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1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
a. \(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}=2x+\dfrac{2x+1}{2}\)
\(\Leftrightarrow2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
\(\Leftrightarrow12x+10-10x-3=8x+4x+2\)
\(\Leftrightarrow12x-10x-8x-4x=2-10+3\)
\(\Leftrightarrow-10x=-5\Leftrightarrow x=\dfrac{1}{2}\)
b. \(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+6x+6\)
\(\Leftrightarrow6x^2+2=6x^2+6x+6\)
\(\Leftrightarrow6x^2-6x^2-6x=6-2\Leftrightarrow-6x=4\)
\(\Leftrightarrow x=\dfrac{-2}{3}\)
c. \(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\left(\dfrac{x+2}{13}+1\right)+\left(\dfrac{2x+45}{15}-1\right)=\left(\dfrac{3x+8}{37}+1\right)+\left(\dfrac{4x+69}{9}-1\right)\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}-\dfrac{3\left(x+15\right)}{37}-\dfrac{4\left(x+15\right)}{9}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
Vì \(\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)>0\)
\(\Leftrightarrow x+15=0\Leftrightarrow x=-15\)
a)\(\dfrac{1}{2}\)(x+1)+\(\dfrac{1}{4}\)(x+3)=3-\(\dfrac{1}{3}\)(x+2)
\(\Leftrightarrow\)\(\dfrac{1}{2}\)x+\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)x+\(\dfrac{3}{4}\)=3-\(\dfrac{1}{3}\)x-\(\dfrac{2}{3}\)
\(\Leftrightarrow\)\(\dfrac{1}{2}\)x+\(\dfrac{1}{4}\)x+\(\dfrac{1}{3}\)x=-\(\dfrac{1}{2}\)-\(\dfrac{3}{4}\)+3-\(\dfrac{2}{3}\)
\(\Leftrightarrow\)\(\dfrac{13}{12}\)x=\(\dfrac{13}{12}\)
\(\Leftrightarrow\)x=1
Vậy nghiệm của pt là x=1
b)\(\dfrac{x+2}{98}\)+\(\dfrac{x+4}{96}\)=\(\dfrac{x+6}{94}\)+\(\dfrac{x+8}{92}\)
\(\Leftrightarrow\)\(\dfrac{x+2}{98}\)+\(\dfrac{x+4}{96}\)-\(\dfrac{x+6}{94}\)-\(\dfrac{x+8}{92}\)=0
\(\Leftrightarrow\)(\(\dfrac{x+2}{98}\)+1)+(\(\dfrac{x+4}{96}\)+1)-(\(\dfrac{x+6}{94}\)+1)-(\(\dfrac{x+8}{92}\)+1)=0
\(\Leftrightarrow\)\(\dfrac{x+2+98}{98}\)+\(\dfrac{x+4+96}{96}\)-\(\dfrac{x+6+94}{94}\)-\(\dfrac{x+8+92}{92}\)=0
\(\Leftrightarrow\)\(\dfrac{x+100}{98}\)+\(\dfrac{x+100}{96}\)-\(\dfrac{x+100}{94}\)-\(\dfrac{x+100}{92}\)=0
\(\Leftrightarrow\)(x+100)(\(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\))=0
\(\Leftrightarrow\)x+100=0(vì\(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\)\(\ne\)0)
\(\Leftrightarrow\)x=-100
Vậy nghiệm của pt là x=-100
\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\)\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}=\dfrac{3\left(x+15\right)}{37}+\dfrac{4\left(x+15\right)}{9}\)\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}-\dfrac{3\left(x+15\right)}{37}-\dfrac{4\left(x+15\right)}{9}=0\)\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}+\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow x+15=0\)
\(\Leftrightarrow x=-15\)
Vậy x = -15.
