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a) \(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+315\)
\(\Leftrightarrow84x+63-90x+30=175x+140+315\)
\(\Leftrightarrow84x-90x-175x=140+315-63-30\)
\(\Leftrightarrow-181x=362\)
\(\Leftrightarrow x=-\dfrac{362}{181}=-2\)
Vậy: Tập ngiệm của phương trình là: \(S=\left\{-2\right\}\)
b) \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)
\(\Leftrightarrow6x+24-30x-120=10x-15x+30\)
\(\Leftrightarrow6x-30x-10x+15x=30-24+120\)
\(\Leftrightarrow-19x=126\)
\(\Leftrightarrow x=-\dfrac{126}{19}\)
Vậy: Tập ngiệm của phương trình là: \(S=\left\{-\dfrac{126}{19}\right\}\)
c) \(\dfrac{x+2}{3}+\dfrac{3\left(2x-1\right)}{4}-\dfrac{5x-3}{6}=x+\dfrac{5}{12}\)
\(\Leftrightarrow4\left(x+2\right)+9\left(2x-1\right)-2\left(5x-3\right)=12x+5\)
\(\Leftrightarrow4x+8+18x-9-10x+6=12x+5\)
\(\Leftrightarrow4x+18x-10x-12x=5-8+9-6\)
\(\Leftrightarrow0x=0\)
Vậy: Tập ngiệm của phương trình là: \(S=\left\{R\right\}\)
_Chúc bạn học tốt_
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 \(S=\left\{-1\right\}\)
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 ^ ^
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
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/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)
=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)
=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
=>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: \(\Leftrightarrow4x+4+9\left(2x+1\right)=2\left(5x+3\right)+12x+7\)
=>4x+4+18x+9=10x+6+12x+7
=>22x+13=22x+13(luôn đúng)
b: \(\Leftrightarrow\left(\dfrac{x+1}{94}+1\right)+\left(\dfrac{x+2}{93}+1\right)+\left(\dfrac{x+3}{92}+1\right)=\left(\dfrac{x+4}{91}+1\right)+\left(\dfrac{x+5}{90}+1\right)+\left(\dfrac{x+6}{89}+1\right)\)
=>x+95=0
=>x=-95
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=\(\left\{1\right\}\)
ĐK: \(x\ne-2;-3;-4;-5\)
\(1+\dfrac{1}{x+2}-\left(1+\dfrac{1}{x+3}\right)=1+\dfrac{1}{x+4}-\left(1+\dfrac{1}{x+5}\right)\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=\left(x+4\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+5x+6=x^2+9x+20\)
\(\Leftrightarrow4x=-14\Rightarrow x=-\dfrac{7}{2}\)
b/ ĐK: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{x+7}{x+2}-\dfrac{12}{x^2-4}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}-\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}-\dfrac{12}{x^2-4}=0\)
\(\Leftrightarrow x^2+3x+2-\left(x^2+5x-14\right)-12=0\)
\(\Leftrightarrow-2x+4=0\Rightarrow x=2\) (ko t/m)
Vậy pt vô nghiệm