Rút gọn biểu thức:
a/ A=100^2-99^2+98^2-97^2+...+2^2-1^2
b/ B=3(2^2+1)(2^4+1)...!2^64+1)+1
K
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24 tháng 8 2016
a)M=2100-299+298-...+22-2
22M=2102-2101+2100-...+22-2
4M-M=2102-2101+2100-...+22-2-2100+299-...-22+2
3M=2102-2101
M=\(\frac{2^{102}-2^{101}}{3}\)
5 tháng 1 2022
\(a,\dfrac{3x+21}{x^2-9}+\dfrac{2}{x+3}-\dfrac{3}{x-3}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)
\(b,\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x+3}{x^2-1}\\ =\dfrac{3x^2+4x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+4x+1-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2+2x-3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{2x^2+6x-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x^2+3x\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x+3}{\left(x-1\right)^2}\)
LP
3
9 tháng 8 2016
a) 5A = 5 + 5^2 + 5^3 + 5^4 +...+ 5^51
=> 5A - A = 4A = 5^51 - 1
=> A = \(\frac{5^{51}-1}{4}\)
b) 3B = 3^100 - 3^99 -...- 3
=> 3B - B = 2B = 3^100 - 2.3^99 + 1
=> B = \(\frac{3^{100}-2\times3^{99}+1}{2}\)
9 tháng 8 2016
a, 1+5+52+.....+550
=> 5(1+5+52+.....+550)=5+52+53.....+551
=>4(1+5+52+.....+550)=551-1
=>1+5+52+.....+550=(551-1):4
b,399-398-...-3-1
=399-(398+...+3+1)
=399-(399-1):2
13 tháng 4 2023
a: \(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{2\left(x-3\right)}{2-x}\)
\(=\dfrac{4+4x+x^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{2-x}{2\left(x-3\right)}\)
\(=\dfrac{5x^2+4x+4-4+4x-x^2}{\left(2+x\right)}\cdot\dfrac{1}{2\left(x-3\right)}\)
\(=\dfrac{4x^2+8x}{x+2}\cdot\dfrac{1}{2\left(x-3\right)}=\dfrac{4x\left(x+2\right)}{2\left(x+2\right)}\cdot\dfrac{1}{x-3}=\dfrac{2x}{x-3}\)
b: |x-2|=2
=>x-2=2 hoặc x-2=-2
=>x=0(nhận) hoặc x=4(nhận)
Khi x=0 thì \(A=\dfrac{2\cdot0}{0-3}=\dfrac{-2}{3}\)
Khi x=4 thì \(A=\dfrac{2\cdot4}{4-3}=8\)
c: A>0
=>x/x-3>0
=>x>3 hoặc x<0
=>x>3
TH
14 tháng 1 2018
a) A= -a+b-c + a+b+c = 2b
b) Vì giá trị của A = không phụ thuộc vào a hay c nên A=2b=2.(-1)= -2
14 tháng 1 2018
a, Rút gọn
A = ( - a - b - c ) - ( - a - b - c )
= - a - b - c - a - b - c
= 2b
1 tháng 8 2016
a)
\(A=\frac{6^3+3.6^3+3^3}{-13}=\frac{3^3.2^3+3^3.2^2+3^3}{-13}=\frac{3^3\left(8+4+1\right)}{-13}=\frac{27.13}{-13}=-27\)
b)
A=1+5+52+53+...+550
5A=5+52+53+...551
5A-A=(5+52+53+...+551)-(1+5+52+...+550)
4A=551-1
A=\(\frac{5^{51}-1}{4}\)
c)
A=2100-299+298-...+22-2
2A=2101-2100+299-...+23-22
2A+A=(2101-2100+...+23-22)+(2100-299+...+22-2)
3A=2101-2
A=\(\frac{2^{101}-2}{3}\)
TN
1 tháng 8 2016
b.
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
\(5A=5+5^2+5^3+...+5^{50}+5^{51}\)
\(5A-A=\left(5+5^2+5^3+...+5^{50}+5^{51}\right)-\left(1+5+5^2+..+5^{50}\right)\)
\(4A=5^{51}-1\)
\(A=\frac{5^{51}-1}{4}\)
30 tháng 10 2018
Xét mẫu số của phân số:
\(\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}\)
\(=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+\left(\dfrac{99}{1}-98\right)\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+1\)
\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\)
Ta thấy mẫu số gấp tử số 100 lần. Vậy phân số đó có giá trị bằng \(\dfrac{1}{100}\)
\(A=100^2-99^2+98^2-97^2+....+2^2-1^2\)
\(=\left(100-99\right).\left(100+99\right)+\left(98-97\right).\left(98+97\right)+....+\left(2-1\right).\left(2+1\right)\)
\(=1+2+....+97+98+99+100=\frac{100.\left(100+1\right)}{2}=5050\)
\(B=3\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)+1=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)......\left(2^{64}+1\right)+1=\left(2^8-1\right).....\left(2^{64}+1\right)+1\)
Tiếp tục rút gọn như vậy,ta đc \(B=\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1+1=2^{128}\)