7x=3x+8
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KT
1
TA
28 tháng 5 2021
ĐK: ` x \ne 2/7`
`(2x+3)((3x+8)/(2-7x)+1)=(x-5)((3x+8)/(2-7x)+1)`
`<=> ((3x+8)(2-7x)+1)(2x+3-x+5)=0`
`<=> ((3x+8)/(2-7x)+1)(x+8)=0`
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x+8}{2-7x}=-1\\x+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)
Vậy `S={5/2 ; -8}`.
IL
1
20 tháng 4 2020
\(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ: x \(\ne\) \(\frac{2}{7}\))
\(\Leftrightarrow\) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\) \(\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)
\(\Leftrightarrow\) \(\left(\frac{10-4x}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\) \(\frac{10-4x}{2-7x}=0\) hoặc x + 8 = 0
\(\Leftrightarrow\) x = \(\frac{5}{2}\) và x = -8
Vậy S = {\(\frac{5}{2}\); -8}
Chúc bn học tốt!!
22 tháng 1 2021
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
DH
1
28 tháng 6 2015
1)80-63x-7x-4x=10
80-[(63-7-4)x)]=10
80-52x=10
52x=80-10
52x=70
x=70/52
x=35/26
25 tháng 1 2021
a) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: S={-5;2}
b) Ta có: \(3x^2-7x+1=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)
c) Ta có: \(3x^2-7x+8=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)
Vậy: \(x\in\varnothing\)
NY
2
21 tháng 1 2017
a, \(\left(3x-7\right)-\left(-5-7x\right)=4x+9\)
\(\Leftrightarrow3x-7+5+7x=4x+9\)
\(\Leftrightarrow3x+7x-4x=9+7-5\)
\(\Leftrightarrow\left(3+7-6\right)x=11\Rightarrow4x=11\Rightarrow x=\frac{11}{4}\notin Z\)
Vậy : \(x\in\varnothing\)
b, \(\left(3x-7\right)-\left(-5-7x\right)=5x+8\)
\(\Leftrightarrow3x-7+5+7x=5x+8\)
\(\Leftrightarrow3x+7x-5x=8+7-5\)
\(\Leftrightarrow\left(3+7-5\right)x=10\Rightarrow5x=10\Rightarrow x=10\div5=2\in Z\)
Vậy : x = 2
c, \(\left(3x-2\right)^3=125\Leftrightarrow\left(3x-2\right)^3=5^3\)
\(\Rightarrow3x-2=5\Rightarrow3x=5+2\Rightarrow3x=7\)
\(\Rightarrow x=7\div3\Rightarrow x=\frac{7}{3}\notin Z\)
Vậy \(x\in\varnothing\)
TL
21 tháng 1 2017
a, (3x-7)-(-5-7x)=4x+9
3x-7+5+7x=4x+9
3x+7x-4x=9+7-5
6x=11
x=11/6(thuộc Z t/m)
Vậy...
b, (3x-7)-(-5-7x)=5x+8
3x-7+5+7x=5x+8
3x+7x-5x=8+7-5
5x=10
x=2(thuộc Z t/m)
vây...
c, (3x-2)^3=125
(3x-2)^3=5^3
3x-2=5
x=7/3(thuộc Z thỏa mãn)
Vậy...
11 tháng 2 2019
\(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
ĐK: \(x\ne\dfrac{2}{7}\)
\(\Leftrightarrow\left(2x+3\right)\left(3x+8+2-7x\right)=\left(x-5\right)\left(3x+8+2-7x\right)\\ \Leftrightarrow\left(2x+3\right)\left(10-4x\right)=\left(x-5\right)\left(10-4x\right)\\ \Leftrightarrow\left(10-4x\right)\left(2x+3-x+5\right)=0\\ \Leftrightarrow\left(10-4x\right)\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}10-4x=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(TM\right)\\x=-8\left(TM\right)\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{5}{2};-8\right\}\)
3 tháng 6 2021
Lần này hết lag :))
`3x^2-7x+8=0`
`<=>x^2-7/3x+8/3=0`
`<=>x^2-2.x. 7/6+49/36+47/36=0`
`<=>(x-7/6)^2+47/36=0` vô lý vì `(x-7/6)^2+47/36>=47/36>0`
Vậy pt vô nghiệm
=>7x-3x=8
=>4x=8
=>x=8:4
=>=2
t i c k mình nha cảm ơn
\(7\cdot x=3\cdot x+8\)
\(\left(7-3\right).x=8\)
\(4.x=8\)
\(x=8:4=2.\)