Giải phương trình:\(\dfrac{5}{x^2+1}+\dfrac{7}{x^2+3}+\dfrac{9}{x^2+5}=\dfrac{4x^2+26}{x^2+10}\)
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\(Đk:\) \(x\ne1,x\ne2,x\ne3\)
\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)
\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Rightarrow0x-14=x-10\)
\(\Rightarrow x=-4\left(tmđk\right)\)
\(\dfrac{x+1}{39}+\dfrac{x+2}{38}+\dfrac{x+3}{37}=0\)
\(\Leftrightarrow\dfrac{x+1}{39}+1+\dfrac{x+2}{38}+1+\dfrac{x+3}{37}+1-3=0\)
\(\Leftrightarrow\dfrac{x+40}{39}+\dfrac{x+40}{38}+\dfrac{x+40}{37}=3\)
\(\Leftrightarrow\left(x+40\right)\left(\dfrac{1}{39}+\dfrac{1}{38}+\dfrac{1}{37}\right)=3\)
\(\Leftrightarrow\left(x+40\right).\dfrac{4331}{54834}=3\)
\(\Leftrightarrow x+40=\dfrac{164502}{4331}\)
\(\Leftrightarrow x=\dfrac{-8738}{4331}\)
-Vậy \(S=\left\{\dfrac{-8738}{4331}\right\}\)
\(\dfrac{2x+1}{3x+2}=\dfrac{x-1}{x-2}\) (đk: x≠ 2; \(-\dfrac{2}{3}\) )
⇔ \(\left(x-2\right)\left(2x+1\right)=\left(x-1\right)\left(3x+2\right)\)
⇔ \(2x^2+x-4x-2=3x^2+2x-3x-2\)
⇔ \(3x^2-x-2-2x^2+3x+2=0\)
⇔ \(x^2+2x=0\)
⇔ \(x\left(x+2\right)=0\)
⇒ \(\left[{}\begin{matrix}x=0\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0;-2\right\}\)
\(\Leftrightarrow3x^2-3x+2x-2=2x^2-4x+x-2\)
\(\Leftrightarrow x^2+2x=0\)
=>x(x+2)=0
=>x=0 hoặc x=-2
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
1)
ĐKXĐ: x>4
Ta có: \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)
\(\Leftrightarrow x^2+8x+15=x^2-6x+8\)
\(\Leftrightarrow8x+6x=8-15\)
\(\Leftrightarrow14x=-7\)
hay \(x=-\dfrac{1}{2}\)(loại)
2) Ta có: \(\sqrt{4x^2-9}=3\sqrt{2x-3}\)
\(\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)
⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\)
⇔ 2x + 2 - 5 - 2x = 12 -16x
⇔ 16x = 15
⇔ x = 15/16
b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)
⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)
⇔ 8 - 6x - 4 + x = 5x + 10
⇔ 10x = -6
⇔ x = -6/10
Câu 1:
x + 1/4 - 5 + 2x/8 = 3 - 4x/2
<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8
<=> 2x + 2 - 5 - 2x = 12 - 16x
<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16
Câu 2:
4 - 3x/5 - 4 - x/10 = x + 2/2
<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10
<=> 8 - 6x - 4 + x = 5x + 10
<=> 4 - 5x = 5x + 10
<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
\(\Leftrightarrow\dfrac{5}{x^2+1}+\dfrac{7}{x^2+3}+\dfrac{9}{x^2+5}-\dfrac{4x^2+26}{x^2+10}=0\)
\(\Leftrightarrow\dfrac{5}{x^2+1}-1+\dfrac{7}{x^2+3}-1+\dfrac{9}{x^2+5}-1-\dfrac{4x^2+26}{x^2+10}+3=0\)
\(\Leftrightarrow\dfrac{4-x^2}{x^2+1}+\dfrac{4-x^2}{x^2+3}+\dfrac{4-x^2}{x^2+5}-\dfrac{x^2-4}{x^2+10}=0\)
\(\Leftrightarrow\left(4-x^2\right)\left(\dfrac{1}{x^2+1}+\dfrac{1}{x^2+3}+\dfrac{1}{x^2+5}+\dfrac{1}{x^2+10}\right)=0\)
\(\Leftrightarrow4-x^2=0\)(vì \(\dfrac{1}{x^2+1}+\dfrac{1}{x^2+3}+\dfrac{1}{x^2+5}+\dfrac{1}{x^2+10}>0\))
\(\Leftrightarrow x=\pm2\)