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18 tháng 4 2017
1.
\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\left(ĐKXĐ:x\ne1\right)\\ \Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\\ \Leftrightarrow21x-9=2x-2\\ \Leftrightarrow19x=7\\ \Leftrightarrow x=\dfrac{7}{19}\left(TMĐK\right)\)
2.
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\left(ĐKXĐ:x\ne-\dfrac{2}{3};x\ne\dfrac{1}{3}\right)\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\\ \Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\\ \Leftrightarrow-8x+1=-11x-14\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\left(TMĐK\right)\)
3.
\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ \Leftrightarrow\left(\dfrac{1-x}{x+1}+3\right)\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}.\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{4+2x}{x+1}\left(x+1\right)=2x+3\\ \Leftrightarrow4+2x=2x+3\\ \Leftrightarrow4=3\)
Vô nghiệm.
Y
26 tháng 8 2018
a) \(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
\(=\dfrac{5x-1-5x+7}{3x+2-3x+1}\)
\(=\dfrac{-1+7}{2+1}\)
\(=\dfrac{6}{3}\)
\(=2\)
Với \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
12 tháng 5 2017
1) \(2\left(3x-1\right)-3x=10\)
<=> \(6x-2-3x=10\)
<=>\(3x-2=10\)
<=> \(3x=12\)
<=> \(x=4\)
Vậy tập nghiệm của pt S={4}
2) \(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)
ĐKXĐ: x khác 0; x khác 1,-1
<=> \(\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}+\dfrac{x\left(x+1\right)}{x\left(x+1\right)}\)= \(\dfrac{3x^2-x}{x\left(x+1\right)}+\dfrac{1}{x\left(x+1\right)}\)
=> \(\left(x+1\right)^2+x\left(x+1\right)\)= \(3x^2-x+1\)
<=> \(x^2+2x+1+x^2+x=3x^2-x+1\)
<=> \(x^2+x^2+2x+x-3x^2+x\)= \(1-1\)
<=> \(-x^2+4x=0\)
<=>\(4x=x^2\)
<=> \(4=x\) ( TMĐKXĐ)
Vậy tập nghiệm của pt S={4}
c) \(\dfrac{2x+1}{3}-\dfrac{3x-2}{2}>\dfrac{1}{6}\)
<=> \(\dfrac{4x+2}{6}-\dfrac{9x-6}{6}>\dfrac{1}{6}\)
<=> \(\dfrac{4x+2-9x+6}{6}-\dfrac{1}{6}>0\)
<=> \(\dfrac{-5x+7}{6}>0\)
Mà 6>0 . Nên \(-5x+7>0\)
Ta có \(-5x+7>0\)
<=> \(-5x>-7\)
<=> \(x< \dfrac{7}{5}\)
Vậy tập nghiệm của bất phương trình S={x thuộc R| \(x< \dfrac{7}{5}\)}
PP
12 tháng 5 2017
1)2.(3x-1)-3x=10
6x-2-3x =10
6x-3x =10+2
3x =12
x =4
Vậy S=4
2) \(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)
Đkxđ: \(x\ne0\) và \(x\ne-1\)
MTC;x(x+1)
\(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)
\(\Leftrightarrow\)\(\dfrac{\left(x+1\right)\left(x+1\right)+x\left(x+1\right)}{x\left(x+1\right)}=\dfrac{x\left(3x-1\right)+1}{x\left(x+1\right)}\)
\(\Leftrightarrow\)(x+1) (x+1)+x(x+1) = x (3x-1)+1
\(\Leftrightarrow\)x2+x+x+1+x2+x =3x2-x+1
\(\Leftrightarrow\)x2+x+x+1+x2+x-3x2+x-1=0
\(\Leftrightarrow\)-x24x=0
\(\Leftrightarrow\)4x-x2=0
\(\Leftrightarrow\)x(4-x)=0
\(\Leftrightarrow\)x=0 hoặc 4-x=0
\(\Leftrightarrow\)x=0 hoặc x =4
3)\(\dfrac{2x+1}{3}-\dfrac{3x-2}{2}>\dfrac{1}{6}\)
\(\Leftrightarrow\)\(\dfrac{2x+1}{3}6-\dfrac{3x-2}{2}6>\dfrac{1}{6}\)6
\(\Leftrightarrow\)2(2x+1)-3(3x-2)>1
\(\Leftrightarrow\)4x+2-9x+6>1
\(\Leftrightarrow\)4x-9x>1-2-6
\(\Leftrightarrow\)-5x>-7
\(\Leftrightarrow\)-5x.\(\dfrac{1}{-5}>-7.\dfrac{1}{-5}\)
\(\Leftrightarrow x>\dfrac{7}{5}\)
CM
14 tháng 12 2018
a) \(\dfrac{2}{3x+9}-\dfrac{x-3}{3x^2+9x}\)
\(=\dfrac{2}{3\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)
\(=\dfrac{2x}{3x\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)
