Tinh gia tri bieu thuc: A=1/2 + 1/6 + 1/12 + 1/20 +1/30 .......... + 1/9900
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\(\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{6\times7}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)= 1 - 1/7 = 6/7
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Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
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a) 85 . 127 + 5 . 127 . 3
= (85 + 15) . 127
= 100 . 127
= 12700
a) 85 . 127 + 5 . 127 . 3
= (85 + 15) . 127
= 100 . 127
= 12700
b) 1/2 + 5/6 + 11/12 +19/20 + 29/30 + 41/42 + 55/56 + 71/72 + 89/90
1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
1-1/10
9/10
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\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{7}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}\)
\(A=\frac{7}{60}\)
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12 x 4 ( a x 1 + a : 1 )
= 12 x 4 ( a + a )
= 12 x 4 x 2 x a
= 96 x a
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\(A=x\left(x+4\right)-6\left(x-1\right)\left(x+1\right)+\left(2x-1\right)^2\)
\(A=x^2+4x-6\left(x^2-1\right)+\left(4x^2-4x+1\right)\)
\(A=x^2+4x-6x^2+6+4x^2-4x+1\)
\(A=-x^2+7\)
Để A có giá trị bằng 3 thì :
\(-x^2+7=3\)
\(-x^2=-4\)
\(x^2=4\)
\(x\in\left\{\pm2\right\}\)
Vậy..........
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Mọi người tk mình đi mình đang bị âm nè!!!!!!
Ai tk mình mình tk lại nha !!!
A = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{9900}\)
A = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{99x100}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
A = \(1-\frac{1}{100}=\frac{99}{100}\)
\(A=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{1}-\frac{1}{100}\)
\(A=\frac{99}{100}\)