H24

a) ( x- 5)2 = ( 3+2x)2

b) 1 – 9x2 = ( 3x+1)2

c) 27x3 – 54 x2 + 36x = 9

d) x2 – 10x = -25

e) 4x2 – 4x = - 1

f) ( x – 2)3 + ( 5 – 2x)3 = 0

Giúp mị zới

H24
4 tháng 11 2021 lúc 15:32

a) \(\Leftrightarrow\left[{}\begin{matrix}x-5=3+2x\\x-5=-3-2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\)

b) \(\Leftrightarrow\left(1-3x\right)\left(1+3x\right)-\left(3x+1\right)^2=0\\ \Leftrightarrow\left(3x+1\right)\left(1-3x-3x-1\right)=0\\ \Leftrightarrow-6x\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{3}\end{matrix}\right.\)

c) \(=\left(3x-3\right)^3=0\\ \Leftrightarrow\left(x-1\right)^3=0\\ \Leftrightarrow x=1\)

d) \(\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)

e) \(\Leftrightarrow\left(2x-1\right)^2=0\\ \Leftrightarrow x=\dfrac{1}{2}\)

 

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NM
4 tháng 11 2021 lúc 15:32

\(a,\Rightarrow\left(x-5\right)^2-\left(2x+3\right)^2=0\\ \Rightarrow\left(x-5-2x-3\right)\left(x-5+2x+3\right)=0\\ \Rightarrow\left(-x-8\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\\ b,\Rightarrow\left(1-3x\right)\left(1+3x\right)-\left(3x+1\right)^2=0\\ \Rightarrow\left(3x+1\right)\left(1-3x-3x-1\right)=0\\ \Rightarrow6x\left(3x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{3}\end{matrix}\right.\\ c,\Rightarrow3x^3-6x^2+4x-1=0\\ \Rightarrow3x^3-3x^2-3x^2+3x+x-1=0\\ \Rightarrow\left(x-1\right)\left(3x^2-3x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Rightarrow x=1\)

\(d,\Rightarrow x^2-10x+25=0\\ \Rightarrow\left(x-5\right)^2=0\\ \Rightarrow x=5\\ e,\Rightarrow4x^2-4x+1=0\\ \Rightarrow\left(2x-1\right)^2=0\\ \Rightarrow x=\dfrac{1}{2}\\ f,\Rightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}3-x=0\\7x^2-33x+39=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\7\left(x-\dfrac{33}{14}\right)^2+\dfrac{3}{28}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Rightarrow x=3\)

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