\(a)\frac{1}{x+3}+\frac{x}{x^2-6x+9}\)
\(b)\frac{2x}{x^2-9}-\frac{x-1}{x+3}\)
\(c)\frac{2x+1}{x-2}.\frac{2-x}{2x+1}\)
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PA
30 tháng 11 2016
\(MTC:\left(x-3\right)^2\left(x^2+3x+9\right)\)
\(\frac{x}{x^3-27}=\frac{x}{\left(x-3\right)\left(x^2+3x+9\right)}=\frac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\frac{2x}{x^2-6x+9}=\frac{2x}{\left(x-3\right)^2}=\frac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\frac{1}{x^2+3x+9}=\frac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(MTC:2\left(x-1\right)\left(x+1\right)\)
\(\frac{x-1}{2x+2}=\frac{x-1}{2\left(x+1\right)}=\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(\frac{x+1}{2x-2}=\frac{x+1}{2\left(x-1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(\frac{1}{1-x^2}=-\frac{1}{\left(x-1\right)\left(x+1\right)}=-\frac{2}{2\left(x-1\right)\left(x+1\right)}\)
\(MTC:2\left(x+1\right)\left(x^2-x+1\right)\)
\(\frac{1}{x^3+1}=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{2}{2\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{3}{2x+2}=\frac{3}{2\left(x+1\right)}=\frac{3\left(x^2-x+1\right)}{2\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2}{x^2-x+1}=\frac{4\left(x+1\right)}{2\left(x+1\right)\left(x^2-x+1\right)}\)
18 tháng 11 2019
a)có khả năng sai đề bài
b)Liệu có sai đề bài không
c)\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)(phân số cuối có âm vì (1-x)=-(x-1)
\(=\frac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)(Hơi tắt)
\(=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{1}{x^2+x+1}\)
d)\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy+4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2+4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x-2y}\)
PN
22 tháng 5 2021
\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)
\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)
\(< =>4x-12-4x+2=10x+10+5\)
\(< =>10x=-10-10-5=-25\)
\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)
PN
22 tháng 5 2021
\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)
\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)
\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)
29 tháng 11 2019
Làm ngắn gọn thôi nhé :v
\(A=\frac{2x}{x^2-3x}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)
\(A=\frac{x^5-3x^4-3x^3+11x^2-6x}{x^5-8x^2+22x^2-24x+9}\)
\(A=\frac{x^4-3x^3-3x^2+11x-6}{x^4-8x^3+22x^2-24x+9}\)
\(A=\frac{\left(x-1\right)\left(x-1\right)\left(x+2\right)\left(x-3\right)}{\left(x-1\right)\left(x-1\right)\left(x-3\right)\left(x-3\right)}\)
\(A=\frac{x+2}{x-3}\)
\(B=\frac{x}{x+2}+\frac{2}{x-2}-\frac{4x}{4-x^2}\)
\(B=\frac{-x^4-4x^3+16x+16}{-x^4+8x^2-16}\)
\(B=\frac{\left(-x-2\right)\left(x+2\right)\left(x+2\right)\left(x-2\right)}{\left(-x-2\right)\left(x-2\right)\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{x+2}{x-2}\)
\(C=\frac{1+x}{3-x}-\frac{1-2x}{3+x}-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{1+x}{3-x}-\left(\frac{1-2x}{3+x}\right)-\frac{x\left(1-x\right)}{9-x^2}\)
\(C=\frac{10x}{-x^2+9}\)
\(D=\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)
\(D=\frac{5}{2x^2+6x}-\left(\frac{4-3x^2}{x^2-9}\right)-3\)
\(D=\frac{51x^2+138x-45}{2x^4+6x^2-18x^2-54x}\)
\(D=\frac{3\left(17x-5\right)\left(x+3\right)}{2x\left(x+3\right)\left(x+3\right)\left(x-2\right)}\)
\(D=\frac{51x-15}{2x^3-18x}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)
\(E=\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\left(\frac{3x-2}{x^2+2x+1}\right)\)
\(E=\frac{10x^4-10}{x^6-3x^4+3x^2-1}\)
\(E=\frac{10\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x+1\right)\left(x-1\right)\left(x-1\right)\left(x-1\right)}\)
\(E=\frac{10x^2+10}{x^4-2x+1}\)
23 tháng 2 2017
a) \(\frac{x+3}{x-2}-\frac{2x+3}{x+2}=\frac{2x^2+5x+12}{x^2-4}\)
ĐKXĐ: \(\left\{\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)-\left(2x+3\right)\left(x-2\right)=2x^2+5x+12\)
\(\Leftrightarrow x^2+2x+3x+6-2x^2+4x-3x+6-2x^2-5x-12=0\)
\(\Leftrightarrow-3x^2+4x=0\)
\(\Leftrightarrow3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\3x=4\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\left(tmđk\right)\\x=\frac{4}{3}\left(tmđk\right)\end{matrix}\right.\)
Vậy: \(x=0;\frac{4}{3}\)
_Chúc bạn học tốt_
23 tháng 2 2017
b) Ta có: \(\frac{2x+5}{x-3}+\frac{x-1}{x+3}=\frac{x^2+6x+18}{x^2-9}\)
ĐKXĐ: \(\left\{\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
\(\Leftrightarrow\frac{\left(2x+5\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{x^2+6x+18}{\left(x+3\right)\left(x-3\right)}\)
\(\Rightarrow\left(2x+5\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)=x^2+6x-18\)
\(\Leftrightarrow2x^2+6x+5x+15+x^2-3x-x+3-x^2-6x-18=0\)
\(\Leftrightarrow2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\2x=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x=0;-\frac{1}{2}\)
_Chúc bạn học tốt_
24 tháng 1 2022
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
a) \(\frac{1}{x+3}+\frac{x}{x^2-6x+9}\left(x\ne\pm3\right)\)
\(=\frac{1}{x+3}+\frac{x}{\left(x-3\right)^2}=\frac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)^2}+\frac{x^2+3x}{\left(x+3\right)\left(x-3\right)^2}\)
\(=\frac{x^2-6x+9-x^2+3x}{\left(x-3\right)^2\left(x+3\right)}=\frac{-3x+9}{\left(x-3\right)^2\left(x+3\right)}=\frac{-3\left(x-3\right)}{\left(x-3\right)^2\left(x+3\right)}=\frac{-3}{\left(x-3\right)\left(x+3\right)}\)
anhdun_•Ŧ๏áйツɦọς• giải a r nha , tớ giải b+c cho
\(b,\frac{2x}{x^2-9}-\frac{x-1}{x+3}\)
\(\frac{2x}{x^2-3^2}-\frac{x-1}{x+3}\)
\(\frac{2x}{\left(x+3\right)\left(x-3\right)}-\frac{x-1}{x+3}\)
\(\frac{2x-\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(\frac{2x-x^2+3x+x-3}{\left(x+3\right)\left(x-3\right)}\)
\(\frac{\left(2x+3x+x\right)-x^2-3}{\left(x+3\right)\left(x-3\right)}\)
\(\frac{6x-x^2-3}{\left(x+3\right)\left(x-3\right)}\)