\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\\ \Leftrightarrow\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\\ \Leftrightarrow\dfrac{x+15}{13}+\dfrac{2x+30}{15}=\dfrac{3x+45}{37}+\dfrac{4x+60}{9}\)
\(\Leftrightarrow\left(x+15\right)\dfrac{1}{13}+\left(x+15\right)\dfrac{2}{15}=\left(x+15\right)\dfrac{3}{37}+\left(x+15\right)\dfrac{4}{9}\\ \Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
vì:\(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\ne0\) nên:
x+15=0 =>x=-15
vậy phương trình có tập nghiệm là S={-15}
giải pt sau
g) 11+8x-3=5x-3+x
\(\Leftrightarrow\) 8x + 8 = 6x - 3
<=> 8x-6x = -3 - 8
<=> 2x = -11
=> x=-\(\dfrac{11}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{11}{2}\)}
h)4-2x+15=9x+4-2x
<=> 19 - 2x = 7x + 4
<=> -2x - 7x = 4 - 19
<=> -9x = -15
=> x=\(\dfrac{15}{9}=\dfrac{5}{3}\)
Vậy tập nghiệm của pt là : S={\(\dfrac{5}{3}\)}
g)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
<=> \(\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2+6.2x}{6}\)
<=> 9x + 6 - 3x + 1 = 10 + 12x
<=> 6x + 7 = 10 + 12x
<=> 6x -12x = 10-7
<=> -6x = 3
=> x= \(-\dfrac{1}{2}\)
Vậy tập nghiệm của PT là : S={\(-\dfrac{1}{2}\)}
\(h,\dfrac{x+4}{5}-x+4=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{x+4-5\left(x+4\right)}{5}=\dfrac{4x+2-5.5}{5}\)
<=> x + 4 - 5x - 20 = 4x + 2 - 25
<=> x - 5x - 4x = 2-25-4+20
<=> -8x = -7
=> x= \(\dfrac{7}{8}\)
Vậy tập nghiệm của PT là S={\(\dfrac{7}{8}\)}
\(i,\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
<=> \(\dfrac{21\left(4x+3\right)}{105}\)-\(\dfrac{15\left(6x-2\right)}{105}\)=\(\dfrac{35\left(5x+4\right)+3.105}{105}\)
<=> 84x + 63 - 90x + 30 = 175x + 140 + 315
<=> 84x - 90x - 175x = 140 + 315 - 63 - 30
<=> -181x = 362
=> x = -2
Vậy tập nghiệm của PT là : S={-2}
K) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)-150}{30}\)
<=> 25x + 10 - 80x - 10 = 24x + 12 - 150
<=> -55x = 24x - 138
<=> -55x - 24x = -138
=> -79x = -138
=> x=\(\dfrac{138}{79}\)
Vậy tập nghiệm của PT là S={\(\dfrac{138}{79}\)}
m) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
<=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
<=> 6x - 3 - 5x + 10 = x+7
<=> x + 7 = x+7
<=> 0x = 0
=> PT vô nghiệm
Vậy S=\(\varnothing\)
n)\(\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{2}\left(x+1\right)-\dfrac{1}{3}\left(x+2\right)\)
<=> \(\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
<=> \(\dfrac{1}{4}x+\dfrac{1}{2}x+\dfrac{1}{3}x=3-\dfrac{1}{2}-\dfrac{2}{3}-\dfrac{3}{4}\)
<=> \(\dfrac{13}{12}x=\dfrac{13}{12}\)
=> x= 1
Vậy S={1}
p) \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-6\)
<=> \(\dfrac{2x-2x+1}{6}=\dfrac{x-36}{6}\)
<=> 2x -2x + 1= x-36
<=> 2x-2x-x = -37
=> x = 37
Vậy S={37}
q) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
<=> \(\dfrac{4\left(2+x\right)-20.0,5x}{20}=\dfrac{5\left(1-2x\right)+20.0,25}{20}\)
<=> 8 + 4x - 10x = 5 - 10x + 5
<=> 4x-10x + 10x = 5+5-8
<=> 4x = 2
=> x= \(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
g) \(11+8x-3=5x-3+x\)
\(\Leftrightarrow8+8x=6x-3\)
\(\Leftrightarrow8x-6x=-3-8\)
\(\Leftrightarrow2x=-11\)
\(\Leftrightarrow x=-\dfrac{11}{2}\)
h, \(4-2x+15=9x+4-2x\)
\(\Leftrightarrow-2x-9x+2x=4-4-15\)
\(\Leftrightarrow-9x=-15\)
\(\Leftrightarrow x=\dfrac{-15}{-9}=\dfrac{5}{3}\)
Giải phương trình
\(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\)\(\dfrac{x+2}{13}+1+\dfrac{2x+45}{15}-1=\dfrac{3x+8}{37}+1+\dfrac{4x+69}{9}-1\)
\(\Leftrightarrow\)\(\dfrac{x+2}{13}+\dfrac{13}{13}+\dfrac{2x+45}{15}-\dfrac{15}{15}=\dfrac{3x+8}{37}+\dfrac{37}{37}+\dfrac{4x+69}{9}-\dfrac{9}{9}\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2x+30}{15}=\dfrac{3x+45}{37}+\dfrac{4x+60}{9}\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}=\dfrac{3\left(x+15\right)}{37}+\dfrac{4\left(x+15\right)}{9}\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)=\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}\right)-\left(x+15\right)\left(\dfrac{3}{37}+\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=0\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-15\\\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\ne0\end{matrix}\right.\)
Do đó: \(x=-15\)
Vậy \(S=\left\{-15\right\}\)
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, Ta có : \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
=> \(\dfrac{6x-3-5x+10}{15}=\dfrac{x+7}{15}\)
=>\(\dfrac{x+7}{15}=\dfrac{x+7}{15}\)
Vậy phương trình thỏa mãn với mọi x
b, Ta có :\(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
=>\(\dfrac{5x+2-2\left(8x+1\right)}{6}=\dfrac{4x+2-25}{5}\)
=>\(\dfrac{5x+2-16x+2}{6}=\dfrac{4x-23}{5}\)
=>\(\dfrac{-11x+4}{6}=\dfrac{4x-23}{5}\)
=> 6(4x-23)= 5(-11x+4) => 24x-138=-55x+20 => 79x =158 =x=2
Vậy x=2
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
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