bài 1:
7x(2x-1)=14x^2-7x
bài 2
a, x^2+2x=x(x+2)
b, x^2+7x-xy-7y
=x(x+7)-y(x+7)
=(x+7)(x-y)
bài 3
1. 2018^2-2018.4034+2017^2
=(2018-2017)^2
2.9(x+5)-3x(x+5)=0
=>(x+5)(9-3x)=0
=>x+5=0=>x=-5
9-3x=0=>x=3
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Bài 1:
\(7x\left(2x-1\right)\)
\(=14x^2-7x\)
Bài 2:
a,\(x^2+2x\)
\(=x\left(x+2\right)\)
b,\(x^2+7x-xy-7y\)
\(=\left(x^2-xy\right)+\left(7x-7y\right)\)
\(=x\left(x-y\right)+7\left(x-y\right)\)
\(=\left(x+7\right)\left(x-y\right)\)
Bài 3:
1,\(2018^2-2018.4034+2017^2\)
\(=2018.2018-2018.4034+2017.2017\)
\(=2018\left(2018-4034\right)+2017.2017\)
\(=2018.\left(-2016\right)+2018.2016+1\)
\(=2018\left(-2016+2016\right)+1\)
\(=1\)
2.\(9\left(x+5\right)-3x\left(x+5\right)=0\)
\(\left(9-3x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}9-3x=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=9\\x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy...
Bài 2:
a, \(x^2+2x=x\left(x+2\right)\)
b, \(x^2+7x-xy-7y\)
\(=\left(x^2-xy\right)+\left(7x-7y\right)\)
\(=x\left(x-y\right)+7\left(x-y\right)\)
\(=\left(x+7\right)\left(x-y\right)\)
Bài 1:
\(7x\left(2x-1\right)=14x^2-7x\)
Bài 2:
\(a,x^2+2x=x\left(x+2\right)\)
\(b,x^2+7x-xy-7y\)
\(=x\left(x+7\right)-y\left(x+7\right)\)
\(=\left(x+7\right)\left(x-y\right)\)
Bài 3 :
1) \(2018^2-2018.4034+2017^2\)
\(=2018^2-2.2018.2017+2017^2\)
\(=\left(2018-2017\right)^2=1^2=1\)
2) \(9\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(9-3x\right)=0\)
⇒ \(3\left(x+5\right)\left(3-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
Bài 1:
\(7x\left(2x-1\right)\)
\(=7x.2x-7x.1\)
\(=14x^2-7x\)
