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a) \(\dfrac{3}{x-4}-\dfrac{2}{4-x}\)
\(=\dfrac{3}{x-4}+\dfrac{2}{x-4}\)
\(=\dfrac{3+2}{x-4}\)
\(=\dfrac{5}{x-4}\)
b) \(\dfrac{7}{x-3}-\dfrac{4}{3-x}\)
\(=\dfrac{7}{x-3}+\dfrac{4}{x-3}\)
\(=\dfrac{7+4}{x-3}\)
\(=\dfrac{11}{x-3}\)
c) \(\dfrac{3}{x-5}-\dfrac{2}{x+2}\) MTC: \(\left(x-5\right)\left(x+2\right)\)
\(=\dfrac{3\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}-\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x+2\right)-2\left(x-5\right)}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{3x+6-2x+10}{\left(x-5\right)\left(x+2\right)}\)
\(=\dfrac{x+16}{\left(x-5\right)\left(x+2\right)}\)
d) \(\dfrac{9}{x-5}-\dfrac{6}{x^2-25}\)
\(=\dfrac{9}{x-5}-\dfrac{6}{\left(x-5\right)\left(x+5\right)}\) MTC: \(\left(x-5\right)\left(x+5\right)\)
\(=\dfrac{9\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{6}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{9\left(x+5\right)-6}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{9x+45-6}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{9x+39}{\left(x-5\right)\left(x+5\right)}\)
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
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}
\(a,\dfrac{3\left(5x-2\right)}{4}-2=\dfrac{7x}{3}-5\left(x-7\right)\)
\(\Leftrightarrow\dfrac{15x-6-8}{4}=\dfrac{7x-15\left(x-7\right)}{3}\)
\(\Leftrightarrow\dfrac{15x-14}{4}=\dfrac{7x-15x+105}{3}\)
\(\Leftrightarrow\dfrac{45x-42}{12}=\dfrac{-32x+420}{12}\)
\(\Leftrightarrow45x+32x=420+42\)
\(\Leftrightarrow77x=462\)
\(\Leftrightarrow x=6\)
\(b,\dfrac{x+5}{2}+\dfrac{3-2x}{4}=x-\dfrac{7+x}{6}\)
\(\Leftrightarrow\dfrac{2x+10+3-2x}{4}=\dfrac{6x-7-x}{6}\)
\(\Leftrightarrow\dfrac{13}{4}=\dfrac{5x-7}{6}\)
\(\Leftrightarrow2\left(5x-7\right)=3.13\)
\(\Leftrightarrow10x-14=39\)
\(\Leftrightarrow10x=53\)
\(\Leftrightarrow x=5,3\)
\(c,\dfrac{x-3}{11}+\dfrac{x+1}{3}=\dfrac{x+7}{9}-1\)
\(\Leftrightarrow\dfrac{3x-9+11x+11}{33}=\dfrac{x+7-9}{9}\)
\(\Leftrightarrow\dfrac{14x+2}{33}=\dfrac{x-2}{9}\)
\(\Leftrightarrow33\left(x-2\right)=9\left(14x+2\right)\)
\(\Leftrightarrow33x-66=126x+18\)
\(\Leftrightarrow-93x=84\)
\(\Leftrightarrow x=-\dfrac{28}{31}\)
\(d,\dfrac{3x-0,4}{2}+\dfrac{1,5-2x}{3}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{3\left(3x-0,4\right)+2\left(1,5-2x\right)}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{9x-1,2+3-4x}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{5x+1,8}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow5\left(5x+1,8\right)=6\left(x+0,5\right)\)
\(\Leftrightarrow25x+9=6x+3\)
\(\Leftrightarrow19x=-6\)
\(\Leftrightarrow x=-\dfrac{6}{19}\)
\(\Leftrightarrow77x=378\)
\(\Leftrightarrow x=\dfrac{54}{11}\)
a. \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{9}{x^2-9}\) (ĐKXĐ: \(x\ne\pm3\))
\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=9\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=9\)
\(\Leftrightarrow12x=9\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)
\(\Rightarrow S=\left\{\dfrac{3}{4}\right\}\)
b. \(\dfrac{x+2}{4}-x+3=\dfrac{1-x}{8}\)
\(\Leftrightarrow2\left(x+2\right)-8\left(x-3\right)=1-x\)
\(\Leftrightarrow2x+4-8x+24=1-x\)
\(\Leftrightarrow2x-8x+x=1-4-24\)
\(\Leftrightarrow-3x=-27\Leftrightarrow x=9\)
\(\Rightarrow S=\left\{9\right\}\)
-Mệt -.-
4)a)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)(1)
ĐKXĐ:\(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
(1)\(\Rightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\)
\(\Leftrightarrow x^2+10x+25-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow x^2-x^2+10x+10x=-25+25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\left(nh\text{ậ}n\right)\)
S=\(\left\{1\right\}\)
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)
=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)
=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
=>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\)
=>x=0
b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\)
=>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\)
=>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\)
Đến đây thì bạn giải giống câu a
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
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
a)\(\frac{3}{x-4}-\frac{2}{4-x}=\frac{3}{x-4}+\frac{2}{x-4}=\frac{5}{x-4}\)
câu b làm tương tự nha bạn
c)\(\frac{3}{x+5}-\frac{2}{x+2}=\frac{3x+6-2x-10}{\left(x+5\right)\left(x+2\right)}=\frac{x-4}{\left(x+5\right)\left(x+2\right)}\)
d)\(\frac{9}{x-5}-\frac{6}{x^2-25}=\frac{9x+45-6}{x^2-25}=\frac{9x+39}{x^2-25}\)
mik làm hơi tắt bạn thông cảm nha