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\(E=163^2+74\times163+37^2=163^2+2\times163\times37+37^2=\left(163+37\right)^2=200^2\)
\(F=147^2-94\times147+47^2=147^2-2\times147\times47+47^2=\left(147-47\right)^2=100^2\)
\(\frac{E}{F}=\frac{200^2}{100^2}=\left(\frac{200}{100}\right)^2=2^2=4\)
\(E=4F\)
\(A=163^2+74.163+37^2\)
\(=163^2+2.37.163+37^2\)
\(=\left(163+37\right)^2=200^2\)
\(B=147^2-94.147+47^2\)
\(=147^2-2.47.147+47^2\)
\(=\left(147-47\right)^2=100^2\)
Vậy A > B
Bài 9:
a) Ta có: \(A=\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)
\(=4x^2+4xy+y^2-4x^2+y^2-xy-y^2\)
\(=3xy-y^2\)
\(=3\cdot\left(-2\right)\cdot3-3^2=-18-9=-27\)
b) Ta có: \(B=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{31}{2}\)
Bài 7:
a) \(498^2=\left(500-2\right)^2=250000-2000+4=248004\)
b) \(93\cdot107=100^2-7^2=10000-49=9951\)
c) \(163^2+74\cdot163+37^2=\left(163+37\right)^2=200^2=40000\)
d) \(1995^2-1994\cdot1996=1995^2-1995^2+1=1\)
e) \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=18^8-18^8+1=1\)
f) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)
Làm dễ hiểu chút
\(A=\left(2^2+4^2+...+100^2\right)-\left(1^2+3^2+...+99^2\right)\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100-99\right)\left(99+100\right)\)
\(=3+7+...+199\)
\(B=3^8.7^8-\left(21^4-1\right)\left(21^4+1\right)\)
\(=21^8-\left(21^8-1\right)=1\)
Vậy A > B
a: \(=1995^2-\left(1995^2-1\right)=1995^2-1995^2+1=1\)
b: \(=18^8-18^8+1=1\)
c: \(=\left(163+37\right)^2=200^2=40000\)
Lời giải:
\(A=2018^2-2017.2019=2018^2-(2018-1)(2018+1)\)
\(=2018^2-(2018^2-1^2)=1\)
\(B=9^8.2^8-(18^4-1)(18^4+1)\)
\(=(9.2)^8-[(18^4)^2-1^2]\)
\(=18^8-(18^8-1)=1\)
\(C=163^2+74.163+37^2=163^2+2.37.163+37^2\)
\(=(163+37)^2=200^2=40000\)
\(D=\frac{2018^3-1}{2018^2+2019}=\frac{(2018-1)(2018^2+2018+1)}{2018^2+2019}\)
\(=\frac{2017(2018^2+2019)}{2018^2+2019}=2017\)
Sử dụng công thức \((a-b)(a+b)=a^2-b^2\)
\(E=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^8-1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^{16}-1)(2^{16}+1)-2^{32}\)
\(=(2^{32}-1)-2^{32}=-1\)