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1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
b)
ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)
Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Leftrightarrow2x^2-14=2x^2+x-10\)
\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(nhận)
Vậy: S={-4}
a: =>1+3x-6=-x+3
=>3x-5=-x+3
=>4x=8
=>x=2(loại)
b: \(\Leftrightarrow\dfrac{3\left(x-3\right)+2\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
=>3x-9+2x-4=x-1
=>5x-13=x-1
=>4x=12
=>x=3(loại)
c: =>x^2-2x+4+x^3+8=12
=>x^3+x^2-2x=0
=>x(x^2+x-2)=0
=>x(x+2)(x-1)=0
=>x=0 hoặc x=1
\(1,\) thiếu đề
\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-55x+20=24x-138\)
\(\Leftrightarrow24x-138+55x-20=0\)
\(\Leftrightarrow79x-158=0\)
\(\Leftrightarrow x=2\)
\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x=0\)
a) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{120}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+5x=30-144\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: S={6}
b) Ta có: \(\dfrac{4-5x}{6}=\dfrac{2\left(-x+1\right)}{2}\)
\(\Leftrightarrow2\cdot\left(4-5x\right)=12\left(-x+1\right)\)
\(\Leftrightarrow2-10x=-12x+12\)
\(\Leftrightarrow2-10x+12x-12=0\)
\(\Leftrightarrow2x-10=0\)
\(\Leftrightarrow2x=10\)
hay x=5
Vậy: S={5}
c) Ta có: \(\dfrac{-\left(x-3\right)}{2}-2=\dfrac{5\left(x+2\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3-x\right)}{4}-\dfrac{8}{4}=\dfrac{5\left(x+2\right)}{4}\)
\(\Leftrightarrow6-2x-8=5x+10\)
\(\Leftrightarrow-2x+2-5x-10=0\)
\(\Leftrightarrow-7x-8=0\)
\(\Leftrightarrow-7x=8\)
hay \(x=-\dfrac{8}{7}\)
Vậy: \(S=\left\{-\dfrac{8}{7}\right\}\)
d) Ta có: \(\dfrac{7-3x}{2}-\dfrac{5+x}{5}=1\)
\(\Leftrightarrow\dfrac{5\left(7-3x\right)}{10}-\dfrac{2\left(x+5\right)}{10}=\dfrac{10}{10}\)
\(\Leftrightarrow35-15x-2x-10-10=0\)
\(\Leftrightarrow-17x+15=0\)
\(\Leftrightarrow-17x=-15\)
hay \(x=\dfrac{15}{17}\)
Vậy: \(S=\left\{\dfrac{15}{17}\right\}\)
a) Ta có: x+45−x+4=x3−x−22x+45−x+4=x3−x−22
⇔6(x+4)30−30x30+12030=10x30−15(x−2)30⇔6(x+4)30−30x30+12030=10x30−15(x−2)30
⇔6x+24−30x+120=10x−15x+30⇔6x+24−30x+120=10x−15x+30
⇔−24x+144=−5x+30⇔−24x+144=−5x+30
⇔−24x+5x=30−144⇔−24x+5x=30−144
⇔−19x=−114⇔−19x=−114
hay x=6
Vậy: S={6}
b) Ta có: 4−5x6=2(−x+1)24−5x6=2(−x+1)2
⇔2⋅(4−5x)=12(−x+1)⇔2⋅(4−5x)=12(−x+1)
⇔2−10x=−12x+12⇔2−10x=−12x+12
⇔2−10x+12x−12=0⇔2−10x+12x−12=0
⇔2x−10=0⇔2x−10=0
⇔2x=10⇔2x=10
hay x=5
Vậy: S={5}
c) Ta có: −(x−3)2−2=5(x+2)4−(x−3)2−2=5(x+2)4
⇔2(3−x)4−84=5(x+2)4⇔2(3−x)4−84=5(x+2)4
⇔6−2x−8=5x+10⇔6−2x−8=5x+10
⇔−2x+2−5x−10=0⇔−2x+2−5x−10=0
⇔−7x−8=0⇔−7x−8=0
⇔−7x=8⇔−7x=8
hay x=−87x=−87
Vậy: S={−87}S={−87}
d) Ta có: 7−3x2−5+x5=17−3x2−5+x5=1
⇔5(7−3x)10−2(x+5)10=1010⇔5(7−3x)10−2(x+5)10=1010
⇔35−15x−2x−10−10=0⇔35−15x−2x−10−10=0
⇔−17x+15=0⇔−17x+15=0
⇔−17x=−15⇔−17x=−15
hay x=1517x=1517
Vậy: S={1517}
a) Ta có: x+45−x+4=x3−x−22x+45−x+4=x3−x−22
⇔6(x+4)30−30x30+12030=10x30−15(x−2)30⇔6(x+4)30−30x30+12030=10x30−15(x−2)30
⇔6x+24−30x+120=10x−15x+30⇔6x+24−30x+120=10x−15x+30
⇔−24x+144=−5x+30⇔−24x+144=−5x+30
