2.(x-3)+3x+0.5=\(\dfrac{3}{4}\)
4x+2+4x=272
(1,2-5x).(2\(\dfrac{1}{8}\) +1/2 x)=0
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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 4 2022
\(\lim\limits_{x\rightarrow-\infty}\dfrac{x-5x^2+1}{x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}-5+\dfrac{1}{x^2}}{1-\dfrac{1}{x^2}}=\dfrac{-5}{1}=-5\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^3\left(2-x^2\right)^3\left(4x^2+1\right)^2}{4x^{13}+x^2-6}=\lim\limits_{x\rightarrow+\infty}\dfrac{5\left(\dfrac{2}{x^2}-1\right)^3\left(4+\dfrac{1}{x^2}\right)^2}{4+\dfrac{1}{x^{11}}-\dfrac{6}{x^{13}}}=\dfrac{5.\left(-1\right)^3.4^2}{4}=-20\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{4x-\sqrt{9x^2+x}}{3-x}=\lim\limits_{x\rightarrow+\infty}\dfrac{4-\sqrt{9+\dfrac{1}{x}}}{\dfrac{3}{x}-1}=\dfrac{4-3}{-1}=-1\)
3 tháng 3 2022
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
28 tháng 8 2021
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
11 tháng 8 2021
1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)
\(\Leftrightarrow5x+1-2x+4=4\)
\(\Leftrightarrow3x=-1\)
hay \(x=-\dfrac{1}{3}\)
2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
\(\Leftrightarrow9x+27+12-36x=-2x+2\)
\(\Leftrightarrow-27x+2x=2-39\)
hay \(x=\dfrac{37}{25}\)
3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x+6-10x=4-4x\)
\(\Leftrightarrow-7x+4x=4-6=-2\)
hay \(x=\dfrac{2}{3}\)
11 tháng 8 2021
4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
\(\Leftrightarrow5x-15-x-1=2x-4\)
\(\Leftrightarrow4x-2x=-4+16=12\)
hay x=6
5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
\(\Leftrightarrow12x+3-9x+5+4x-8=0\)
\(\Leftrightarrow7x=0\)
hay x=0
18 tháng 2 2021
b) Ta có: \(3+\left(x-5\right)=2\left(3x-2\right)\)
\(\Leftrightarrow3+x-5=6x-4\)
\(\Leftrightarrow x-2-6x+4=0\)
\(\Leftrightarrow-5x+2=0\)
\(\Leftrightarrow-5x=-2\)
\(\Leftrightarrow x=\dfrac{2}{5}\)
Vậy: \(S=\left\{\dfrac{2}{5}\right\}\)
c) Ta có: \(2\left(x-0.5\right)+3=0.25\left(4x-1\right)\)
\(\Leftrightarrow2x-1+3=x-\dfrac{1}{4}\)
\(\Leftrightarrow2x+2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x+\dfrac{9}{4}=0\)
\(\Leftrightarrow x=-\dfrac{9}{4}\)
Vậy: \(S=\left\{-\dfrac{9}{4}\right\}\)
d) Ta có: \(2\left(x-\dfrac{1}{4}\right)-4=-6\left(-\dfrac{1}{3}x+0.5\right)+2\)
\(\Leftrightarrow2x-\dfrac{1}{2}-4=2x-3+2\)
\(\Leftrightarrow2x-\dfrac{9}{2}=2x-1\)
\(\Leftrightarrow2x-2x=-1+\dfrac{9}{2}\)
\(\Leftrightarrow0x=\dfrac{7}{2}\)(vô lý)
Vậy: \(S=\varnothing\)
AH
Akai Haruma
Giáo viên
13 tháng 12 2021
Lời giải:
a.
\(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)
\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)
\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)
b.
\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)
\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)
c.
\(\frac{4x^2-3x+5}{x^3-1}\)
\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)
\(-2=\frac{-2(x^3-1)}{x^3-1}\)
S
22 tháng 3 2021
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
22 tháng 3 2021
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
11 tháng 8 2021
1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow2x-8+12x=4x-2\)
\(\Leftrightarrow10x=6\)
hay \(x=\dfrac{3}{5}\)
2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
\(\Leftrightarrow15x-6-30=10-20x\)
\(\Leftrightarrow35x=46\)
hay \(x=\dfrac{46}{35}\)
3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
\(\Leftrightarrow3x-6-4=6x-6\)
\(\Leftrightarrow-3x=4\)
hay \(x=-\dfrac{4}{3}\)
6 tháng 2 2018
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
\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)
a) \(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\)
\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\Rightarrow5x-6=\dfrac{1}{4}\)
\(\Rightarrow5x=\dfrac{1}{4}+6\)
\(\Rightarrow5x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:5\)
\(\Rightarrow x=\dfrac{5}{4}\)
b) \(4^{x+2}+4^x=272\)
\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)
\(\Rightarrow4^x\cdot\left(16+1\right)=272\)
\(\Rightarrow4^x\cdot17=272\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)