Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{a^2+ax+ab+bx}{a^2+ax-ab-bx}\)
\(=\frac{a\left(a+b\right)+x\left(a+b\right)}{a\left(a-b\right)+x\left(a-b\right)}\)
\(=\frac{\left(a+b\right)\left(a+x\right)}{\left(a-b\right)\left(a+x\right)}\)
\(=\frac{a+b}{a-b}\)
Thay \(a=5;b=2\) vào A ta có:
\(A=\frac{5+2}{5-2}=\frac{7}{3}\)
Vậy tại \(a=5;b=2\) thì A=7/3
c) \(\frac{a\left(a^2-ab+b^2\right)}{b\left(a+b\right)\left(a^2-ab+b^2\right)}\)
=\(\frac{a}{b\left(a+b\right)}\)
a/
\(\frac{a+b+c}{\left(a+b\right)^2-c\left(a+b\right)}.\frac{2a+2b}{a^2+2ab-c^2+b^2}\)
\(=\frac{a+b+c}{\left(a+b\right)\left(a+b-c\right)}.\frac{2\left(a+b\right)}{\left(a+b+c\right)\left(a+b-c\right)}\)
\(=\frac{2}{\left(a+b-c\right)^2}\)
b/ \(\frac{3x+3y}{x^2+y^2-2xy}:\frac{6x+6y}{ax-by+bx-ay}\)
\(=\frac{3\left(x+y\right)}{\left(x-y\right)^2}.\frac{\left(x-y\right)\left(a-b\right)}{6\left(x+y\right)}\)
\(=\frac{a-b}{2\left(x-y\right)}\)
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
B=\(\frac{5\left(x-y\right)-3\left(x-y\right)}{10\left(x-y\right)}\)
B=\(\frac{\left(x-y\right)\left(5-3\right)}{10\left(x-y\right)}\)
B= \(\frac{\left(x-y\right)2}{10\left(x-y\right)}\)
B= 5
vậy B=5
\(P=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow xyz=1\Rightarrow P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(P\ge\frac{\left(x+y+z\right)^2}{y+z+x+z+x+y}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\Leftrightarrow a=b=c=1\)
Cần cách khác thì nhắn cái
1, b) \(\frac{x^2+y^2-4+2xy}{x^2-y^2+4+4x}\) = \(\frac{\left(x^2+2xy+y^2\right)-4}{\left(x^2+4x+4\right)-y^2}\) =\(\frac{\left(x+y\right)^2-2^2}{\left(x+2\right)^2-y^2}\)= \(\frac{\left(x+y+2\right)\left(x+y-2\right)}{\left(x+2+y\right)\left(x+2-y\right)}\) = \(\frac{x+y-2}{x+2-y}\)
2, A= \(\frac{a^2+ax+ab+bx}{a^2+ax-ab-bx}\) = \(\frac{\left(a^2+ax\right)+\left(ab+bx\right)}{\left(a^2+ax\right)-\left(ab+bx\right)}\) = \(\frac{a\left(a+x\right)+b\left(a+x\right)}{a\left(a+x\right)-b\left(a+x\right)}\)= \(\frac{\left(a+x\right)\left(a+b\right)}{\left(a+x\right)\left(a-b\right)}\)= \(\frac{a+b}{a-b}\)
THANKS BN