Nhanh lên! Ai xong trước 9h30

\(\Leftrightarrow-\dfrac{8}{17}\left(\dfrac{5}{2}-3x\right)=\dfrac{5}{3}\left(2x+\dfrac{8}{5}\right)\)

\(\Leftrightarrow x\cdot\dfrac{24}{17}-\dfrac{20}{17}=\dfrac{10}{3}x+\dfrac{8}{3}\)

\(\Leftrightarrow x\cdot\dfrac{-98}{51}=\dfrac{196}{51}\)

hay x=-2

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

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

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

<=> x²+2x-x+2=2

<=> x²+x=2-2

<=> x²+x=0

<=>x(x+1)=0

<=>x=0 hoặc x+1=0

<=>x=0 hoặc x = -1

a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)

<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

<=> x-3 =10x-15

<=> x-10x= -15+3

<=> -9x = -12

<=> x = \(\frac{-12}{-9}\)

<=> x = \(\frac{4}{3}\)

a) Qui đồng rồi khử mẫu ta được:

   3(3x+2)-(3x+1)=2x.6+5.2

<=> 9x+6-3x-1 = 12x+10

<=> 9x-3x-12x  = 10-6+1

<=> -6x            = 5

<=> x               = -5/6

Vậy ....

b) ĐKXĐ: \(x\ne\pm2\)

Qui đồng rồi khử mẫu ta được:

   (x+1)(x+2)+(x-1)(x-2) = 2(x²+2)

<=> x²+3x+2+x²-3x+2 = 2x²+4

<=> x²+x²-2x²+3x-3x = 4-2-2

<=> 0x             = 0

<=> x vô số nghiệm

Vậy x vô số nghiệm với x khác 2 và x khác -2

c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)

\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)

Vậy ...... 

d) (x+1)²-4(x²-2x+1) = 0

<=> x²+2x+1-4x²+8x-4 = 0

<=> -3x²+10x-3 = 0

giải phương trình

\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)  ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)

\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

\(\Leftrightarrow x-3=10x-15\)

\(\Leftrightarrow x-10x=3-15\)

\(\Leftrightarrow-9x=-12\)

\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)

KL :....

\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)   ĐKXĐ : \(x\ne0;2\)

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

\(\Leftrightarrow x^2+2x-x+2=2\)

\(\Leftrightarrow x^2+x=2-2\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

KL ::

$\textbf{A}$

$A=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{(2^2\cdot3)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{(125\cdot7)^3+5^9\cdot14^3}$

$=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+2^3\cdot5^9\cdot7^3}$

$=\dfrac{2^{12}\cdot3^4(3-1)}{2^{12}\cdot3^5(3+1)}-\dfrac{5^{10}\cdot7^3(1-7)}{5^9\cdot7^3(1+8)}$

$=\dfrac{2}{12}-\dfrac{-30}{9}$

$=\dfrac16+\dfrac{10}{3}$

$=\boxed{\dfrac72}.$

$\textbf{B}$

$B=\dfrac{\left(-\dfrac12\right)^3-\left(\dfrac34\right)^3\cdot(-2)^2}{2\cdot(-1)^5+\left(\dfrac34\right)^2-\dfrac38}$

$=\dfrac{-\dfrac18-\dfrac{27}{64}\cdot4}{-2+\dfrac9{16}-\dfrac38}$

$=\dfrac{-\dfrac18-\dfrac{27}{16}}{-2+\dfrac3{16}}$

$=\dfrac{-\dfrac{29}{16}}{-\dfrac{29}{16}}$

$=\boxed{1}.$