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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

 

27 tháng 11 2015

\(=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)

 

 

27 tháng 11 2015

1/ (x+1)(x+2) +1/ (x+2)(x+3) +1/ (x+3)(x+4) +1/ (x+4)(x+5)

=1/x+1 -1/x+2 +1/x+2 -1/x+3 +1/x+3 -1/x+4 +1/x+4 -1/x+5

=1/x+1 -1/x+5

=4/(x+1)(x+5)

16 tháng 12 2020

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}=\frac{-5}{x\left(x-5\right)}\)

16 tháng 12 2020

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}\)

\(=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}\)

\(=\frac{x-5-x}{x\left(x-5\right)}\)

\(=-\frac{5}{x\left(x-5\right)}\)

27 tháng 1 2020

Khó vl , dẹp mẹ điiii

27 tháng 1 2020

a)     \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)

\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)

\(\Leftrightarrow A=4\)

b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)

\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)

\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)

\(\Leftrightarrow B=x^3-20x^2+18x+69\)

c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)

\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)

\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)

\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)

\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)

d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)

\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)

\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)

\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)

Chúc bạn học tốt !