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18, \(\frac{x}{2}+\frac{x^2}{8}=0\Leftrightarrow4x+x^2=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow x=-4;x=0\)
19, \(4-x=2\left(x-4\right)^2\Leftrightarrow\left(4-x\right)-2\left(4-x\right)^2=0\)
\(\Leftrightarrow\left(4-x\right)\left[1-2\left(4-x\right)\right]=0\Leftrightarrow\left(4-x\right)\left(-7+2x\right)=0\Leftrightarrow x=4;x=\frac{7}{2}\)
20, \(\left(x^2+1\right)\left(x-2\right)+2x-4=0\Leftrightarrow\left(x^2+1\right)\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3>0\right)=0\Leftrightarrow x=2\)
21, \(x^4-16x^2=0\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\Leftrightarrow x=0;x=\pm4\)
22, \(\left(x-5\right)^3-x+5=0\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\Leftrightarrow\left(x-5\right)\left(x-6\right)\left(x-4\right)=0\Leftrightarrow x=4;x=5;x=6\)
23, \(5\left(x-2\right)-x^2+4=0\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\Leftrightarrow x=2;x=3\)
x^2 - x - y^2 - y
= x^2 - y^2 - x - y
= ( x - y ) ( x + y ) - ( x + y )
= ( x + y ) ( x - y - 1 )
x^2 - 2xy + y^2 - z^2
= ( x- y ) ^2 - z^2
= ( x - y - z ) ( x - y + z )
Câu 1:
Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Thay \(a+b+c=0\) vào biểu thức ta được:
\(a^3+b^3+c^3-3abc=0\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow a^3+b^3=3abc\left(đpcm\right)\)
Vậy \(a^3+b^3=3abc\) khi \(a+b+c=0\)
Câu 3:
\(\text{a) }x^2+x+1\\ =x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left[x^2+2\cdot\dfrac{1}{2}x+\left(\dfrac{1}{4}\right)^2\right]+\dfrac{3}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ \text{Ta có : }\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ \text{ Vậy biểu thức luôn nhận giá trị dương}\text{ }\forall x\\ \)
\(\text{b) }2x^2+2x+1\\ =2x^2+2x+\dfrac{1}{2}+\dfrac{1}{2}\\ =2\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{2}\\ =2\left[x^2+2\cdot\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{1}{2}\\ =2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\\ \text{Ta có: }2\left(x+\dfrac{1}{2}\right)^2\forall x\\ 2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\forall x\\ \text{Vậy giá trị của biểu thức luôn nhận giá trị dương }\forall x\\ \)
Bài 2:
a: \(2\left(x-4\right)-x+3=0\)
\(\Leftrightarrow2x-8-x+3=0\)
hay x=5
b: \(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)