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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) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)
a: \(=1995^2-\left(1995^2-1\right)=1995^2-1995^2+1=1\)
b: \(=18^8-18^8+1=1\)
c: \(=\left(163+37\right)^2=200^2=40000\)
1) A=19952-1994.1996
=19952-(1995-1)(1995+1)
=19952-(19952-1)
=1
2) B=98.28-(184-1)(184+1)
=(9.2)8-[(184)2-1]
= 188-188+1
=1
3) C=1632+74.163+372
=1632+2.37.163+372
=1632+2.163.37+372
=(163+37)2.2
=80000
1) \(2x^2-5x+3=2x^2-2x-3x+3=2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x-3\right)\left(x-1\right)=\left(2x+2-5\right)\left(x+1-2\right)=\left(2\left(x+1\right)-5\right)\left(x+1-2\right)\)
\(=\left(2y-5\right)\left(y-2\right)\)
Answer:
\(A=127^2+146.127+73^2\)
\(=127^2+2.127.73+73^2\)
\(=\left(127+73\right)^2\)
\(=200^2\)
\(=40000\)
\(B=9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=\left(9.2\right)^8-[\left(18^4\right)^2-1]\)
\(=18^8-18^8+1\)
\(=1\)
\(C=\left(20^2+18^2+16^2+...+4^2+2^2\right)-\left(19^2+17^2+15^2+...+3^2+1^2\right)\)
\(=20^2+18^2+16^2+...+4^2+2^2-19^2-17^2-15^2-...-3^2-1^2\)
\(=\left(20^2-19^2\right)+\left(18^2-17^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...+\left(2-1\right)+\left(2+1\right)\)
\(=1.39+1.35+...+1.3\)
\(=39+35+...+3\)
Số số hạng \(\frac{39-3}{4}+1=10\) số hạng
Tổng \(\frac{\left(39+3\right).10}{2}=210\)
Giải:
a) Sửa đề: 1272 + 146.127 + 732
\(127^2+146.127+73^2=\left(127+7\right)^2=200^2=40000\)
b) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)=18^8-\left(18^4-1\right)^2=18^8-18^8-1=-1\)
c) \(20^2+18^2+16^2+...+4^2+2^2-\left(19^2+17^2+...+3^2+1\right)\)
\(=20^2+18^2+16^2+...+4^2+2^2-19^2-17^2-...-3^2-1\)
\(=\left(20^2-19^2\right)+\left(18^2-17^2\right)+\left(16^2-15^2\right)+...+\left(4^2-3^2\right)+\left(2^2-1\right)\)
\(=20+19+18+17+16+15+...+4+3+2+1\)
\(=\dfrac{\left(20+1\right).20}{2}=210\)
Chúc bạn học tốt!
1272 + 146.127 + 732
= 1272 + 2 . 73 .127 + 732
= (127 + 73 ) 2
= 200 2
Bài 1:
a,\(127^2+146.127+73^2=127^2+2.127.73+73^2\)\(=\left(127+73\right)^2=200^2=40000\)
b,\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(18^8-\left(18^8-1\right)=1\)
\(c,100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=199+195+...+3\)
áp dụng công thức Gauss ta đc đáp án là:10100
d, mk khỏi ghi đề dài dòng:
\(\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560000}{40000}=14\)Bài 2:
\(A=\left(2-1\right)\left(2+1\right)\)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)Cứ tiếp tục ta đc \(A=2^{32}-1< B=2^{32}\)
\(\left(3-1\right)C=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^2+16\right)\)giải như câu a đc:\(\left(3-1\right)C=3^{32}-1\)
\(\Rightarrow C=\dfrac{3^{32}-1}{3-1}=\dfrac{3^{32}-1}{2}< D=3^{32}-1\)
1c,
\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)
MẤY BẠN GIÚP MÌNH VỚI MÌNH CẦN GẤP LẮM
A = 19952 - ( 1995-1) (1995+1)
= 19952 - (19952 - 12)
= 19952 - 19952 +1
= 1