\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)

\(=x^3+27-x^3-54\)

=-27

2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)​

\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)

a: Ta có: \(\left(3x+5\right)^2-4x^2=0\)

\(\Leftrightarrow\left(3x+5+2x\right)\left(3x+5-2x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

a, `(8x^3-4x^2): 4x -(4x^2-5x) : 2x + (2x)^2`

`=4x (2x^2-x) : 4x - 2x(2x-5/2 ) :2x + 4x^2`

`=2x^2-x-2x+5/2+4x^2`

`=6x^2-3x+5/2`

b, `(3x^3-x^2y) :x^2 -(xy^2+x^2y) :xy + 2x(x+1)`

`=x^2 (3x-y) :x^2 -xy(y+x) + (2x^2+2x)`

`=3x-y-y-x+2x^2+2x`

`=2x^2+4x-2y`

a) 2(x+1)=2x-1

<=> 2x+2=2x-1

<=> 2x+2-2x+1=0

<=>1=0

=>Pt vô nghiệm

a) \(\text{5x(x-2)+(2-x)=0}\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\text{x(2x-5)-10x+25=0}\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)

\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)

\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)

a) \(5x\left(x-2\right)+\left(2-x\right)=0\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(x\left(2x-5\right)-10x+25=0\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(2x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{2}\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(x-\dfrac{9}{8}\right)\left(x+\dfrac{1}{8}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{9}{8}=0\\x+\dfrac{1}{8}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{1}{8}\end{matrix}\right.\)

d) \(x^4+2x^2-8=0\)

\(\Rightarrow\left(x^4+2x^2+1\right)-9=0\)

\(\Rightarrow\left(x^2+1\right)^2-3^2=0\)

\(\Rightarrow\left(x^2+1-3\right)\left(x^2+1+3\right)=0\)

\(\Rightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\end{matrix}\right.\) \(\Rightarrow x^2=2\) \(\Rightarrow x=\pm\sqrt{2}\)

a.

\(\left(4x^2+4x+1\right)-y^2=\left(2x+1\right)^2-y^2=\left(2x+1-y\right)\left(2x+1+y\right)\)

b.

\(\Leftrightarrow2x^2+2x-x-1-2x^2-3x+1=0\)

\(\Leftrightarrow-2x=0\)

\(\Leftrightarrow x=0\)