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bài 1:
a,\(\left(x+1\right)^3-\left(x+3\right)^2\cdot\left(x+1\right)+4x^2=\)-12
\(\Rightarrow\left(x+1\right)\cdot[\left(x+1\right)^2-\left(x+3\right)^2]+4x^2=-12\)
\(\Rightarrow\left(x+1\right)\cdot[\left(x+1+x+3\right)\cdot\left(x+1-x-3\right)]+4x^2=-12\)
\(\Rightarrow\left(x+1\right)\cdot\left(2x+4\right)\cdot\left(-2\right)+4x^2=-4\cdot3\)
\(\Rightarrow\left(x+1\right)\cdot2\cdot\left(x+2\right)\cdot\left(-2\right)+4x^2=-4\cdot3\)
\(\Rightarrow\left(x+1\right)\cdot\left(x+2\right)\cdot\left(-4\right)+4x^2=-4\cdot3\)
\(\Rightarrow\left(x+1\right)\cdot\left(x+2\right)-x^2=3\)
\(\Rightarrow x^2+2x+x+2-x^2=3\)
\(\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)
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
Ta có:\(77^2+23^2+77.46\)
\(=77^2+2.23.77+23^2\)
\(=\left(77+23\right)^2\)
\(=100^2=10000\)
=77*77+23*23+77*46
=77*(77+46)+23*23
=77*123+23*23
=9471+529
=1000
Lời giải:
$A=64^2-54^2=(64-54)(64+54)=10.118=1180$
$B=76^2+2.76.23+23^2=(76+23)^2=99^2=(100-1)^2=100^2-2.100+1$
$=10000-200+1=9801$
$C=88^2+62^2-38^2-12^2=(88^2-38^2)+(62^2-12^2)$
$=(88-38)(88+38)+(62-12)(62+12)$
$=50.126+50.74=50(126+74)=50.200=10000$
\(202^2-54^2+256.352\)
\(=\left(202-54\right)\left(202+54\right)+256.352\)
\(=\left(202-54\right).256+256.352\)
\(=256\left(202-54+352\right)=256.300=76800\)
\(202^2-54^2+256.352\)
\(\Leftrightarrow\left(202-54\right).\left(202+54\right)+256.352\)
\(\Leftrightarrow\left(202-54\right).256+256.352\)
\(\Leftrightarrow256.\left(202-54+352\right)=128000\)
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)
không biết đang bí