Giúp mình câu a b c vơiơ ạ
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\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
6a.
$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$
$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$
Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$
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a: \(25x^2-16=0\)
=>(5x-4)(5x+4)=0
=>x=4/5 hoặc x=-4/5
b: \(-3x^2+18x=0\)
\(\Leftrightarrow3x^2-18x=0\)
=>3x(x-6)=0
=>x=0 hoặc x=6
\(25x^2-16=0\Leftrightarrow\left(5x\right)^2=16\)
\(\Leftrightarrow5x=\sqrt{4^2}=\left|4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=4\Leftrightarrow x=\dfrac{4}{5}\\5x=-4\Leftrightarrow x=\dfrac{-4}{5}\end{matrix}\right.\)
Vậy \(x=\pm\dfrac{4}{5}\)
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b: =>(x+1)(x-1)-(x+3)(x-3)=2x^2+6x
=>2x^2+6x=x^2-1-x^2+9=8
=>2x^2+6x-8=0
=>x^2+3x-4=0
=>(x+4)(x-1)=0
=>x=-4 hoặc x=1(loại)
a: =>x^3+2x-2x(x^2+1)=0
=>x^3+2x-2x^3-2x=0
=>-x^3=0
=>x=0(nhận)
c: =>(x-2)(x+2)-(x+5)^2=x^2-8
=>x^2-4-x^2-10x-25=x^2-8
=>x^2-8=-10x-29
=>x^2+10x+21=0
=>(x+3)(x+7)=0
=>x=-3 hoặc x=-7
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a: Khi x=3 thì \(A=\dfrac{3\cdot3}{3-2}=9\)
b: C=A+B
\(=\dfrac{3x}{x-2}-\dfrac{6}{x-2}-\dfrac{x^2+4x+4}{x^2-4}\)
\(=\dfrac{3x-6}{x-2}-\dfrac{x+2}{x-2}\)
\(=\dfrac{3x-6-x-2}{x-2}=\dfrac{2x-8}{x-2}\)
c: Để C nguyên thì 2x-4-4 chia hết cho x-2
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6\right\}\)
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