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a: \(\Rightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
=>x+36=0
=>x=-36
b: \(\Leftrightarrow\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}-1\right)+\left(\dfrac{x-6}{1998}-1\right)+\left(\dfrac{x-4}{2000}-1\right)+\left(\dfrac{x-2}{2002}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)+\left(\dfrac{x-1996}{8}-1\right)+\left(\dfrac{x-1994}{10}-1\right)\)
=>x-2004=0
=>x=2004
a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)
<=> \(25x+10-80x+10=24x+12-30\)
<=> \(25x-80x-24x=12-30-10-10\)
<=> \(-79x=-38\)
<=> \(x=\dfrac{-38}{-79}\)
\(x=\dfrac{38}{79}\)
b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)
<=> \(30x-12x+30+5x+40=210+10x-10\)
<=> \(30x-12x+5x-10x=210-10-30-40\)
<=> \(13x=130\)
<=> \(x=\dfrac{130}{13}\)
\(x=10\)
c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)
<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)
<=> \(28x+28+60x+120+105x+420+2520=0\)
<=> \(28x+60x+105x=-28-120-420-2520\)
<=> \(193x=-3088\)
<=> \(x=\dfrac{-3088}{193}\)
\(x=-16\)
d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)
<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)
<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)
<=> \(22968x=8199576\)
<=> \(x=\dfrac{8199576}{22968}\)
\(x=357\)
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
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=\(\left\{1\right\}\)
\(\text{a) }\left|2-5x\right|=\left|3x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}2-5x=3x+1\\2-5x=-3x-1\end{matrix}\right. \Leftrightarrow\left[{}\begin{matrix}-5x-3x=1-2\\-5x+3x=-1-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-8x=-1\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{1}{8};\dfrac{3}{2}\right\}\)
\(\text{b) }\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)
ĐXKĐ của phương trình \(:x\ne\pm5\)
\(\text{Ta có }:\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{2\left(25-x^2\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x+5\right)\left(x-5\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{9\left(x+5\right)}{12\left(x+5\right)\left(x-5\right)}-\dfrac{90}{12\left(x+5\right)\left(x-5\right)}+\dfrac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\\ \Rightarrow9x+45-90+14x-70=0\\ \Leftrightarrow23x=115\\ \Leftrightarrow x=5\left(KTM\right)\)
Vậy phương trình vô nghiệm
\(\text{c) }\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\\ \Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\\ \Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}=0\\ \Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\\ \Leftrightarrow x+60=0\left(\text{Vì }\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\\ \Leftrightarrow x=-60\)
Vậy \(x=-60\) là nghiệm của phương trình
a: \(x< -9:\dfrac{3}{2}=-9\cdot\dfrac{2}{3}=-6\)
b: 2/3x>-2
hay x>-2:2/3=-3
c: \(2x>\dfrac{9}{5}-\dfrac{4}{5}=1\)
hay x>1/2
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}>6-4=2\)
hay x>2:3/5=2x5/3=10/3
a) \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{36}{x^2-9}\)
\(\Rightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\dfrac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\left(x+3\right)^2-\left(x-3\right)^2=36\)
\(\Rightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=36\)
\(\Rightarrow x^2+6x+9-x^2+6x-9=36\)
\(\Rightarrow12x=36\)
\(\Rightarrow x=\dfrac{36}{12}\)
Vậy x = 3
b) \(x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\)
\(\Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{6x-3-5x+10}{15}=\dfrac{x+17}{15}\)
... Phần còn lại cũng tương tự như vậy thôi
\(\Leftrightarrow\)\(\dfrac{x}{7}-\dfrac{20}{7}+\dfrac{x}{33}-\dfrac{46}{33}+\dfrac{x}{8}-\dfrac{5}{8}+\dfrac{x}{19}+\dfrac{6}{19}=0\)
\(\dfrac{5016x}{35112}+\dfrac{1064x}{35112}+\dfrac{4389x}{35112}+\dfrac{1848x}{35112}=\dfrac{100320}{35112}+\dfrac{48994}{35112}-\dfrac{21945}{35112}+\dfrac{11088}{35112}\)
\(\dfrac{12317x}{35112}=\dfrac{160121}{35112}\)
\(12317x=160121\)
\(x=13\)
Làm cách này nhưng tớ thấy nó vẫn không tiến triển gì, thôi, qua cái này:
\(\dfrac{x-20}{7}+\dfrac{x-46}{33}+\dfrac{x-5}{8}+\dfrac{x+6}{19}=0\)
\(\dfrac{5016\left(x-20\right)+1064\left(x-46\right)+4389\left(x-5\right)+1848\left(x+6\right)}{35112}=0\)
\(5016x+1064x+4389x+1848x=100320+48944+21945-11088\)
\(12317x=160121\)
\(x=13\)
\(\Leftrightarrow\frac{x-20}{7}+1+\frac{x-46}{33}+1+\frac{x-5}{8}-1+\frac{x+6}{19}-1=0\)
\(\Leftrightarrow\frac{x-13}{7}+\frac{x-13}{33}+\frac{x-13}{8}+\frac{x-13}{19}=0\)
\(\Leftrightarrow\left(x-13\right)\left(\frac{1}{7}+\frac{1}{33}+\frac{1}{8}+\frac{1}{19}\right)=0\)
\(\Leftrightarrow x-13=0\)
\(\Leftrightarrow x=13\)