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Bài 1:
a) \(x^2-7x=x\left(x-7\right)\)
b) \(5x-10y=5\left(x-2y\right)\)
c) \(5x^2-15xy=5x\left(x-3y\right)\)
d) \(3\left(x-y\right)-y+x=3\left(x-y\right)+x-y=4\left(x-y\right)\)
e) \(5\left(x-y\right)-x^2+xy=5\left(x-y\right)-x\left(x-y\right)=\left(x-y\right)\left(5-x\right)\)
Bài 2 :
a, \(75.20,9+5^2.20,9=20,9\left(75+25\right)=20,9.100=2090\)
b, \(85.15+150.1,4=85.15+15.14=15.99=1485\)
c, \(93.32+14.16=186.16+14.16=16.200=3200\)
d, \(98,6.199-990.9,86=98,6.199-99.98,6=98,6.100=9860\)
e) \(8x^3-12x^2+6x-1=\left(2x-1\right)^3\)
f) \(x^3-6x^2+12x-8=\left(x-2\right)^3\)
g) \(\left(x-5\right)^2-\left(2x+3\right)^2=\left(x-5-2x-3\right)\left(x-5+2x+3\right)=\left(-x-8\right)\left(3x-2\right)\)
h) \(\left(2x-2\right)^2-\left(3x+6\right)^2=\left(2x-2-3x-6\right)\left(2x-2+3x+6\right)=\left(-x-8\right)\left(5x+4\right)\)
b: Ta có: \(x\left(x+1\right)-\left(2x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
d: Ta có: \(\left(x-1\right)^2-4\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x-1-2x-4\right)\left(x-1+2x+4\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)
\(a,=3\left(x-y\right)\left(x+y\right)\\ b,=2x\left(x^2-25\right)=2x\left(x-5\right)\left(x+5\right)\\ c,=5x\left(x^2-2x+1\right)=5x\left(x-1\right)^2\\ d,=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)=\left(x-y\right)\left(x+y+2\right)\\ e,=x\left(y-3\right)+y\left(y-3\right)=\left(y-3\right)\left(x+y\right)\\ f,=\left(x+2\right)^2-16y^2=\left(x-4y+2\right)\left(x+4y+2\right)\)
2) Tìm x:
a) \(x^2+5x+6=0\)
⇒\(x^2+2x+3x+6=0\)
⇒\(\left(x^2+2x\right)+\left(3x+6\right)=0\)
⇒\(\left(x.x+2.x\right)+\left(3.x+3.2\right)=0\)
⇒\(x.\left(x+2\right)+3.\left(x+2\right)=0\)
⇒\(\left(x+2\right).\left(x+3\right)=0\)
⇒\(x+2=0\) \(hoặc\) \(x+3=0\)
\(+\))\(x+2=0\) \(+\))\(x+3=0\)
⇔\(x=-2\) ⇔\(x=-3\)
\(Vậy\) \(x\in\left\{-3;-2\right\}\)
c)\(x^2+6x+8=0\)
⇒ \(x^2+4x+2x+8=0\)
⇒ \(\left(x^2+4x\right)+\left(2x+8\right)=0\)
⇒ \(\left(x.x+4.x\right)+\left(2.x+2.4\right)=0\)
⇒ \(x.\left(x+4\right)+2.\left(x+4\right)=0\)
⇒ \(\left(x+4\right).\left(x+2\right)=0\)
⇒\(x+4=0\) \(hoặc\) \(x+2=0\)
\(+\)) \(x+4=0\) \(+\)) \(x+2=0\)
⇔\(x=-4\) ⇔\(x=-2\)
\(Vậy\) \(x\in\left\{-4;-2\right\}\)
a) x2+5x+6=0
⇒x2+2x+3x+6=0
⇒(x2+2x)+(3x+6)=0
⇒x(x+2)+3(x+2)=0
⇒(x+2)(x+3)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
b) 9x3=x
⇒9x3-x=0
⇒9x(x2-1)=0
⇒\(\left[{}\begin{matrix}9x=0\\x^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
c) x2+6x+8=0
⇒ x2+2x+4x+8=0
⇒ (x2+2x)+(4x+8)=0
⇒ x(x+2)+4(x+2)=0
⇒ (x+2)(x+4)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
d) x(x-3)-(x-2)2=x+1
⇒x2-3x-x2+4x-4=x+1
⇒x2-3x-x2+4x-4-x-1=0
⇒-5=0(vô lí)
e) (x+2)(x+3)-(x+1)2=0
⇒x2+5x+6-x2-2x-1=0
⇒3x+5=0
⇒3x=-5
⇒x=\(-\dfrac{5}{3}\)
f)x(x+1)-(x+1)2=0
⇒(x-x-1)(x+1)=0
⇒-1(x+1)=0
⇒x+1=0
⇒x=-1
g) (x-2)2-4(x+3)2=0
⇒x2-4x+4-4(x2+6x+9)=0
⇒x2-4x+4-4x2-24x-36=0
⇒-3x2-28x-32=0
đến đây mik chx bt lm
a: \(x^2-4x-5=\left(x-5\right)\left(x+1\right)\)
b: \(x^2-3x+2=\left(x-2\right)\left(x-1\right)\)
d: \(2x^2-3x+1=\left(x-1\right)\left(2x-1\right)\)
k: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
a) 2x2-3x=0
\(\Rightarrow\)x(2x-3)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
c) (x+1)(x-2)-(x-3)2=x+1
⇒(x+1)(x-2)-(x-3)2-(x+1)=0
\(\Rightarrow\)x2-2x+x-2-(x2-6x+9)-x-1=0
⇒x2-2x+x-2-x2+6x-9-x-1=0
\(\Rightarrow\)4x-12=0
\(\Rightarrow\)4x=12
\(\Rightarrow\)x=3
`a)87^2-13^2`
`=(87-13)(87+13)`
`=74.100=7400`
`b)53^2-2.53.3+3^2`
`=(53-3)^2`
`=50^2=2500`
`c)99^2`
`=(100-1)^2`
`=100^2-2.100+1`
`=10000-100+1`
`=9900+1`
`=9901`
a) Ta có: \(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b) Ta có: \(\left(x+1\right)^2-\left(2x+3\right)^2=0\)
\(\Leftrightarrow\left(x+1-2x-3\right)\left(x+1+2x+3\right)=0\)
\(\Leftrightarrow\left(-x-2\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-2=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-4}{3}\end{matrix}\right.\)
c) Ta có: \(27x^3-27x+9x-1=0\)
\(\Leftrightarrow\left(3x-1\right)^3=0\)
\(\Leftrightarrow3x-1=0\)
hay \(x=\dfrac{1}{3}\)