Bài 1: Thực hiện phép tính a) [ $\frac{1}{\left(2x-y\right)^2}$ $\frac{2}{4x^2-y^2}$ $\frac{1}{\left(2x y\right)^2}$ ] : $\frac{16x}{4x^2 4xy y^2}$ b) $\frac{3}{3x 3}$ \(\frac{10}{5-...

TN

thungan nguyen

8 tháng 8 2020

Bài 1: Thực hiện phép tính a) [ $\frac{1}{\left(2x-y\right)^2}$ + $\frac{2}{4x^2-y^2}$ + $\frac{1}{\left(2x+y\right)^2}$ ] : $\frac{16x}{4x^2+4xy+y^2}$ b) $\frac{3}{3x+3}$ + $\frac{10}{5-5x}$ + $\frac{5x-1}{x^2-1}$ c) A = ( x$^4$ - x$^2$ + 2x - 1) : ( x$^2$ + x - 1) - ( x$^2$ - x ) ...

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#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

8 tháng 8 2020

Bài 1:

a) Ta có: $\left(\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right):\frac{16x}{4x^2+4xy+y^2}$

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

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

$=\frac{\left(4x\right)^2}{\left(2x-y\right)^2}\cdot\frac{1}{16x}$

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

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

b) Ta có: $\frac{3}{3x+3}+\frac{10}{5-5x}+\frac{5x-1}{x^2-1}$

$=\frac{1}{x+1}-\frac{2}{x-1}+\frac{5x-1}{x^2-1}$

$=\frac{x-1}{\left(x+1\right)\left(x-1\right)}-\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{5x-1}{\left(x-1\right)\left(x+1\right)}$

$=\frac{x-1-2\left(x+1\right)+5x-1}{\left(x-1\right)\left(x+1\right)}$

$=\frac{x-1-2x-2+5x-1}{\left(x-1\right)\left(x+1\right)}$

$=\frac{4x-4}{\left(x-1\right)\left(x+1\right)}=\frac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}$

$=\frac{4}{x+1}$

c) Ta có: $A=\left(x^4-x^2+2x-1\right):\left(x^2+x-1\right)-\left(x^2-x\right)$

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

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

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

$=x^2-x+1-x^2+x$

=1

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QA

Quốc An

24 tháng 8 2016

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

#Toán lớp 8

2

NT

Ngô Tấn Đạt

24 tháng 8 2016

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

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

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

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VD

Võ Đông Anh Tuấn

24 tháng 8 2016

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

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

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

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NC

Nguyễn Cảnh Tùng

29 tháng 2 2020 - olm

Rút gon biểu thức

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

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

#Toán lớp 8

3

AZ

Agatsuma Zenitsu

29 tháng 2 2020

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

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

$=\frac{1}{ab}$

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

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

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

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

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LT

Lê Thị Nhung

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

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

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

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

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

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

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S

Scorpio

6 tháng 1 2016 - olm

$\frac{4xy}{y^2-x^2}:\left(\frac{1}{x^2+2xy+y^2}-\frac{1}{x^2-y^2}\right)$

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

#Toán lớp 9

0

NT

Nguyễn Thùy Trang

20 tháng 2 2020

Rút gọn các biểu thức sau:

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

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

#Toán lớp 8

1

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)

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

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

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

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

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PJ

Park Jimin

16 tháng 8 2019 - olm

2) Giải phương trìnha) $\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}$b) $\left(2x+3\right).\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right).\left(\frac{3x+8}{2-7x}+1\right)$3) Rút...

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#Toán lớp 8

2

BH

Bui Huyen

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

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PJ

Park Jimin

16 tháng 8 2019

bạn có thể giải ra giúp mik đc ko?

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