4x^2+8x+2=0
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8x2+30x+7=0
8x2+16x+14x+7=0
8x(x+2) +7(x+2)=0
(8x+7)(x+2)=0
=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)
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a) x3 - 16x = 0
x(x2 - 16) = 0
=> x = 0 hoặc x2 - 16 = 0
x = 4
Vậy x = 0 hoặc x = 4
b) x4 -2x3 + 10x2 - 20x = 0
x3 (x - 2) + 10x(x - 2) = 0
(x - 2)(x3 + 10x) = 0
=> x - 2 = 0 hoặc x3 + 10x = 0
x = 2 x(x2 + 10) = 0
+ TH1: x = 0
+ TH2: x2 + 10 = 0
x2 = -10 (vô lí)
Vậy x = 2 hoặc x = 0
c) (2x - 3)2 = (x + 5)2
(2x)2 + 2 . 2x . 3 + 32 = x2 + 2.x.5 + 52
4x2 + 12x + 9 = x2 + 10x + 25
4x2 + 12x - x2 - 10x = 25 - 9
3x2 + 2x = 16
x(3x + 2) = 16
Đến đây bạn làm nốt câu c nhé!
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\(4x^2+8x+2=0\)
\(\Delta=8^2-4.4.2=64-32=32>0\)
=> pt có 2 nghiệm phân biệt
\(\left\{{}\begin{matrix}x_1=\dfrac{-8+\sqrt{32}}{8}=\dfrac{-2+\sqrt{2}}{2}\\x_2=\dfrac{-8-\sqrt{32}}{8}=\dfrac{-2-\sqrt{2}}{2}\end{matrix}\right.\)
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\(PT\Leftrightarrow\left(x^2-4x+1\right)^2+2\left(x^2-4x+1\right)-3=0\\ \Leftrightarrow\left(x^2-4x+1\right)^2-\left(x^2-4x+1\right)+3\left(x^2-4x+1\right)-3=0\\ \Leftrightarrow\left(x^2-4x+1\right)\left(x^2-4x\right)+3\left(x^2-4x\right)=0\\ \Leftrightarrow\left(x^2-4x\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=2\end{matrix}\right.\)
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\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\\ \Leftrightarrow x^6-2x^5+x^4-4x^2+8x-4=0\\ \Leftrightarrow\left(x-1\right)^2\left(x^2-2\right)\left(x^2+2\right)=0;x^2+2>0\forall x\\ \Rightarrow\left[{}\begin{matrix}x^2-2=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
\(4x^2+8x+2=0\)
\(\Leftrightarrow2\left(2x^2+4x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)
<=>4x(x+2)=-2
<=>4x=0 hay x+2=0
<=>X=0 hay x=-2