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b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
Đặt $A=\dfrac1{(x+y)^3}\left(\dfrac1{x^3}+\dfrac1{y^3}\right)+\dfrac3{(x+y)^4}\left(\dfrac1{x^2}+\dfrac1{y^2}\right)+\dfrac6{(x+y)^5}\left(\dfrac1x+\dfrac1y\right).$
$=\dfrac{(x+y)^2(x^3+y^3)+3xy(x+y)(x^2+y^2)+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)^2(x+y)(x^2-xy+y^2)+3xy(x+y)(x^2+y^2)+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)\left[(x+y)^2(x^2-xy+y^2)+3xy(x^2+y^2)\right]+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)\left[x^4+x^3y+x^2y^2+xy^3+y^4+3x^3y+3xy^3\right]+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)\left(x^4+4x^3y+x^2y^2+4xy^3+y^4\right)+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)^4-6x^2y^2}{x^3y^3(x+y)^4}+\dfrac{6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)^5}{x^3y^3(x+y)^5}.$
$=\dfrac1{x^3y^3}.$
a) sửa đề: \(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
=\(\frac{-x^2\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-y^2\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(-\frac{\left\lbrace x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
xét tử số:
Tử=\(x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)
=\(x^2\left(y-z\right)-x\left(y^2-z^2\right)+yz\left(y-z\right)\)
=\(x^2\left(y-z\right)-x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\)
=\(\left(y-z\right)\left\lbrace x^2-x\left(y+z\right)+yz\right\rbrace\)
=\(\left(y-z\right)\left\lbrace x\left(x-y\right)-z\left(x-y\right)\right\rbrace\)
=\(\left(y-z\right)\left(x-y\right)\left(x-z\right)\)
=\(-\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
thay lại vào biểu thức cũ:
\(\Rightarrow-\frac{\left\lbrace-\left(x-y\right)\left(y-z\right)\left(z-x\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(1\)
b) \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)
=\(\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
A = \(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right).\left(x+1\right)}-\frac{x+3}{2\left(x+2\right)}\right).\frac{4x^2-4}{5}\)
A = \(\left(\frac{\left(x+1\right)^2+3.2-\left(x+3\right).\left(x-1\right)}{2\left(x-1\right).\left(x+1\right)}\right).\frac{4x^2-4}{5}\)
A = \(\left(\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right).\left(x+1\right)}\right).\frac{4\left(x^2-1\right)}{5}\)
A = \(\frac{10}{2\left(x-1\right).\left(x+1\right)}.\frac{4\left(x-1\right).\left(x+1\right)}{5}\)
A = 4
Ta có:
\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)
\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)
\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)
\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)
\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)
b/ Thế vô rồi tính nhé
Đoạn gần cuối thay y-x= 1 luôn
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)
\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\) giờ mới thay không biết đã tối giản chưa