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Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Bài 2: Rút gọn biểu thức sau một cách nhanh nhất:
a, A=(6x-2)2+(2-5x)2+2.(6x-2)(2-5x)
\(=\left(6x-2\right)^2+2\left(6x-2\right)\left(2-5x\right)+\left(2-5x\right)^2\)
\(\text{(Hằng đẳng thức số 2)}\)
\(=\left(6x-2+2-5x\right)\)
\(=x\)
\(B=\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a^2+1\right)^2\)
\(=\left(2a^2+1+2a\right)\left(2a^2+1-2a\right)-\left(2a^2+1\right)^2\)
\(=\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2\)
\(=-4a^2\)
Làm dễ hiểu chút
\(A=\left(2^2+4^2+...+100^2\right)-\left(1^2+3^2+...+99^2\right)\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100-99\right)\left(99+100\right)\)
\(=3+7+...+199\)
\(B=3^8.7^8-\left(21^4-1\right)\left(21^4+1\right)\)
\(=21^8-\left(21^8-1\right)=1\)
Vậy A > B
\(A=163^2+74.163+37^2\)
\(=163^2+2.37.163+37^2\)
\(=\left(163+37\right)^2=200^2\)
\(B=147^2-94.147+47^2\)
\(=147^2-2.47.147+47^2\)
\(=\left(147-47\right)^2=100^2\)
Vậy A > B
\(2020^2-2019^2+2018^2-2017^2+...+2^2-1^2\)
\(=\left(2020-2019\right)\left(2020+2019\right)+\left(2018-2017\right)\left(2018+2017\right)++\left(2-1\right)\left(2+1\right)\)
\(=2019+2018+2017+...+2+1\)
\(=\frac{\left(2019+1\right)2019}{2}\)
\(=2039190\)
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