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`a, 126^2 - 152 . 126 + 5776.`
`= 126^2 - 2 . 76 . 126 + 76^2`
`= (126+76)^2 = 202^2 = 40804`
`b, 3^8 . 5^8 - (15^4-1)(15^4+1)`
`= 15^8 - 15^8 + 1`
`= 1`
\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Câu 2:
a: \(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow x^3-4x-x^3-8=4\)
hay x=-3
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
a) \(\left(x+2\right)\left(x^2-2x+4\right)+\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4+x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{23}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(N\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}>0\left(L\right)\end{matrix}\right.\)
Vậy \(S=\left\{-2\right\}\)
b) \(9x^2-4-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2-3x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\cdot4=0\)
\(\Leftrightarrow3x-2=0\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(S=\left\{\dfrac{2}{3}\right\}\)
a)
`4(x-2)^2 =4`
`<=>(x-2)^2 =1`
`<=>x-2=1` hoặc `x-2=-1`
`<=>x=3` hoặc `x=1`
b)
`5(x^2 -6x+9)=5`
`<=>(x-3)^2 =1`
`<=>x-3=1`hoặc `x-3=-1`
`<=>x=4` hoặc `x=2`
c)
`4x^2 +4x+1=0`
`<=>(2x+1)^2 =0`
`<=>2x+1=0`
`<=>x=-1/2`
d)
`9x^2 +6x+1=2`
`<=>(3x+1)^2 =2`
\(< =>\left[{}\begin{matrix}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{\sqrt{2}-1}{3}\\x=\dfrac{-\sqrt{2}-1}{3}\end{matrix}\right.\)
b)x2-2x+1=4
⇔(x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
c)x2-4x+4=9
⇔ (x-2)2=9
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
d)4x2-4x+1=4
⇔ (2x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
e)x2-2x-8=0
⇔ x2-4x+2x-8=0
⇔ x(x-4)+2(x-4)=0
⇔(x-4)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
f)9x2-6x-8=0
⇔ 9x2-12x+6x-8=0
⇔ 3x(3x-4)+2(3x-4)=0
⇔ (3x-4)(3x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
a) Ta có: \(A=\left(4-x\right)\left(16+4x+x^2\right)-\left(4-x\right)^3\)
\(=64-x^3+\left(x-4\right)^3\)
\(=64-x^3+x^3-12x^2+48x-64\)
\(=-12x^2+48x\)
b) Ta có: \(B=\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(9x^2+6x+4\right)\)
\(=27x^3+8-27x^3+8\)
=16
c) Ta có: \(C=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)^2\)
\(=x^3+1-x\left(x^2+2x+1\right)\)
\(=x^3+1-x^3-2x^2-x\)
\(=-2x^2-x+1\)