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NT
2
TQ
27 tháng 1 2018
<=>\(\dfrac{\left(3x-0,4\right)15}{30}+\dfrac{\left(1,5-2x\right)10}{30}=\dfrac{\left(x+0,5\right)6}{30}\)
=>\(\left(3x-0,4\right)15+\left(1,5-2x\right)10=\left(x+0,5\right)6\)
<=>\(45x-6+15-20x=6x+3\)
<=>\(45x-20x-6x=6+3-15\)
<=>\(19x=-6\)
<=>\(x=\dfrac{-6}{19}\)
NN
2
22 tháng 1 2018
ĐKXĐ x≠3 ; x≠-3
\(\dfrac{2x-1}{x+3}=\dfrac{2x+1}{x-3}\)
=> (2x-1)(x-3)=(2x+1)(x+3)
⇔2x2-6x-x+3=2x2+6x+x+3
⇔2x2-2x2-7x-6x=3-3
⇔ -13x=0
⇔x=0 (tm)
vậy phương trình trên có tập no S={0}
PP
1
3 tháng 5 2017
ĐKXĐ: x\(\ne2\), \(x\ne1\)
\(\dfrac{2x-5}{x-2}-\dfrac{3x-5}{x-1}=-1\)
<=> \(\dfrac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}-\dfrac{\left(3x-5\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{-1.\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}\)
=> 2x2-2x-5x+5-3x2+6x+5x-10= -x2+2x-2+x
<=> 2x2-2x-5x+5-3x2+6x+5x-10+x2-2x+2-x=0
<=> x-3=0
<=> x=3 (thỏa mãn ĐKXĐ)
Vậy S=\(\left\{3\right\}\)
MT
2
DM
23 tháng 4 2017
Pt trên có MSC là \(\left(x-1\right)\left(x^2+x+1\right)\)
Quy đồng mẫu số :
\(\dfrac{1}{x-1}+\dfrac{7x-10}{x^3-1}-\dfrac{3}{x^2+x+1}=0\)
( ĐKXĐ \(x\ne1\))
\(\Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{7x-10}{x^3-1}-\dfrac{3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+x+1+7x-10-3x+3}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\) \(\dfrac{x^2+5x-6}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\left(KTMĐK\right)\\x=-6\left(TMĐK\right)\end{matrix}\right.\)
Vậy \(S=\left\{-6\right\}\)
TS
23 tháng 4 2017
ĐKXĐ: \(x\ne1\); \(x\ne-1\)
\(\dfrac{1}{x-1}+\dfrac{7x-10}{x^3-1}-\dfrac{3}{x^2+x+1}=0\)
\(\Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{7x-10}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Rightarrow x^2+x+1+7x-10-3x+3=0\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow x-1=0\) ; \(x+6=0\)
+) \(x-1=0\)
\(\Leftrightarrow x=1\) (Không thỏa mãn ĐKXĐ)
+) \(x+6=0\)
\(\Leftrightarrow x=-6\) (Thỏa mãn ĐKXĐ)
Tập nghiệm: \(S=\left\{-6\right\}\)
18 tháng 6 2022
a: \(\Leftrightarrow\left(\left|x\right|\right)^2-5\left|x\right|-6=0\)
\(\Leftrightarrow\left(\left|x\right|-6\right)\left(\left|x\right|+1\right)=0\)
\(\Leftrightarrow\left|x\right|-6=0\)
=>x=6 hoặc x=-6
b: \(\dfrac{x}{x-2}+\dfrac{5}{\left|x+2\right|}=1\)
Trường hợp 1: x>-2 và x<>2
Pt sẽ là \(\dfrac{x}{x-2}+\dfrac{5}{x+2}=1\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=x\left(x+2\right)+5\left(x-2\right)\)
\(\Leftrightarrow x^2+2x+5x-10=x^2-4\)
=>7x=6
hay x=6/7(nhận)
TRường hợp 2: x<-2
Pt sẽ là \(\dfrac{x}{x-2}-\dfrac{5}{x+2}=1\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=x\left(x+2\right)-5\left(x-2\right)\)
\(\Leftrightarrow x^2+2x-5x+10=x^2-4\)
=>-3x=-14
hay x=14/3(loại)
10 tháng 8 2018
\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\) ( ĐK : \(x\ne-2\) )
\(\Leftrightarrow\dfrac{12}{x^3+2^3}=1+\dfrac{1}{x+2}\)
\(\Leftrightarrow\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}+\dfrac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\Leftrightarrow12=\left(x+2\right)\left(x^2-2x+4\right)+x^2-2x+4\)
\(\Leftrightarrow x^3+8+x^2-2x+4=12\)
\(\Leftrightarrow x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=1\left(N\right)\\x=-2\left(L\right)\end{matrix}\right.\)
Vậy \(S=\left\{0;1\right\}\)
12 tháng 4 2018
a) \(15x-3\left(3x-2\right)=45-5\left(2x-5\right)\)
\(\Leftrightarrow15x-9x+6=45-10x+25\)
\(\Leftrightarrow15x-9x+10x=45+25-6\)
\(\Leftrightarrow16x=64\)
\(\Leftrightarrow x=4\)
b) \(x^2-9+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\x+7=0\Leftrightarrow x=-7\end{matrix}\right.\)
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow x+4+x^2-4x+2x-8=5x-4\)
\(\Leftrightarrow x^2+x-4x+2x-5x=-4+8-4\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\Leftrightarrow x=6\end{matrix}\right.\)
KN
12 tháng 4 2018
a) 15x - 3(3x - 2) = 45 - 5(2x - 5)
\(\Leftrightarrow\) 15x - 9x + 6 = 45 - 10x + 25
\(\Leftrightarrow\) 6x + 10x = 70 - 6
\(\Leftrightarrow\) 16x = 64
\(\Leftrightarrow\) x = 4
Vậy.......................
b) x2 - 9 + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3) + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3 + 4) = 0
\(\Leftrightarrow\) (x - 3)(x + 7) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=3\end{matrix}\right.\)
Vậy........................
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\) (đk: x\(\ne\pm\)4)
\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}+\dfrac{\left(x+2\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5x-4}{\left(x+4\right)\left(x-4\right)}\)
\(\Leftrightarrow\) x + 4 + x2 - 4x + 2x - 8 = 5x - 4
\(\Leftrightarrow\) x2 - x - 5x - 4 + 4 = 0
\(\Leftrightarrow\) x2 - 6x = 0
\(\Leftrightarrow\) x(x - 6) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=6\left(tmđk\right)\end{matrix}\right.\)
Vậy...............
18 tháng 6 2022
\(\Leftrightarrow\left(\dfrac{5}{x}+1\right)+\left(\dfrac{4}{x+1}+1\right)=\left(\dfrac{3}{x+2}+1\right)+\left(\dfrac{2}{x+3}+1\right)\)
=>x+5=0
hay x=-5
15 tháng 5 2022
a: \(\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow x^2+30x+25=x^2+25x\)
=>5x=-25
hay x=-5(loại)
b: \(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
=>2x+7=10
hay x=3/2
\(x=126\)
x/60 + x/45 + 1,5 = 32/5
3x + 4x + 270 = 1152
7x + 270 = 1152
7x = 1152 − 270
7x = 882
x = 882/7
x = 126