\(=\dfrac{2x-x+3}{3x\left(x+3\right)}\)
\(=\dfrac{x+3}{3x\left(x+3\right)}\)
\(=\dfrac{1}{3x}\)
b) \(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}.\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\)
\(=\dfrac{x}{\left(x-1\right).3}\)
\(=\dfrac{x}{3x-3}\)
c) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+99}-\dfrac{1}{x+100}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+100}\)
\(=\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
\(=\dfrac{x+100-x}{x\left(x+100\right)}\)
\(=\dfrac{100}{x\left(x+100\right)}\)
14 tháng 12 2018
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
18 tháng 4 2017
a) ĐKXĐ: \(x\ne-1,x\ne0\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
<=> \(\dfrac{x\left(x+3\right)+\left(x-2\right)\left(x+1\right)-2x\left(x+1\right)}{x\left(x+1\right)}=0\)
<=> \(\dfrac{x^2+3x+x^2-x-2-2x^2-2x}{x\left(x+1\right)}=0\)
<=> \(\dfrac{-2}{x\left(x+1\right)}=0\) (vô lí)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne3,x\ne-2\)
ta có:\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\)
<=> \(\dfrac{\left(x+2\right)\left(3-x\right)+x\left(x+2\right)-5x-2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\)
<=> \(\dfrac{x-x^2+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\)
<=> \(\dfrac{0}{\left(x+2\right)\left(3-x\right)}=0\) (luôn đúng)
Vậy pt trên luôn đúng với mọi x khác 3 và -2
LM
18 tháng 4 2017
a) \(\dfrac{x+3}{x+1}\)+\(\dfrac{x-2}{x}\)=2
(đk: x\(\ne\); x\(\ne\)-1)
<=> \(x^2\)+3x + \(x^2\)-x -2 =\(2x^2\)+2x
<=> 2x -2 =2x
<=>0x=2
=>Pt vô nghiệm.
b) 1+ \(\dfrac{x}{3-x}\)= \(\dfrac{5x}{\left(x+2\right)\left(3-x\right)}\)+\(\dfrac{2}{x+2}\)
(đk:x\(\ne\)3; x\(\ne\)-2)
<=> 3x +6=3x+6
<=>0x=0
=> Pt vô số no.
c)\(\dfrac{3x+2}{3x-2}\)-\(\dfrac{6}{2+3x}\)=\(\dfrac{9x^2}{9x^2-4}\)
(đk: x\(\ne\)\(\pm\)\(\dfrac{2}{3}\))
<=>\((3x+2)^2\)-6(3x-2)=\(9x^2\)
<=>\(9x^2 \)+12x +4 -18x+12=\(9x^2\)
<=>16-6x=0
<=>6x=16
<=> x=\(\dfrac{8}{3}\)(t/m)
Vậy pt có no duy nhất là x=\(\dfrac{8}{3}\)
TT
20 tháng 12 2018
a.
\(\dfrac{x+3}{x-2}+\dfrac{4+x}{2-x}\\ =\dfrac{x+3}{x-2}-\dfrac{4+x}{x-2}\\ =\dfrac{x+3-4-x}{x-2}\\ =-\dfrac{1}{x-2}\)
b. \(\dfrac{x+1}{2x+6}+\dfrac{2x+3}{x^2+3x}\)
\(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x}{2x\left(x+3\right)}+\dfrac{4x+6}{2x\left(x+3\right)}=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x^2+3x+2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x\left(x+3\right)+2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{x+2}{2x}\)
c. \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
d. \(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)
\(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}:\dfrac{-x\left(x+3\right)}{3x-1}\)
\(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}.\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}\)
\(=-\dfrac{2}{x^2}\)
23 tháng 2 2019
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 \(S=\left\{-1\right\}\)
|:-))
\(\dfrac{3x+1}{x+2}\le1\)
\(\Leftrightarrow\dfrac{3x+1}{x+2}-1\le0\)
\(\Leftrightarrow\dfrac{3x+1-x-2}{x+2}\le0\)
\(\Leftrightarrow\dfrac{2x-1}{x+2}\le0\)
Đặt \(n\left(x\right)=\dfrac{2x-1}{x+2}\)
Ta có bảng xét dấu:
Với: \(x=\dfrac{1}{2}\Rightarrow f\left(x\right)=0\)
\(-2< x< \dfrac{1}{2}\Rightarrow f\left(x\right)< 0\)
\(\Rightarrow\dfrac{3x+1}{x+2}\le1\)
\(\Leftrightarrow-2< x\le\dfrac{1}{2}\)