Tính Nhanh
A= 1995^2 -1994.1996
B= 9^8.2^8-(18^4-1).(18^4+1)
C=163^2+74.163+37^2
D=127^2+146.127+73^2
Bài 7. Tính nhanh
a/ 498mũ 2
b/ 93. 107
c/ 163 mũ 2+ 74.163 + 37mũ 2
d/ 1995 mũ 2 – 1994.1996
e/ 9 mũ 8.2 mũ 8 – (18mũ 4 – 1)(18 mũ 4+ 1)
f/ 125 mũ 2 - 2. 125. 25 + 25 mũ 2
Bài 8. Rút gọn các biểu thức sau
a/ (x mũ 2+ 3x+ 1)mũ 2 + (3x – 1) mữ 2 – 2(x mũ 2+ 3x+ 1)(3x– 1)
b/ (3x mũ 3+ 3x + 1)(3x mũ 3– 3x +1) – (3xmũ 3+1)mũ 2
c/ (2xmũ2+ 2x + 1)(2xmũ2 – 2x + 1) – (2xmũ 2+ 1)mũ 2
Bài 9. Rút gọn rồi tính giá trị biểu thức
a/ A = (2x + y)mũ 2 - (2x + y) (2x - y)+ y(x - y) vì x= - 2; y= 3.
b/ B = (a - 3b)mũ 2 - (a + 3b)mũ 2 - (a -1)(b -2 ) vì a =1/2; b = -3.
MN GIÚP MIK VS MIK CẦN GẤP
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) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)
Bài 8:
a) Ta có: \(\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x+1-3x+1\right)^2\)
\(=\left(x^2+2\right)^2\)
\(=x^4+4x^2+4\)
b) Ta có: \(\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)
\(=-9x^2\)
c) Ta có: \(\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1\right)^2-4x^2-\left(2x^2+1\right)^2\)
\(=-4x^2\)
tính nhanh các kết quả sau
a)1995^2-1994.1996
b)9^8.2^8-(18^4-1)(18^4+1)
c)163^2+74.163+37^2
d)\(\frac{1993^2-1}{1996^2+1997}\)
a: \(=1995^2-\left(1995^2-1\right)=1995^2-1995^2+1=1\)
b: \(=18^8-18^8+1=1\)
c: \(=\left(163+37\right)^2=200^2=40000\)
tính nhanh:
1) A=19952 - 1994.1996
2) B=98.28-(184-1).(184+1)
3) C= 1632+74.163+372
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
Tính hợp lí:
a) A= 1995^3+1/1995^2-1994
b) B= 3^8.7^8-(21^4-1).(21^4+1)
c) C= 163^2+74.163+37^2
d) E= 147^2-94.147+47^2
e) F= 3^24-(27^4+1).(9^6-1)
f) D= (2^2+4^2+...+100^2)-(1^2-3^2+...+99^2)
cho đa thức 2x^2-5x+3 Viết đa thức dưới dạng 1 đa thức của biến y trong đó y=x+1
tính nhanh
a) 1272+146.127+732
b) 9^2-2^8-(18^4-1)(18^4+1)
1) \(2x^2-5x+3=2x^2-2x-3x+3=2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x-3\right)\left(x-1\right)=\left(2x+2-5\right)\left(x+1-2\right)=\left(2\left(x+1\right)-5\right)\left(x+1-2\right)\)
\(=\left(2y-5\right)\left(y-2\right)\)
Tính:
a,A=127^2+146×127+73^2
b,B=9^8×2^8-(18^4-1)×(18^4+1)+(a+b)^2
c, C=(20^2+18^2+...+4^2+2^2)-(19^2+17^2+...+3^2+1)
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 \(\frac{\left(39+3\right).10}{2}=210\)
Tính nhanh:
a) 1272 + 146.127 + 722
b) 98.28 - (184 - 1)(184 + 1)
c) 202 + 182 + 162 + .... + 42 + 22 - (192 + 172 + .... + 32 + 1)
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!
\(I\)Tính nhanh
\(a.127^2+146.127+73^2\)
\(b.9^8.2^8\left(18^4-1\right)\left(18^4+1\right)\)
\(c.100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(d.\frac{780^2-220^2}{125^2+150.125+75^2}\)
\(II.\)Rút gọn các biểu thức
\(x^2\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
1272 + 146.127 + 732
= 1272 + 2 . 73 .127 + 732
= (127 + 73 ) 2
= 200 2
Bài 1: Tính nhanh:
a) \(127^2+146.127+73^2\)
b) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
c) \(100^2-99^2+98^2-97^2+...+2^2-1\)
d) \(\dfrac{780^2-220^2}{125^2+150.125+75^2}\)
Bài 2 : So sánh:
a) \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)và \(B=2^{32}\)
b) \(C=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)và \(D=3^{32}-1\)
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,
\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)