảm :)

Câu a và câu c bn kia làm rồi nên mk làm câu b thôi nhé....

b) y² + 4^x + 2y - 2^x+1 + 2 = 0

\(\Leftrightarrow\) (y² + 2y + 1) + 4^x - 2^x.2 + 1 = 0

\(\Leftrightarrow\) (y + 1)² + [(2^x)² - 2.2^x.1 + 1] = 0

\(\Leftrightarrow\) (y + 1)² + (2^x - 1)² = 0

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

Vậy...................

a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)

=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)

=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22

=>-x^2+59x+14-8x^2+5x+22=0

=>-9x^2+54x+36=0

=>x^2-6x-4=0

=>\(x=3\pm\sqrt{13}\)

b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)

=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)

=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32

=>x^2+6x+19=x^2+4x-32

=>2x=-51

=>x=-51/2

24: 

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

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

\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)

\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)

=>(x+10)(x-2)=0

=>x=-10 hoặc x=2

25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)

\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)

\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)

\(\Leftrightarrow x+5=0\)

hay x=-5

\(\Leftrightarrow\dfrac{x^2+2x+1-1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)

\(\Leftrightarrow x+1-\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)

\(\Leftrightarrow2x+5-\dfrac{1}{x+1}+\dfrac{4}{x+4}=2x+5+\dfrac{2}{x+2}+\dfrac{3}{x+3}\)

=>-x-4+4x+4=2x+6+3x+6

=>3x=5x+12

=>-2x=12

hay x=-6(nhận)