tính nhanh:
a)A=75^2+50*75+25^2
b)123^2+23^2-46*123
c)(3^4-1)*(3^4+1)-9^4
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a) \(A=123\left(123+154\right)+77^2\)
\(A=123^2+\left(123.154\right)+77^2=\left(123+77\right)^2=200^2=400\)
b) \(B=3^{24}-\left(2^{47}+1\right)\left(9^6-1\right)\)
\(B=3^{24}-\left(3^{12}-1\right)\left(3^{12}+1\right)\)
\(B=3^{24}-3^{24}+1=1\)
c) \(C=85^2+75^2+65^2+55^2-45^2-35^2-25^2-15^2\)
\(C=\left(85^2-15^2\right)+\left(75^2-25^2\right)+\left(65^2-35^2\right)+\left(55^2-45^2\right)\)
\(C=\left(85+15\right)\left(85-15\right)+\left(75+25\right)\left(75-25\right)+\left(65+35\right)\left(65-35\right)\left(55+45\right)\left(55-45\right)\)
\(C=100\left(60+50+40+30+20+10\right)\)
\(C=100.210=21000\)
B=\(3^{24}-\left(27^4+1\right)\left(9^6+1\right)=-1062883\)A=123(123+154)+772=123.277+772=34843
C=\(85^2+75^2+65^2+55^2-45^2-35^2-25^2-15^2=16000\)
a)\(-7-25+27+5=\left(27-7\right)+\left(5-25\right)\)
\(=20-20=0\)
b)\(-8+16-\left(-8\right)=-8+16+8=\left(8-8\right)+16=16\)
c)\(1+1-\left(-8\right)=1+1+8=10\)(Vì số nào mũ 0 cũng bằng 1(ngoại trừ số 0))
d)\(65.23+65.35-35.65+35.23\)
\(=\left(65.35-65.35\right)+23.\left(65+35\right)\)
\(=0+23.100=2300\)
f) \(8+\left[1800-\left(64-54\right)^3\right].8-5\)
\(=8+\left(1800-10^3\right).8-5\)
\(=8+800.8-5=6403\)
h)\(-123-46+46+123=\left(123-123\right)+\left(46-46\right)\)
\(=0+0=0\) (phá ngoặc ra thôi nhé)
a) Ta có:
\(A=\dfrac{-68}{123}\cdot\dfrac{-23}{79}=\dfrac{68}{123}\cdot\dfrac{23}{79}\)
\(B=\dfrac{-14}{79}\cdot\dfrac{-68}{7}\cdot\dfrac{-46}{123}=-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)\)
\(C=\dfrac{-4}{19}\cdot\dfrac{-3}{19}\cdot...\cdot\dfrac{0}{19}\cdot...\cdot\dfrac{3}{19}\cdot\dfrac{4}{19}=0\)
Suy ra A là số hữu tỉ dương, B là số hữu tỉ âm và C là 0.
Vậy A > C > B.
b) Ta có:
\(\dfrac{B}{A}=\dfrac{-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)}{\dfrac{68}{123}\cdot\dfrac{23}{79}}=-\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\cdot\dfrac{123}{68}\cdot\dfrac{79}{23}\)
\(\dfrac{B}{A}=-\dfrac{14\cdot68\cdot46\cdot123\cdot79}{79\cdot7\cdot123\cdot68\cdot23}=-\left(2\cdot2\right)=-4\)
Vậy B : A = -4
Lời giải:
a)
\(A=75^2+50.75+25^2=75^2+2.25.75+25^2\)
\(=(75+25)^2=100^2=10000\)
b) \(123^2+23^2-46.123=123^2-2.23.123+23^2\)
\(=(123-23)^2=100^2=10000\)
c) \((3^4-1)(3^4+1)-9^4\)
\(=[(3^4)^2-1^2]-9^4=(9^4-1)-9^4=-1\)