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28 tháng 11 2016

Ta có:

\(A=\frac{x^3-53x+88}{\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)+16}=\frac{\left(x^3+8x^2\right)-\left(8x^2+64x\right)+\left(11x+88\right)}{\left[\left(x-1\right)\left(x-7\right)\right]\left[\left(x-3\right)\left(x-5\right)\right]+16}=\frac{x^2\left(x+8\right)-8x\left(x+8\right)+11\left(x+8\right)}{\left[x^2-8x+7\right]\left[x^2-8x+15\right]+16}=\frac{\left(x+8\right)\left(x^2-8x+11\right)}{\left[x^2-8x+7\right]\left[x^2-8x+15\right]+16}\)

Gọi \(x^2-8x+11=y\)

\(\Rightarrow A=\frac{y\left(x-8\right)}{\left(y-4\right)\left(y+4\right)+16}=\frac{y\left(x-8\right)}{y^2-16+16}=\frac{y\left(x-8\right)}{y^2}=\frac{x-8}{x^2-8x+11}\)

 

28 tháng 11 2016

tôi không thấy hết dk cách làm bạn ak

 

17 tháng 7 2019

A = 5(x + 3)(x - 3) + (2x + 3)3 + (x - 6)2

A = 5(x + 3)(x - 3) + 4x2 + 12x + 9 + x2 - 12x + 36

A = 5x2 - 45x + 4x2 + 12x + 9 + x- 12x + 36

A = 10x2 (1)

Thay x = -1/5 vào (1), ta có:

A = 10x2 = 10.(-1/5)2 = 2/5

A = 2/5

Vậy:...

NM
12 tháng 10 2021

ta có:

undefined

12 tháng 10 2021

Thầy làm chi tiết giúp em được ko ạ ?

31 tháng 3 2020

\(\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{2}{\left(x+2\right)^2}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{\left(x+3\right)\left(x+2-2x-2\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{\left(x+3\right)\left(-x\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{-x^2-3x+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

1 tháng 4 2020

ĐKXD: x\(\ne\)-1,-2,-3

Ta có

\(\frac{1}{\left(x+1\right)\left(x+2\right)}\)-\(\frac{2}{\left(x+2\right)^2}\)+\(\frac{1}{\left(x+2\right)\left(x+3\right)}\)

=\(\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{\left(x+2\right)\left(x+3+x+1\right)-2\left(x^2+4x+3\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{\left(x+2\right)\left(2x+4\right)-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{2x^2+8x+8-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

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