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\(a,=\left(x+y+x-y\right)\left(x+y-x+y\right)=4xy\\ b,=\left(x+y+x-y\right)^2=4x^2\\ c,=\left(x-y+z\right)^2+\left(z-y\right)^2-2\left(x-y+z\right)\left(z-y\right)\\ =\left(x-y+z-z+y\right)^2=x^2\)
\(1,\\ a,\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow\left(2x+3\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\\ 2,\\ a,\Leftrightarrow\left(x+5\right)^2=0\Leftrightarrow x=-5\\ b,\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\\ c,\Leftrightarrow\left(x-9\right)^2=0\Leftrightarrow x=9\\ d,\Leftrightarrow\left(x-3\right)^3=0\Leftrightarrow x=3\\ e,\Leftrightarrow3x\left(x^2-2x+3\right)=0\\ \Leftrightarrow3x\left(x^2-2x+1+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2+2=0\left(vô.nghiệm\right)\end{matrix}\right.\\ \Leftrightarrow x=0\)
\(f,\Leftrightarrow3x\left(x^2-4x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Bài 1:
a) \(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
b) \(\Rightarrow3x\left(x-4\right)-\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-7\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow\left(x+5\right)^2=0\Rightarrow x=-5\)
b) \(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
c) \(\Rightarrow\left(x-9\right)^2=0\Rightarrow x=9\)
d) \(\Rightarrow\left(x-3\right)^3=0\Rightarrow x=3\)
e) \(\Rightarrow3x\left(x^2-6x+9\right)=0\)
\(\Rightarrow3x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
f) \(\Rightarrow3x\left(x^2-4x+4\right)=0\)
\(\Rightarrow3x\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Bài 3:
a: \(M=x^2-4x+5\)
\(=x^2-4x+4+1\)
\(=\left(x-2\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=2
b: \(N=y^2-y-3\)
\(=y^2-2\cdot y\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{13}{4}\)
\(=\left(y-\dfrac{1}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\forall y\)
Dấu '=' xảy ra khi \(y=\dfrac{1}{2}\)
`3)(x+4)/(x-3)-(x-3)/(x+4)=(x^2+18x+7)/(x^2+x-12)`
`đk:x ne 3,x ne -4`
Nhân 2 vế với `(x-3)(x+4) ne 0` ta có:
`(x+4)^2-(x-3)^2=x^2+18x+7`
`<=>x^2+8x+16-x^2+6x-9=x^2+18x+7`
`<=>14x+7=x^2+18x+7`
`<=>x^2+4x=0`
`<=>x(x+4)=0`
Vì `x ne -4=>x+4 ne 0`
`<=>x=0`
Vậy `S={0}`
\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)
\(a,=4x^3-4x^2+10x\\ b,=3x^2-6x+6x-2x^2+5=x^2+5\)