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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=502 - 492 +482 -472 +...+22 -12
=(502 - 492)+(482 -472 )+.....+(22-1)(
= (50 - 49)(50 + 49) + (48 – 47)(48 + 47) + ... +
(2 + 1)(2 – 1)
=(50+49).1+(48+47).1+.....+(2+1).1
= 50 + 49 + 48 + 47 + ... + 2 + 1
= (50 + 1) + (49 + 2) + ... + (25 +26)
= 51 . 25 = 1275
- \(\Leftrightarrow\left(a^2+4\right)x=3a^2-48\Leftrightarrow x=\frac{3a^2-48}{a^2+4}\)
- \(\Leftrightarrow\left(a^2+5\right)x=a^2\Leftrightarrow x=\frac{a^2}{a^2+5}\)
a) \(1001^2=\left(1000+1\right)^2=1000^2+2.1000.1+1^2=1002001\)
b) \(29,9\times30,1=\left(30-0,1\right).\left(30+0,1\right)=30^2-\left(0,1\right)^2=899,99\)
c) \(\left(31,8\right)^2-2.31,8.21,8+\left(21,8\right)^2=\left(31,8-21,8\right)^2=10^2=100\)
\(A=a^3+b^3+3ab=\left(a+b\right)\left(a^2+b^2-ab\right)+3ab=a^2+b^2-ab+3ab\)
\(A=a^2+b^2+2ab\)
\(A=\left(a+b\right)^2=1\)