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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: \(2016\cdot2018-2017^2\)
\(=\left(2017-1\right)\left(2017+1\right)-2017^2\)
\(=2017^2-1-2017^2\)
\(=-1\)
1
a,
=(202+54).(202-54)+256.352
=37888+256.352
=37888+90112
=128000
b,
=621-769.373-21904
=621-286837-21904
=-308120
c,
42^2-10^2/(36,5)^2-(27,5)^2
=(42-10).(42+10)/(36,5-27,5).(27,5+36,5)
=1664/576=2(8)
1. Tính nhanh :
a) \(202^2-54^2+256.352\)
\(=\left(202-54\right).\left(202+54\right)+256.352\)
\(=148.256+256.352\)
\(=256.\left(148+252\right)=256.400=102400\)
\(x^2+7x+12=x^2+3x+4x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)=0\)
\(x^2+3x-18=0\Leftrightarrow x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x+6\right)\left(x-3\right)=0\)
b) \(x^2-10x+16\)
\(=x^2-2x-8x+16\)
\(=x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x-8\right)\)
c) \(x^2+6x+8\)
\(=x^2+2x+4x+8\)
\(=x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
d) tươn tự c
e) \(x^2+3x-18=0\)
\(\Leftrightarrow x^2+6x-3x-18=0\)
\(\Leftrightarrow x\left(x+6\right)-3\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+6=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=3\end{cases}}}\)
Vậy ,...
\(621^2-769.373-148^2\)
\(=385641-286837-21904\)
\(=98804-76900\)
\(=21904\)
6212 - 769 . 373 - 1482
= 385641 - 769 . 373 - 21904
= 385641 - 286837 - 21904
= 76900
số hơi dài nha
~hok tốt~
a,=(202-54)(202+54)+256*352=248.*256+256*352=256*(248+352)=256*600=256*6*100=153600
b. làm tương tự
c,=5/(1+2+...+10)=5.\(\frac{10.\left(10+1\right)}{2}\)=275
(ta có công thức 1+2+...+n=\(\frac{n.\left(n+1\right)}{2}\) dễ dàng chứng minh)
2022 - 542 + 256.352
= (202 - 54).(202 + 54) - 256.352
= 148.256 + 256.352
= 256.(148 + 352)
= 256.500
= 128.2.500
= 128.1000
= 128000
c) Ta có: \(33\cdot55+33\cdot67+45\cdot33+67\cdot67\)
\(=33\left(55+45\right)+67\left(33+67\right)\)
\(=33\cdot100+67\cdot100\)
\(=100\cdot100=10000\)