Câu a bạn sửa lại đề 11→1

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

\(b,=\left[\dfrac{\left(1-x\right)\left(x^2+x+1\right)}{1-x}-x\right]\cdot\dfrac{\left(1+x\right)\left(1-x^2\right)}{1+x}\\ =\dfrac{\left(x^2+1\right)\left(1+x\right)\left(1-x^2\right)}{1+x}=\left(x^2+1\right)\left(1-x^2\right)=VP\)

C

B

\(y'=3\left(x^2-\dfrac{2}{x}\right)^2.\left(x^2-\dfrac{2}{x}\right)'=3\left(x^2-\dfrac{2}{x}\right)^2\left(2x+\dfrac{2}{x^2}\right)\)

\(=6\left(x+\dfrac{1}{x^2}\right)\left(x^2-\dfrac{2}{x}\right)^2\)

ngonhuminh

a) PT \(\Leftrightarrow\dfrac{x^2-x+2}{\left(x-1\right)^3}=\dfrac{A+B\left(x-1\right)+C\left(x-1\right)^2}{\left(x-1\right)^3}\)

\(\Leftrightarrow x^2-x+2=A+Bx-B+Cx^2-2Cx+C\)

\(\Leftrightarrow x^2-x+2=Cx^2+x\left(B-2C\right)+\left(A+C-B\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}C=1\\B-2C=-1\\A+C-B=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=2\\B=1\\C=1\end{matrix}\right.\)

b: \(\Leftrightarrow\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{A\cdot x^2+A+\left(Bx+C\right)\left(x-1\right)}{\left(x^2+1\right)\left(x-1\right)}\)

\(\Leftrightarrow x^2\cdot A+A+x^2\cdot B-x\cdot B+x\cdot C-C=x^2+2x-1\)

\(\Leftrightarrow x^2\left(A+B\right)+x\left(-B+C\right)+A-C=x^2+2x-1\)

=>A+B=1; -B+C=2; A-C=-1

=>A+C=3; A-C=-1; A+B=1

=>A=1; C=2; B=1-A=0

a)Ta thấy:

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

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

\(=\dfrac{a}{x\left(x+a\right)}\)

\(\Rightarrowđpcm\)

b)Ta thấy:

\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)

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

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

\(=\dfrac{\left(x+2\right)-x}{x\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\Rightarrowđpcm\)

c)Ta thấy:

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

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

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

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

\(=\dfrac{\left(x-x\right)+a}{x\left(x+a\right)}\) hay \(\dfrac{a}{x\left(x+a\right)}\)

\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\left(đpcm\right)\)

\(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)

\(\Leftrightarrow\dfrac{a\left(y+z\right)}{abc}=\dfrac{b\left(z+x\right)}{abc}=\dfrac{c\left(x+y\right)}{abc}\)

\(\Leftrightarrow\dfrac{\left(x+y\right)-\left(z+x\right)}{ab-ac}=\dfrac{y-z}{a\left(b-c\right)}\)

\(\Leftrightarrow\dfrac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{\left(z+x\right)-\left(y+z\right)}{ac-bc}=\dfrac{x-y}{c\left(a-b\right)}\)

\(\Rightarrow\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\left(đpcm\right)\)