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\(a,\Leftrightarrow4x^2+4x+1-4x^2-12x=9\\ \Leftrightarrow-8x=8\Leftrightarrow x=-1\\ b,\Leftrightarrow\left(x-6\right)^2=0\Leftrightarrow x=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(12x-9x^2-3=0\)
\(9x^2-12x+3=0\)
\(3\left(3x^2-4x+1\right)=0\)
\(3\cdot\left(3x-1\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)
\(12x-9x^2-3=0\)
\(-9x^2+12x-3=0\)
\(-3\left(3x^2-4x+1\right)=0\)
\(3x^2-4x+1=0\)
\(3x^2-3x-x+1=0\)
\(3x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(3x-1\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)
Vậy...
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a) Ta có P = ( 4 x 2 − 1 ) ( 2 x + 1 ) − ( 2 x − 1 ) − ( 4 x 2 − 1 ) ( 2 x + 1 ) ( 2 x − 1 ) = 3 − 4 x 2
b) Ta có Q = 3 x ( x + 3 ) . ( x + 3 ) ( x − 3 ) − x = 9 − 3 x x + 3
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\(a,=x^2-8x+16+1=\left(x-4\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow x=4\)
\(b,=\left(4x^2-12x+9\right)+4=\left(2x-3\right)^2+4\ge4\)
Dấu \("="\Leftrightarrow x=\dfrac{3}{2}\)
\(c,=\left(9x^2-2\cdot3\cdot\dfrac{1}{3}x+\dfrac{1}{9}\right)+\dfrac{26}{9}=\left(3x-\dfrac{1}{3}\right)^2+\dfrac{26}{9}\ge\dfrac{26}{9}\)
Dấu \("="\Leftrightarrow3x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(D=9x^2+12x-1=9x^2+12x+4-3=\left(3x+2\right)^2-3\)
Mà: \(D=\left(3x+2\right)^2-3\le-3\forall x\)
Dấu "=" xảy ra
\(\Leftrightarrow\left(3x+2\right)^2=0\Leftrightarrow3x+2=0\Leftrightarrow3x=-2\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy \(D_{min}=-3\Leftrightarrow x=-\dfrac{2}{3}\)
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a: 49x^2-25=0
=>(7x-5)(7x+5)=0
=>7x-5=0 hoặc 7x+5=0
=>x=5/7 hoặc x=-5/7
b: Đề thiếu vế phải rồi bạn
c: (3x-2)^2-9(x+4)(x-4)=2
=>9x^2-12x+4-9(x^2-16)=2
=>9x^2-12x+4-9x^2+144=2
=>-12x+148=2
=>-12x=-146
=>x=146/12=73/6
d: x^3-6x^2+12x-8=0
=>(x-2)^3=0
=>x-2=0
=>x=2
e: x^3-9x^2+27x-27=0
=>(x-3)^3=0
=>x-3=0
=>x=3
a) \(-25+49x^2=0\)
\(\Leftrightarrow49x^2-25=0\)
\(\Leftrightarrow\left(7x\right)^2-5^2=0\)
\(\Leftrightarrow\left(7x-5\right)\left(7x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-5=0\\7x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}7x=5\\7x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-\dfrac{5}{7}\end{matrix}\right.\)
b) \(16x^2-25\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[5\left(x-2\right)\right]^2=0\)
\(\Leftrightarrow\left(4x-5x+10\right)\left(4x+5x-10\right)=0\)
\(\Leftrightarrow\left(10-x\right)\left(9x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\9x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{10}{9}\end{matrix}\right.\)
c) \(\left(3x-2\right)^2-9\left(x+4\right)\left(x+4\right)=2\)
\(\Leftrightarrow9x^2-12x+4-9\left(x^2+8x+16\right)=2\)
\(\Leftrightarrow9x^2-12x+4-9x^2-72x-144=2\)
\(\Leftrightarrow-84x-140=2\)
\(\Leftrightarrow-84x=142\)
\(\Leftrightarrow x=-\dfrac{142}{84}\)
\(\Leftrightarrow x=-\dfrac{71}{42}\)
d) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
e) \(-27+27x-9x^2+x^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
a)\(2x+143=557\)
\(\Leftrightarrow2x=557-143\)
\(\Leftrightarrow2x=414\)
\(\Leftrightarrow x=414\div2\)
\(\Leftrightarrow x=207\)
Vậy x = 207
\(2x^2+12x-23=-41\)
\(\Rightarrow2x^2-12x+18=0\)
\(\Rightarrow2x^2-4x-9x+18=0\)
\(\Rightarrow2x.\left(x-2\right)-9.\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right).\left(2x-9\right)=0\)
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