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TT
20 tháng 12 2018
a.
\(\dfrac{x+3}{x-2}+\dfrac{4+x}{2-x}\\ =\dfrac{x+3}{x-2}-\dfrac{4+x}{x-2}\\ =\dfrac{x+3-4-x}{x-2}\\ =-\dfrac{1}{x-2}\)
b. \(\dfrac{x+1}{2x+6}+\dfrac{2x+3}{x^2+3x}\)
\(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x}{2x\left(x+3\right)}+\dfrac{4x+6}{2x\left(x+3\right)}=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x^2+3x+2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x\left(x+3\right)+2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{x+2}{2x}\)
c. \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
d. \(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)
\(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}:\dfrac{-x\left(x+3\right)}{3x-1}\)
\(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}.\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}\)
\(=-\dfrac{2}{x^2}\)
NN
1 tháng 12 2017
a) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\) MTC: \(2x\left(x+3\right)\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-\left(x-6\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x+6}{4-x^2}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{x^2-4}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\) MTC: \(\left(x-2\right)\left(x+2\right)\)
\(=\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x+6\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8+2x+4-5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)
NN
1 tháng 12 2017
c) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{2x\left(2x-1\right)}\) MTC: \(2x\left(2x-1\right)\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}-\dfrac{3x-2}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{-2x+1}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{-1}{2x}\)
d) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\) MTC: \(\left(x-1\right)\left(x^2+x+1\right)\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x^2+2\right)+2\left(x-2\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-4-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-3+x}{\left(x-1\right)\left(x^2+x+1\right)}\)
23 tháng 2 2019
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
31 tháng 12 2022
a: \(\Leftrightarrow4\left(6-x\right)-3x=6\left(2x+3\right)-12\)
=>24-4x-3x=12x+18-12
=>12x+6=-7x+24
=>19x=18
=>x=18/19
b: \(\Leftrightarrow-210x-6\left(x-3\right)-15x=30x+10\left(2x+1\right)\)
=>-225x-6x+18=30x+20x+10
=>-231x+18-50x-10=0
=>-281x=-8
=>x=8/281
c: \(\Leftrightarrow36-2\left(x+3\right)=-4x+1-x\)
=>36-2x-6=-5x+1
=>3x=1+6-36=5-36=-31
=>x=-31/3
d: \(\Leftrightarrow-30\left(x-3\right)+10\left(2x-7\right)=6\left(6-x\right)\)
=>-30x+90+20x-70=36-6x
=>-10x+20=36-6x
=>-4x=16
=>x=-4
24 tháng 3 2017
bạn nên bổ sung chữ "bất"
1)
\(x-\dfrac{x-1}{3}+\dfrac{x+2}{6}>\dfrac{2x}{5}+5\\ \Leftrightarrow x-\dfrac{x-1}{3}+\dfrac{x+2}{6}-\dfrac{2x}{5}-5>0\\ \Leftrightarrow\dfrac{30x-10\left(x-1\right)+5\left(x+2\right)-2x\cdot6-5\cdot30}{30}>0\\ \Leftrightarrow30x-10x+10+5x+10-12x-150>0\\ \Leftrightarrow30x-10x=5x-12x>-10-10+150\\ \Leftrightarrow13x>130\\ \Leftrightarrow13x\cdot\dfrac{1}{13}>130\cdot\dfrac{1}{13}\\ \Leftrightarrow x>10\)
Vậy tập ngiệm của bât hương trình là {x/x>10}
mình mới học đến đây nên cách giải còn dài, thông cảm nha
24 tháng 3 2017
2)
\(\dfrac{2x+6}{6}-\dfrac{x-2}{9}< 1\\ \Leftrightarrow\dfrac{2\left(x+3\right)}{6}-\dfrac{x-2}{9}< 1\\ \Leftrightarrow\dfrac{x+3}{3}-\dfrac{x-2}{9}-1< 0\\ \Leftrightarrow\dfrac{3\left(x+3\right)-x+2-9}{9}< 0\\ \Leftrightarrow3x+9-x+2-9< 0\\ \Leftrightarrow3x-x< -9+9-2\\ \Leftrightarrow2x< -2\\ \Leftrightarrow2x\cdot\dfrac{1}{2}< -2\cdot\dfrac{1}{2}\Leftrightarrow x< -1\)
Vậy tập nghiệm của bất phương trình là {x/x<-1}
14 tháng 12 2018
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
NN
10 tháng 11 2017
a) \(\dfrac{2}{x+3}+\dfrac{1}{x}\) [ MTC: x(x+3) ]
\(=\dfrac{x.2}{x\left(x+3\right)}+\dfrac{1\left(x+3\right)}{x\left(x+3\right)}\)
\(=\dfrac{2x+x+3}{x\left(x+3\right)}\)
\(=\dfrac{3x+3}{x\left(x+3\right)}\)
\(=\dfrac{3\left(x+1\right)}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\)
\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\) \(\left[MTC:2\left(x-1\right)\left(x+1\right)\right]\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-2x.2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x^2+2x+1\right)-4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x^2-2x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)}{2\left(x+1\right)}\)
11 tháng 11 2017
a) Ta có :
\(\dfrac{2}{x+3}+\dfrac{1}{x}=\dfrac{2x+x+3}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)-2x.2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{-3x+1}{2\left(x-1\right)\left(x+1\right)}\)
c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)
\(=\dfrac{y^2-12+36}{6y\left(y-6\right)}=\dfrac{y^2-24}{6y\left(y-6\right)}\)
d) \(\dfrac{6+x}{x+3x}+\dfrac{3}{2x+6}=\dfrac{6+x}{4x}+\dfrac{3}{2\left(x+3\right)}\)
\(=\dfrac{\left(6+x\right)\left(2x+6\right)+12x}{8x\left(x+3\right)}\)(Đề câu này phải sửa thành\(\dfrac{6+x}{x^2+3x}chứ\)) ???
Đáp án:x+2/2x
bạn cho mình xin các bước làm với