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a) Ta có: \(33\cdot55+33\cdot67+45\cdot33+67^2\)
\(=\left(33\cdot55+33\cdot45\right)+\left(33\cdot67+67^2\right)\)
\(=33\cdot\left(55+45\right)+67\left(33+67\right)\)
\(=33\cdot100+67\cdot100\)
\(=100\cdot\left(33+67\right)\)
\(=100\cdot100\)
\(=10000\)
c) Ta có: \(2016\cdot2018-2017^2\)
\(=\left(2017-1\right)\left(2017+1\right)-2017^2\)
\(=2017^2-1-2017^2\)
\(=-1\)
c) Ta có: \(33\cdot55+33\cdot67+45\cdot33+67\cdot67\)
\(=33\left(55+45\right)+67\left(33+67\right)\)
\(=33\cdot100+67\cdot100\)
\(=100\cdot100=10000\)
\(\left(x+3\right)^2-\left(x+3\right)=0\)
\(\left(x+3\right).\left[\left(x+3\right)-1\right]=0\)
\(\left(x+3\right).\left(x+2\right)=0\)
\(=>\orbr{\begin{cases}x+3=0\\x+2=0\end{cases}=>\orbr{\begin{cases}x=-3\\x=-2\end{cases}}}\)
Vậy ...
P/S: mk mới lớp 7 sai sót mong bỏ qua
\(\dfrac{x+4}{104}+\dfrac{x+2}{102}=\dfrac{x+3}{103}+\dfrac{x+1}{101}\\ \Leftrightarrow\left(\dfrac{x+4}{104}-1\right)+\left(\dfrac{x+2}{102}-1\right)=\left(\dfrac{x+3}{103}-1\right)+\left(\dfrac{x+1}{101}-1\right)\\ \Leftrightarrow\dfrac{x-100}{104}+\dfrac{x-100}{102}-\dfrac{x-100}{103}-\dfrac{x-100}{101}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{104}+\dfrac{1}{102}-\dfrac{1}{103}-\dfrac{1}{101}\right)=0\\ \Leftrightarrow x-100=0\left(vì.\dfrac{1}{104}+\dfrac{1}{102}-\dfrac{1}{103}-\dfrac{1}{101}\ne0\right)\\ \Leftrightarrow x=100\)
Ta có:
\(1999^2=\left(2000-1\right)^2\)
\(=2000^2-2\cdot2000+1\)
\(=4000000-4000+1\)
\(=3996001\)
\(104^2-16=104^2-4^2\)
\(=\left(104-4\right)\left(104+4\right)\)
\(=100\cdot108\)
\(=10800\)
\(87\cdot93=\left(90-3\right)\left(90+3\right)\)
\(=90^2-3^2\)
\(=8100-9\)
\(=8091\)
Học tốt nhé!
a) 10000. b) 875000. c) -1.