K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

AH
Akai Haruma
Giáo viên
12 tháng 8 2019

Lời giải:
a)

ĐKXĐ: \(x\neq 0; x\neq - 1\)

\(M=\frac{(x+2)(x+1)+2.3x-3.3x(x+1)}{3x(x+1)}:\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)

\(=\frac{-8x^2+2}{3x(x+1)}.\frac{x+1}{2-4x}-\frac{3x-x^2+1}{3x}=\frac{2(1-4x^2)}{3x(2-4x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{2(1-2x)(1+2x)}{6x(1-2x)}-\frac{3x-x^2+1}{3x}=\frac{1+2x}{3x}-\frac{3x-x^2+1}{3x}=\frac{x^2-x}{3x}=\frac{x-1}{3}\)

b)

Khi $x=2006\Rightarrow M=\frac{2006-1}{3}=\frac{2005}{3}$

c)

\(M< 0\Leftrightarrow \frac{x-1}{3}< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ suy ra $x< 1; x\neq 0; x\neq -1$

d)

Để \(\frac{1}{M}=\frac{3}{x-1}\in\mathbb{Z}\) thì \(3\vdots x-1\)

\(\Rightarrow x-1\in\left\{\pm 1;\pm 3\right\}\)

\(\Rightarrow x\in\left\{0;2;-2;4\right\}\)

Kết hợp đkxđ suy ra $x\in\left\{-2;2;4\right\}$

AH
Akai Haruma
Giáo viên
12 tháng 8 2019

Lời giải:
a)

ĐKXĐ: \(x\neq 0; x\neq - 1\)

\(M=\frac{(x+2)(x+1)+2.3x-3.3x(x+1)}{3x(x+1)}:\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)

\(=\frac{-8x^2+2}{3x(x+1)}.\frac{x+1}{2-4x}-\frac{3x-x^2+1}{3x}=\frac{2(1-4x^2)}{3x(2-4x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{2(1-2x)(1+2x)}{6x(1-2x)}-\frac{3x-x^2+1}{3x}=\frac{1+2x}{3x}-\frac{3x-x^2+1}{3x}=\frac{x^2-x}{3x}=\frac{x-1}{3}\)

b)

Khi $x=2006\Rightarrow M=\frac{2006-1}{3}=\frac{2005}{3}$

c)

\(M< 0\Leftrightarrow \frac{x-1}{3}< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ suy ra $x< 1; x\neq 0; x\neq -1$

d)

Để \(\frac{1}{M}=\frac{3}{x-1}\in\mathbb{Z}\) thì \(3\vdots x-1\)

\(\Rightarrow x-1\in\left\{\pm 1;\pm 3\right\}\)

\(\Rightarrow x\in\left\{0;2;-2;4\right\}\)

Kết hợp đkxđ suy ra $x\in\left\{-2;2;4\right\}$

26 tháng 6 2018

ĐKXĐ: \(x\ne0;x\ne\pm2\)

a, \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=\left[\frac{3x^2}{3x\left(x-2\right)\left(x+2\right)}-\frac{6x\left(x+2\right)}{3x\left(x-2\right)\left(x+2\right)}+\frac{3x\left(x-2\right)}{3x\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(=\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{-18x}{3x\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}\)

\(=\frac{-3x}{3x\left(x-2\right)}=\frac{-1}{x-2}\)

b, Ta có: \(\left|x\right|=\frac{1}{2}\Rightarrow x=\pm\frac{1}{2}\)

Với \(x=\frac{1}{2}\) thì \(A=\frac{-1}{\frac{1}{2}-2}=\frac{-1}{\frac{-3}{2}}=\frac{2}{3}\)

Với \(x=\frac{-1}{2}\)thì \(A=\frac{-1}{\frac{-1}{2}-2}=\frac{-1}{\frac{-5}{2}}=\frac{2}{5}\)

c, Để A=2 <=> \(\frac{-1}{x-2}=2\Leftrightarrow-1=2x-4\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Vậy x=3/2 thì A=2

d, Để A<0 <=> \(\frac{-1}{x-2}< 0\Leftrightarrow x-2>0\Leftrightarrow x>2\)

