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x^2-25x^4=0
=>x^2-25.x^2.x^2=0
=>x^2.(1-25x^2)=0
=>x^2=0 hoặc 1-25x^2=0
=>x=0 hoặc 25x^2=1=>x^2=1/5=>x=+1/5
vậy x thuộc {-0,2;0;0,2}
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![](https://rs.olm.vn/images/avt/0.png?1311)
x2 - 25x4=0
<=>x2(1-25x2)=0
<=>x2[12-(5x)2]=0
<=>x2(1-5x)(1+5x)=0
=>x2=0 =>x=0
hoặc 1-5x=0 =>1=5x <=>x=0,2
hoặc 1+5x=0 =>1=-5x <=> x=-0,2
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a)Ta có:
\(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)
\(\Rightarrow x-3,5=y-\dfrac{1}{10}=0\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}=0,1\end{matrix}\right.\)
b) Ta có:
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
b: ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^2-25x^4=0
x(x-25x^3)=0
x=0 và x-25x^3=0
x(1-25x^2)=0
x(1-5x)(1+5x)=0
x=0, 1-5x=0,1+5x=0
-5x=-1, 5x=-1
x=1/5 và x=-1/5
vậy x=0,1/5,-1/5
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