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NV
3 tháng 3 2022

\(M=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{900}\right)\)

\(=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{30^2}\right)\)

\(=\left(\dfrac{2^2-1}{2^2}\right)\left(\dfrac{3^2-1}{3^2}\right)\left(\dfrac{4^2-1}{4^2}\right)...\left(\dfrac{30^2-1}{30^2}\right)\)

\(=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{2.4}{3^2}\right)\left(\dfrac{3.5}{4^2}\right)...\left(\dfrac{29.31}{30^2}\right)\)

\(=\left(\dfrac{1.2.3...29}{2.3.4...30}\right).\left(\dfrac{3.4.5...31}{2.3.4...30}\right)=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

NV
2 tháng 3 2022

\(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)

Do đó:

\(M=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{30^2}\right)\)

\(=\dfrac{\left(2-1\right)\left(2+1\right)}{2^2}.\dfrac{\left(3-1\right)\left(3+1\right)}{3^2}.\dfrac{\left(4-1\right)\left(4+1\right)}{4^2}...\dfrac{\left(30-1\right)\left(30+1\right)}{30^2}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{29.31}{30^2}=\dfrac{1.2.3...29}{2.3.4...30}.\dfrac{3.4.5...31}{2.3.4...30}\)

\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

a: Ta có: \(P=\left(\dfrac{1}{a+\sqrt{a}}+\dfrac{1}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)

\(=\dfrac{a+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)

\(=\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\)

11 tháng 7 2018

1) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x\left(x^2-16\right)\)

\(=x^3-16x-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x^3-16x-x^4+1\)

b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)

\(=28xy-7x^2+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)

\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)

\(=-7x^2+7x\)

c) \(\left(3x-1\right)\left(2x-5\right)-4\left(2x^2-5x+2\right)\)

\(=6x^2-17x+5-4\left(2x^2-5x+2\right)\)

\(=6x^2-17x+5-8x^2+20x-8\)

\(=-2x^2+3x-3\)

11 tháng 7 2018

a)  x(x+4)(x-4)-(x2+1)(x2-1)

=>x(x2-42)-(x4-12)

=>x3-16x-x4+1

=>-x4-x3-15x

b)  7x(4y-x)+4y(y-7x)-2(2y2-3.5x)

=>28xy-7x2+4y2-28xy-4y2+30x

=>-7x2+30x

c)  (3x+1)(2x-5)-4(2x2-5x+2)

=>6x2-15x+2x-5-8x2+20x-8

=>-2x2+7x-13

21 tháng 7 2021

`(6x+1)^2-2(1+6x)(6x-1)+(6x-1)^2`

`=(6x+1-6x+1)^2`

`=2^2=4`

 

b: Ta có: \(N=a^3+b^3+3ab\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)

\(=1-3ab+3ab\)

=1