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20 tháng 11 2017

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2 tháng 3 2020

bạn đợi mình tí nhé, khoảng 15 phút sau mình gửi câu trả lời :>>

2 tháng 3 2020

\(\text{Câu 1 thì bạn tự làm chịu khó tí:}y^2-4x^2=\left(y+2x\right)\left(y-2x\right)\)

\(\text{Bài 2:}ab\left(b-a\right)+bc\left(c-b\right)+ac\left(a-c\right)\)

\(=ab\left(b-a\right)-bc\left(b-c\right)+ac\left(a-c\right)\)

\(=ab\left(b-a\right)-bc\left[\left(b-a\right)+\left(a-c\right)\right]+ac\left(a-c\right)\)

\(=\left(ab-bc\right)\left(b-a\right)+\left(ac-bc\right)\left(a-c\right)\)

\(=b\left(a-c\right)\left(b-a\right)+c\left(a-b\right)\left(a-c\right)=\left(b-c\right)\left(b-a\right)\left(a-c\right)\)

8 tháng 7 2016

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

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

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

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

\(P=\frac{1}{2y-x}\)

Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\)

 

8 tháng 7 2016

thanks hihi

8 tháng 7 2016

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

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

    \(C=\frac{x+1}{2x^2+y+2}\)

Ta có: 

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

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

B=\(\frac{\left(2x^2\right)^2+2.2x^2.y+y^2-4}{x^2+xy+x+y}=\frac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}=\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

=>\(P=\left(A:B\right):C\)

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

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

        \(=\frac{1}{2y-x}\)

=>\(P=\frac{1}{2y-x}\)

Thế x=-1,76 và y=3/25 vào P

=>\(P=\frac{1}{2.\frac{3}{25}-1,76}=\frac{1}{2}\)

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

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

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

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

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

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

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

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