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24 tháng 8 2016

\(\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)

\(=\frac{\left(2x+y\right)^22\left(4x^2-y^2\right)+\left(2x-y\right)^2}{\left(2x-y\right)^2\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{16x^2}{16x\left(2x-y\right)^2}=\frac{x}{\left(2x-y\right)^2}\)

24 tháng 8 2016

\(\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-4^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)

\(=\frac{\left(2x+y\right)^22\left(4x^2-y^2\right)+\left(2x-y\right)^2}{\left(2x-y\right)^2\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{16x^2}{16x\left(2x-y\right)^2}=\frac{x}{\left(2x-y\right)^2}\)

29 tháng 2 2020

\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)

\(=\left(\frac{1}{a}+\frac{1}{b}\right)^2.\frac{ab}{\left(a+b\right)^2}\)

\(=\frac{1}{ab}\)

\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+14xy+y^2}{16x}\)

\(=\frac{\left(2x+y\right)^2+2\left(2x+y\right)\left(2x-y\right)+\left(2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{x}{\left(2x-y\right)^2}\)

29 tháng 2 2020

\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)

ĐK: a, b khác 0, a khác -b

\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{a+b}{ab}\right)\right].\frac{ab}{\left(a+b\right)^2}\)

\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}\right].\frac{ab}{\left(a+b\right)^2}=\left(\frac{1}{a}+\frac{1}{b}\right)^2.\frac{ab}{\left(a+b\right)^2}\)

\(A=\frac{\left(a+b\right)^2}{ab}.\frac{ab}{\left(a+b\right)^2}=1\)

 \(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(4x^2-y^2\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16xy}\)

ĐK: xy khác 0, y  \(\ne\pm\)2x

\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(2x-y\right).\left(2x+y\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{\left(2x+y\right)^2}{16xy}\)

\(B=\left[\frac{1}{\left(2x-y\right)}+\frac{1}{\left(2x+y\right)}\right]^2.\frac{\left(2x+y\right)^2}{16xy}\)

\(B=\left(\frac{2x+y+2x-y}{\left(2x-y\right).\left(2x+y\right)}\right)^2.\frac{\left(2x+y\right)^2}{16xy}\)

\(B=\frac{16x^2}{\left(2x-y\right)^2.\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16xy}\)

\(B=\frac{x}{\left(2x-y\right)^2.y}\)

AH
Akai Haruma
Giáo viên
21 tháng 2 2020

Lời giải:

a) ĐK: $a\neq -b\neq 0$

\(A=\left(\frac{a^2+b^2}{a^2b^2}+\frac{2}{a+b}.\frac{a+b}{ab}\right).\frac{ab}{(a+b)^2}\)

\(=\left(\frac{a^2+b^2}{a^2b^2}+\frac{2ab}{a^2b^2}\right).\frac{ab}{(a+b)^2}=\frac{(a+b)^2}{a^2b^2}.\frac{ab}{(a+b)^2}=\frac{1}{ab}\)

b)

\(B=\left[\frac{(2x+y)^2}{(2x-y)^2(2x+y)^2}+\frac{(2x-y)^2}{(2x-y)^2(2x+y)^2}+\frac{2}{(2x-y)(2x+y)}\right].\frac{(2x+y)^2}{16x}\)

\(=\left[\frac{8x^2+2y^2}{(2x-y)^2(2x+y)^2}+\frac{2(2x-y)(2x+y)}{(2x-y)^2(2x+y)^2}\right].\frac{(2x+y)^2}{16x}\)

\(=\frac{8x^2+2y^2+2(4x^2-y^2)}{(2x-y)^2(2x+y)^2}.\frac{(2x+y)^2}{16x}\)

\(=\frac{16x^2}{(2x-y)^2(2x+y)^2}.\frac{(2x+y)^2}{16x}=\frac{x}{(2x-y)^2}\)

16 tháng 8 2019

\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)

Vậy pt có vô số nghiệm

\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)

Mấy câu rút gọn bạn quy đồng nha

16 tháng 8 2019

bạn có thể giải ra giúp mik đc ko?

AH
Akai Haruma
Giáo viên
5 tháng 9 2020

Lời giải:
a)

\(A=\frac{x^2y(y-x)-xy^2(x-y)}{3y^2-2x^2}=\frac{x^2y(y-x)+xy^2(y-x)}{3y^2-2x^2}=\frac{(xy^2+x^2y)(y-x)}{3y^2-2x^2}\)

\(=\frac{xy(x+y)(y-x)}{3y^2-2x^2}=\frac{xy(y^2-x^2)}{3y^2-2x^2}\)

Với $x=-3; y=\frac{1}{2}$ thì:

$xy=\frac{-3}{2}; x^2=9; y^2=\frac{1}{4}$

Do đó $A=\frac{-35}{46}$

b)
\(B=\frac{(8x^3-y^3)(4x^2-y^2)}{(2x+y)(4x^2-4xy+y^2)}=\frac{(2x-y)(4x^2+2xy+y^2)(2x-y)(2x+y)}{(2x+y)(2x-y)^2}\)

\(=4x^2+2xy+y^2=4.2^2+2.2.\frac{-1}{2}+(\frac{-1}{2})^2=\frac{57}{4}\)

30 tháng 6 2017

a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)

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

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

\(=\frac{4x}{\left(x+1\right)^2}\)=VP

b.VT\(=\frac{2+x}{2-x}.\frac{\left(2-x\right)^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{4-2x+x^2}{2-x}\right)\)

=\(\frac{4-x^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{4-x^2}\right)=\frac{4-x^2}{4x^2}.\frac{2\left(2+x\right)-4}{4-x^2}\)

=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP

c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)

\(=\left[\frac{3\left(x+y\right)+3x}{\left(x+y\right)\left(x-y\right)}.\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)

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

\(=x+y=\)VP

Vậy các đẳng thức được chứng minh

=

30 tháng 6 2017

C là xy mà ko phải x+y