A= 2 + 2^2 + 2^3....+ 2^99
Tính giá trị biểu thức
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1, a)
Ta có:
\(x^2+2x+1=\left(x+1\right)^2\)
Thay x=99 vào ta có:
\(\left(99+1\right)^2=100^2=10000\)
b) Ta có:
\(x^3-3x^2+3x-1=\left(x-1\right)^3\)
Thay x=101 vào ta có:
\(\left(101-1\right)^3=100^3=1000000\)
Giải:
\(B=1+2\cdot\left(1+1\right)+3\cdot\left(2+1\right)+...+99\cdot\left(98+1\right)+100\cdot\left(99+1\right)\)
\(B=1+1\cdot2+2\cdot3\cdot3+...+98\cdot99+99+99\cdot100+100\)
\(B=\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)+\left(1+2+3+...+99+100\right)\)
\(B=333300+5050\)
\(B=3338050\)
a)
C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.
b)
B = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.
Đặt A = 1 + 2 + 3 + ... + 97 +98 + 99
A = 99 + 98 + 97 + ... + 3 + 2 + 1
2A = ( 99 + 1 ) + ( 98 + 2 ) + ( 97 + 3 ) + ... + ( 3 + 97 ) + ( 98 + 2 ) + ( 99 + 1 ) ( 99 cặp số )
2A = 100 + 100 + 100 + ... + 100 + 100 + 100 ( 99 số )
2A = 100 . 99
2A = 9900
A = 4950
Vậy A = 4950
Ta có :
1+2+3+....+99
= \(\frac{\left(1+99\right).99}{2}\)
\(=50.99\)
\(=4950\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
ĐẶT : A= \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)\(\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
= \(1-\frac{1}{99}=\frac{98}{99}\)
\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{99.100}\)
=\(9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
=\(9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
=\(9.\left(\frac{1}{1}-\frac{1}{100}\right)\)
=\(9.\frac{99}{100}\)
=\(\frac{891}{100}\)
\(A=\left(2x-3\right)^2-\left(4x-6\right)\left(2x-5\right)+\left(2x-5\right)^2\)
\(=\left(2x-3\right)^2-2\left(2x-3\right)\left(2x-5\right)+\left(2x-5\right)^2\)
\(=\left(2x-3-2x+5\right)^2\)
\(=4\)
Vì giá trị bt trên ko phụ thuộc vào biến nên giá trị của bt luôn là 4
\(1^2-2^2+3^2-4^2+...-100^2+101^2\)
\(\left(1-2\right).\left(1+2\right)+\left(3-4\right)\left(3+4\right)\)\(+...+\left(99-100\right).\left(99+100\right)+101^2\)
\(-3-7-11-...-199+101^2\)
\(101^2-\left(3+7+11+...+199\right)\)
Ta de thay :(3+7+11+ . . .+199) la 1 cap so cong co d=4 ,n=50
\(101^2-\left(199+3\right)\cdot50:2\)
\(=5151\)
\(A=2+2^2+2^3+....+2^{99}\)
\(\Leftrightarrow2A=2^2+2^3+2^4+...+2^{100}\)
\(\Leftrightarrow2A-A=A=\left(2^2+2^3+2^4+....+2^{100}\right)-\left(2+2^2+2^3+....+2^{99}\right)\)
\(\Leftrightarrow A=2^{100}-2\)