⇔−24x+5x=30−144⇔−24x+5x=30−144
⇔−19x=−114⇔−19x=−114
hay x=6
Vậy: S={6}
b) Ta có: 4−5x6=2(−x+1)24−5x6=2(−x+1)2
⇔2⋅(4−5x)=12(−x+1)⇔2⋅(4−5x)=12(−x+1)
⇔2−10x=−12x+12⇔2−10x=−12x+12
⇔2−10x+12x−12=0⇔2−10x+12x−12=0
⇔2x−10=0⇔2x−10=0
⇔2x=10⇔2x=10
hay x=5
Vậy: S={5}
c) Ta có: −(x−3)2−2=5(x+2)4−(x−3)2−2=5(x+2)4
⇔2(3−x)4−84=5(x+2)4⇔2(3−x)4−84=5(x+2)4
⇔6−2x−8=5x+10⇔6−2x−8=5x+10
⇔−2x+2−5x−10=0⇔−2x+2−5x−10=0
⇔−7x−8=0⇔−7x−8=0
⇔−7x=8⇔−7x=8
hay x=−87x=−87
Vậy: S={−87}S={−87}
d) Ta có: 7−3x2−5+x5=17−3x2−5+x5=1
⇔5(7−3x)10−2(x+5)10=1010⇔5(7−3x)10−2(x+5)10=1010
⇔35−15x−2x−10−10=0⇔35−15x−2x−10−10=0
⇔−17x+15=0⇔−17x+15=0
⇔−17x=−15⇔−17x=−15
hay x=1517x=1517
Vậy: S={1517}
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
a ) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)(1)
ĐKXĐ : \(x\ne1;x\ne2\)
(1)\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow4x+7=15\\\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\left(KTMĐKXĐ\right)\)
Vậy pt vô nghiệm .
b ) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( 2 )
ĐKXĐ : \(x\ne3;x\ne-2\)
(2) \(\Leftrightarrow3x-x^2+6-2x+x^2+2x=3x+6-x^2-2x\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=-2\left(KTMĐKXĐ\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S={0}.
c ) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) (3)
ĐKXĐ : \(x\ne1;x\ne3\)
\(\left(3\right)\Leftrightarrow\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow6\left(3-x\right)+4\left(x-1\right)=8\)
\(\Leftrightarrow18-6x+4x-4=8\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy tập nghiệm của phương trình là S={-3}
d ) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (4)
ĐKXĐ : \(x\ne0;x\ne2\)
\(\left(4\right)\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTMĐKXĐ\right)\\x=-1\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S={-1}
a) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\) ( đk: x ≠ -1; x ≠ 2 )
\(\Leftrightarrow\) \(\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\) \(2-x+5\left(x+1\right)=15\)
\(\Leftrightarrow\) \(2-x+5x+5=15\)
\(\Leftrightarrow\)\(4x=8\)
\(\Rightarrow\) \(x=2\) ( KTM )
S = ∅
b) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( đk: x ≠ - 2 ; x ≠ 3 )
\(\Leftrightarrow\) \(\left(x+2\right)\left(3-x\right)+x\left(x+2\right)=5x+2\left(3-x\right)\)
\(\Leftrightarrow\) \(3x-x^2+6-2x+x^2+2x=5x+6-2x\)
\(\Leftrightarrow\) \(3x+6=3x+6\)
\(\Rightarrow\)\(0x=0\) ( TM )
\(\Rightarrow\) Phương trình vô số nghiệm
S = R
c) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) ( đk: x ≠ 1 ; x ≠ 3 )
\(\Leftrightarrow\) \(\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow\)\(6\left(3-x\right)+4\left(x-1\right)=8\)
\(\Leftrightarrow\) \(18-6x+4x-4=8\)
\(\Leftrightarrow\) \(-2x=-6\)
\(\Rightarrow x=3\) ( KTM )
S = ∅
d) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (đk: x ≠ 2; x ≠ 0 )
\(\Leftrightarrow\) \(x\left(x+2\right)-x+2=2\)
\(\Leftrightarrow\) \(x^2+2x-x+2=2\)
\(\Leftrightarrow\) \(x^2+x=0\)
\(\Leftrightarrow\) \(x\left(x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\)
S = \(\left\{2\right\}\)