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DD
1
15 tháng 12 2017
\(Q=\)\(1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(Q=1+\frac{x+3}{x^2+3x+2x+6}:\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right]\)
\(Q=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left[\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right]\)
\(Q=1+\frac{1}{x+2}:\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(Q=1+\frac{1}{x+2}:\left[\frac{2x+4-2x+2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(Q=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(Q=1+\frac{1}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{6}\)
\(Q=1+\frac{x-2}{6}\)
\(Q=\frac{6+x-2}{6}\)
\(Q=\frac{x+4}{6}\)
b) khi \(Q=0\)thì \(\frac{x+4}{6}=0\)
\(\Rightarrow x+4=0\)
\(\Rightarrow x=-4\)
vậy \(x=-4\)khi \(Q=0\)
c) khi \(Q>0\)thì \(\frac{x+4}{6}>0\)
\(\Rightarrow x+4>0\)
\(\Leftrightarrow x>-4\)
vậy \(x>-4\)thì \(Q>0\)
6 tháng 7 2016
Bài 1) A=(8x3+27x3):2x+3y
=[(2x)3+(3y)3]:2x+3y
=(2x)2+(3y)2
=4x2+9y2
B=(x3-27):(x-3)
=(x3-33):(x-3)
=x2-32
=x2-9
AH
Akai Haruma
Giáo viên
7 tháng 2 2020
Lời giải:
ĐKXĐ: $x\neq \pm 2; x\neq -3$
Ta có:
$M=1+\frac{x+3}{(x+2)(x+3)}:\left(\frac{8x}{4x^2(x-2)}-\frac{3x}{3(x-2)(x+2)}-\frac{1}{x+2}\right)$
$=1+\frac{1}{x+2}:\left(\frac{2}{x(x-2)}-\frac{x}{(x-2)(x+2)}-\frac{1}{x+2}\right)$
$=1+\frac{1}{x+2}:\frac{2(x+2)-x^2-x(x-2)}{x(x-2)(x+2)}$
$=1+\frac{1}{x+2}:\frac{-2(x^2-2x-2)}{x(x-2)(x+2)}$
$=1-\frac{x(x-2)}{2x^2-4x-4}=\frac{x^2-2x-4}{2x^2-4x-4}$
NM
2
24 tháng 11 2018
\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)
\(=\frac{2x+5}{3x-1}\)
Còn bài b bạn tự làm nhé
24 tháng 11 2018
Điều kiện: \(x\ne\left\{-1;-2;-5\right\}\)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]}\)
\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\)
Điều kiện: \(x\ne\left\{3;\frac{1}{3}\right\}\)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)
\(=\frac{\left(2x+5\right)\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)}=\frac{2x+5}{3x-1}\)
22 tháng 2 2019
a, \(A=\frac{4x^2\left(x-2\right)+3\left(x-2\right)}{2x\left(x-2\right)+x-2}\)
\(=\frac{\left(x-2\right)\left(4x^2+3\right)}{\left(x-2\right)\left(2x+1\right)}=\frac{4x^2+3}{2x-1}\left(ĐKXĐ:x\ne2;x\ne-\frac{1}{2}\right)\)
b, \(A\in Z\Leftrightarrow\frac{4x^2+3}{2x-1}\in Z\Leftrightarrow2x+1+\frac{4}{2x-1}\in Z\)
\(\Leftrightarrow\frac{4}{2x-1}\in Z\Leftrightarrow4⋮\left(2x-1\right)\)
\(\Rightarrow2x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Mà 2x - 1 là số lẻ nên \(2x-1\in\left\{-1;1\right\}\Rightarrow x\in\left\{0;1\right\}\) (thỏa mãn ĐKXĐ)
6 tháng 1 2018
\(P=\frac{2x^5-x^4-2x+1}{4x^2-1}+\frac{8x^2-4x+2}{ }\)
\(P=\frac{x^4\left(2x-1\right)-\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(P=\frac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2}{2x+1}\)
\(P=\frac{x^4-1}{2x+1}+\frac{2}{2x+1}\)
\(P=\frac{x^4+1}{2x+1}\)
Vậy \(P=\frac{x^4+1}{2x+1}\)
Bài làm
\(x^3-2x^2+3x-4x^2+8x-12\)
\(=x^3-6x^2+11x-12\)