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ta thấy \(\begin{cases}\left(2x-5\right)^{2000}\\\left(3y+4\right)^{2002}\end{cases}\ge0}\)
Theo bài ra ta có (2x-5)2000+(3y+4)2002\(\le\) 0
=> (2x-5)2000+(3y+4)2002=0
=>2x-5=0 => x=2,5
=>3y+4=0=>y=\(\frac{-4}{3}\)
Vì (2x-5)2000 > 0 với mọi x
(3y+4)2002 > 0 với mọi y
=>(2x-5)2000+(3y+4)2002 > 0 ới mọi x;y
Mà (2x-5)2000+(3y+4)2002 < 0 (theo đề)
=>(2x-5)2000+(3y+4)2002=0
=>(2x-5)2000=(3y+4)2002=0
+)(2x-5)2000=0=>2x-5=0=>x=5/2
+)(3y+4)2002=0=>3y+4=0=>y=-4/3
Vậy x=5/2;y=-4/3
Bài 1 :
\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)
\(\Leftrightarrow x=-23\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
Ta có :
\(\left(2x-5\right)^{2000}\ge0\forall x\)
\(\left(3y+4\right)^{2002}\ge0\forall y\)
\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\forall x;y\)
Mà theo GT : \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)
\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}=0\)
Dấu \("="\) xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2000}=0\\\left(3y+4\right)^{2002}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{5}{2};y=-\dfrac{4}{3}\)
\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
\(\Rightarrow2x+1=1\)hoặc\(2x+1=0\)
\(\Rightarrow2x=1-1\) \(2x=0-1\)
\(\Rightarrow2x=0\) \(2x=-1\)
\(\Rightarrow x=0:2\) \(x=-1:2\)(loại)
\(\Rightarrow x=0\)
a.
\(\left[{}\begin{matrix}2x-5=0\\3y+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5\\3y=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{-1}{3}\end{matrix}\right.\)
b.
\(\left[{}\begin{matrix}3x-4=0\\3y-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=4\\3y=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{5}{3}\end{matrix}\right.\)
chúc bạn học tốt 
b) \(\left(3x-2\right)^5=-243\)
\(\Rightarrow\left(3x-2\right)^5=\left(-3\right)^5\)
\(\Rightarrow3x-2=-3\Rightarrow x=\dfrac{-1}{3}\)
c) Vì \(\left(2x-5\right)^{2000}\ge0\forall x;\left(3y+4\right)^{2002}\ge0\forall y\)
\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\forall x,y\)
Mà theo bài ra \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)
\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right........\)