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a) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
b) Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
Ta có: \(\left(x-2\right)^3+\left(5-2x\right)^3=0\)
\(\Leftrightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\)
\(\Leftrightarrow3-x=0\)
hay x=3
Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
e)x3-4x+14x(x-2)=0
⇔ x(x2-4)+14x(x-2)=0
⇔ x(x-2)(x+2)+14x(x-2)=0
⇔ (x-2)(x2+2x+14x)=0
⇔ x(x-2)(x+16)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)
g)x2(x+1)-x(x+1)+x(x-1)=0
⇔ (x+1)(x2-x)+x(x-1)=0
⇔ x(x+1)(x-1)+x(x-1)=0
⇔ x(x-1)(x+2)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a: Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left[2\left(x-2\right)^2+1\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
c: Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^3=0\)
\(\Leftrightarrow\left(x-1.5\right)^6-2\left(x-1.5\right)^3=0\)
\(\Leftrightarrow\left(x-1.5\right)^3\cdot\left[\left(x-1.5\right)^3-2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1.5\\x=\sqrt[3]{2}+1.5\end{matrix}\right.\)
Bn ơi bn có thể giải thích câu đầu tiên đoạn sau giấu <=> đc ko?
a) \(x^3+3x^2+3x=0\Rightarrow x\left(x^2+3x+3\right)=0\Rightarrow x\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\Rightarrow x=0\)
(do \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
b) \(x^3+6x^2+12x=0\Rightarrow x\left(x^2+6x+12\right)=0\Rightarrow x\left[\left(x+3\right)^2+4\right]=0\Rightarrow x=0\)
(do (x+3)2+4≥4>0)
a: Ta có: \(x^3+3x^2+3x=0\)
\(\Leftrightarrow x\left(x^2+3x+3\right)=0\)
hay x=0
b: Ta có: \(x^3+6x^2+12x=0\)
\(\Leftrightarrow x\left(x^2+6x+12\right)=0\)
hay x=0
\(\left(x^2+2x\right)^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+4x^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\Leftrightarrow\left(x-1\right)\left(x+1\right)^2\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)
Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=1\end{matrix}\right.\)