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a: \(=\dfrac{1}{x-1}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}\)
\(=\dfrac{2}{x-1}-\dfrac{1}{x-3}\)
\(=\dfrac{2x-6-x+1}{\left(x-1\right)\left(x-3\right)}=\dfrac{x-5}{\left(x-1\right)\left(x-3\right)}\)
b: \(=\dfrac{x^2-2x+4}{x+2}-\left(x+2\right)\)
\(=\dfrac{x^2-2x+4-x^2-4x-4}{x+2}=\dfrac{-6x}{x+2}\)
c: \(=\dfrac{1-x+2x}{\left(1-x\right)\left(1+x\right)}\cdot\dfrac{1-x}{x}\)
\(=\dfrac{x+1}{x+1}\cdot\dfrac{1}{x}=\dfrac{1}{x}\)
d: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{x^2+2x+1}\)
Bài 1:
a: \(A=y^3-4y^2+2y-8-y^3+4y^2-\dfrac{1}{2}y+2\)
\(=\dfrac{3}{2}y-6=\dfrac{3}{2}\cdot\dfrac{-2}{3}-6=-1-6=-7\)
b: \(x+1=5\)
\(B=x^5-x^4\left(x+1\right)+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-2\)
=x-2
=4-2
=2
c: \(C=\dfrac{1}{229}\cdot\left(6+\dfrac{3}{433}\right)-\dfrac{1}{299}\cdot\dfrac{432}{433}-\dfrac{1}{229}\cdot\dfrac{4}{433}\)
\(=\dfrac{1}{299}\left(6+\dfrac{3}{433}-\dfrac{432}{433}-\dfrac{4}{333}\right)\)
\(=\dfrac{-5}{299}\)
! Bài khó quá à!
Thank you very much!!!!
\(S=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{3\cdot\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\cdot\dfrac{1}{2}}{2a^2+2b^2+2c^2-2ab-2bc-2ac}=\dfrac{3}{2}\)