9^8.2^8-(18^4-1)(18^4+1)
thực hiện phép tính: 9^8.2^8 - (18^4-1) (18^4+1)
Ta có :
\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=18^8-\left(18^8-1\right)\)
\(=18^8-18^8+1\)
\(=1\)
98.28-(184-1)(184+1)= (9.2)8 - [(184)2-12 ] = 188 - 188 +1 =1
tính nhanh
47.53
9^8.2^8-(18^4-1)(18^4+1)
Giải:
a) \(47.53\)
\(=\left(50-3\right).\left(50+3\right)\)
\(=50^2-3^2\)
\(=2500-9\)
\(=2491\)
Vậy ...
b) \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=\left(9.2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)
\(=18^8-\left(18^8-1\right)\)
\(=18^8-18^8+1\)
\(=1\)
Vậy ...
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
MẤY BẠN GIÚP MÌNH VỚI MÌNH CẦN GẤP LẮM
A = 19952 - ( 1995-1) (1995+1)
= 19952 - (19952 - 12)
= 19952 - 19952 +1
= 1
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:
a. \(134^2-68.134+34^2\)
b. \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
c. \(100^2-99^2+98^2-97^2+...+2^2-1\)
a. 134^2 - 68.134 + 34^2 = ( 134 - 34 ) ^2 = 100^2 = 10000
b. 9^8.2^8 - ( 18^4 - 1 )(18^4 + 1 ) = 18^8 - 18^8 + 1 = 1
c. 100^2 - 99^2 + 98^2 - 97^2 + ... + 2^2 - 1
=( 100 - 99 )( 100 + 99 ) + ( 98 - 97 )( 98 + 97 ) + ... + ( 2 - 1 )( 2 + 1 )
= 100 + 99 + 98 + 97 + ... + 2 + 1
=( 100 + 1 ).100:2 = 5050
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: \(\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1\right)^2-4x^2-\left(2x^2+1\right)^2\)
\(=-4x^2\)
\(9^8\).\(2^8\)-(\(18^4\)-1)(\(18^4\)+1)
\(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=\left(9\cdot2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)
\(=18^8-\left(18^8-1\right)\)
\(=18^8-18^8+1\)
\(=1\)
\(\frac{2^{18}.27^2}{9^{13}.4^8.2}\)
TÍNH NHẨM: 9^8-2^8-(18^4-1)(18^4+1)