trannnn

Các bạn giúp mình gấp với ạ

 

Nguyễn Hoàng Minh
25 tháng 10 2021 lúc 7:56

\(a,\Leftrightarrow\left(x+1\right)^2-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x+1-2x+2\right)\left(x+1+2x-2\right)=0\\ \Leftrightarrow\left(3-x\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(3x-6-2x+2\right)\left(3x-6+2x-2\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x-8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{8}{5}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-3\left(4x^2-12x+9\right)=0\\ \Leftrightarrow2x^2-12x^2+36x-27=0\\ \Leftrightarrow10x^2-36x+27=0\Leftrightarrow x=\dfrac{18\pm3\sqrt{6}}{2}\)

\(d,\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\\ e,\Leftrightarrow\left(x-2\right)\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\\ f,\Leftrightarrow\left(7x+5\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{7}\\x=-1\end{matrix}\right.\)

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ILoveMath
25 tháng 10 2021 lúc 8:02

a) \(\left(x+1\right)^2-4\left(x^2-2x+1\right)=0\\ \Rightarrow\left(x+1\right)^2-4\left(x-1\right)^2=0\\ =\left(x+1\right)^2-\left(2x-2\right)^2=0\\ \Rightarrow\left(x+1-2x+2\right)\left(x+1+2x-2\right)=0\\ \Rightarrow\left(3-x\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

b) \(9\left(x-2\right)^2-4\left(x-1\right)^2=0\\ \Rightarrow\left(3x-6\right)^2-\left(2x-2\right)^2=0\\ \Rightarrow\left(3x-6-2x+2\right)\left(3x-6+2x-2\right)=0\\ \Rightarrow\left(x-4\right)\left(5x-8\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{8}{5}\end{matrix}\right.\)

d) \(x^2-4x+3=0\\ \Rightarrow\left(x^2-x\right)-\left(3x-3\right)=0\\ \Rightarrow x\left(x-1\right)-3\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

e) \(x^2+6x-16=0\\ \Rightarrow\left(x^2-2x\right)+\left(8x-16\right)=0\\ \Rightarrow x\left(x-2\right)+8\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x+8\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

f) \(7x^2+12x+5=0\\ \Rightarrow\left(7x^2+7x\right)+\left(5x+5\right)=0\\ \Rightarrow7x\left(x+1\right)+5\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(7x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{7}\end{matrix}\right.\)

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Norad II
25 tháng 10 2021 lúc 8:08

a) \(\left(x+1\right)^2-4\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2=4\left(x-1\right)^2\)

\(\Leftrightarrow x+1=2x-2\)

\(\Leftrightarrow x=3\)

b) \(9\left(x-2\right)^2-4\left(x-1\right)^2=0\)

\(\Leftrightarrow9\left(x-2\right)^2=4\left(x-1\right)^2\)

\(\Leftrightarrow3x-6=2x-2\)

\(\Leftrightarrow x=4\)

c) \(2x^2-3\left(2x-3\right)^2=0\)

\(\Leftrightarrow2x^2=3\left(2x-3\right)^2\)

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

\(\Leftrightarrow\sqrt{2}x=2\sqrt{3}x-3\sqrt{3}\)

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

\(\Leftrightarrow x=\dfrac{3\sqrt{3}}{2\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{3}\left(2\sqrt{3}-\sqrt{2}\right)}{12-2}=\dfrac{18-3\sqrt{6}}{10}\)

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Norad II
25 tháng 10 2021 lúc 8:14

d) \(x^2-4x+3=0\)

\(\Leftrightarrow x^2-x-3x+3=0\)

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

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

\(\Leftrightarrow x=1\) hoặc \(x=3\)

e) \(x^2+6x-16=0\)

\(\Leftrightarrow x^2-2x+8x-16=0\)

\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+8\right)=0\)

\(\Leftrightarrow x=2\) hoặc \(x=-8\)

f) \(7x^2+12x+5=0\)

\(\Leftrightarrow7x^2+7x+5x+5=0\)

\(\Leftrightarrow7x\left(x+1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(7x+5\right)=0\)

\(\Leftrightarrow x=-1\) hoặc \(x=-\dfrac{5}{7}\)

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trannnn
25 tháng 10 2021 lúc 14:26

Cảm ơn mọi người nhiều ạ

 

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