Vậy với x>2 thì A<0

e, Để A thuộc Z <=> x-2 thuộc Ư(-1)={1;-1}

Ta có: x-2=1 => x=3 (t/m)

          x-2=-1 => x=1 (t/m)

Vậy x thuộc {3;1} thì A thuộc Z

26 tháng 6 2018

a)  \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)(ĐKXĐ: x khác 0; + 2)

\(A=\left(\frac{x^2}{x\left(x^2-4\right)}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right)\)

\(A=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\right):\frac{6}{x+2}\)

\(A=\frac{-6x}{x\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}=\frac{-x}{x\left(x-2\right)}=\frac{1}{2-x}.\)

Vậy \(A=\frac{1}{2-x}.\)

b) \(\left|x\right|=\frac{1}{2}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\). Nếu \(x=\frac{1}{2}\)thì \(A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}.\)

Nếu \(x=-\frac{1}{2}\)thì \(A=\frac{1}{2+\frac{1}{2}}=\frac{2}{5}.\)Vậy ...

c) Để A=2 thì \(\frac{1}{2-x}=2\Rightarrow4-2x=1\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}.\)Vậy ...

d) Để A<0 thì \(\frac{1}{2-x}< 0\Rightarrow2-x< 0\Leftrightarrow x>2.\)Vậy ...

e) Để A thuộc Z thì \(\frac{1}{2-x}\in Z\Rightarrow1⋮2-x\). Mà 2-x thuộc Z (Do x thuộc Z)

Nên \(2-x\in\left\{1;-1\right\}\Rightarrow x\in\left\{1;3\right\}.\)(t/m ĐKXĐ)

Vậy x=1 hay x=3 thì A nguyên.

11 tháng 3 2020

ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)

\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)

\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)

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

\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)

\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)

Đề sai à ??

12 tháng 1 2019

a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)

\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)

\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2x+1-3x-1+x^2}{3x}\)

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

\(A=\frac{x\left(x-1\right)}{3x}\)

\(A=\frac{x-1}{3}\)

b) Thay x = 4 ta có :

\(A=\frac{4-1}{3}=\frac{3}{3}=1\)

c) Để A thuộc Z thì \(x-1⋮3\)

\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)

\(\Rightarrow x\in\left\{1;4;7;...\right\}\)

Vậy.....

27 tháng 2 2020

Cho Bt 

a,Tìm điều kiện xác định và rút gọn bt A

b,Tính giá trị bt A tại x=4

c,tìm x thuộc Z để a thuộc Z

13 tháng 6 2016

\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)

\(\left(DK:x\ne0;x\ne-1;x\ne\frac{1}{2}\right)\)

\(=\frac{\left(x+2\right)\left(x+1\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)

\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)

\(=\frac{-8x^2+2}{6x}.\frac{1}{1-2x}+\frac{x^2-3x-1}{3x}=\frac{-2\left(4x^2-1\right)}{6x}.\frac{1}{1-2x}+\)\(\frac{x^2-3x-1}{3x}\)

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

13 tháng 6 2016

a)\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\left(DK:x\ne0;x\ne-1;x\ne\frac{1}{2}\right)\)

\(=\frac{\left(x+2\right)\left(x+1\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)

\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)

\(=\frac{-8x^2+2}{6x}.\frac{1}{1-2x}+\frac{x^2-3x-1}{3x}=\frac{-2\left(4x^2-1\right)}{6x}.\frac{1}{1-2x}+\frac{x^2-3x-1}{3x}\)

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

b) \(\left|x\right|=\frac{1}{3}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\left(x\ge0\right)\\x=-\frac{1}{3}\left(x< 0\right)\end{cases}}\)

Thay vào \(\frac{x-1}{3}\)tính được A.

c) \(A< 0\Rightarrow\frac{x-1}{3}< 0\Rightarrow x-1< 0\Rightarrow x< 1\)

Kết hợp cùng với điều kiện của ở phần rút gọn.

d) \(A\in Z\Rightarrow\frac{x-1}{3}\in Z\Rightarrow x=3k+1\)(\(k\in Z\))

29 tháng 6 2016

M = \(\left(\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

<=> M =