Tìm x:
a) -x/8 = -9/2x
b) x/3 = 10/x+1
c) 2\(\dfrac{5}{6}\)x-1\(\dfrac{2}{3}\)+2\(\dfrac{3}{4}\)=1\(\dfrac{1}{3}\)
d) 2x-1/21 = 3/2x+1
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T
1 tháng 9 2023
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
T
1 tháng 9 2023
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
26 tháng 1
a: \(x-3\left(2x-6\right)=21-\left(5x+3\right)\)
=>\(x-6x+18=21-5x-3\)
=>18=18(luôn đúng)
=>\(x\in R\)
b: \(\left(x-2\right)\left(x+2\right)-\left(x-1\right)^2=2\left(x+1\right)\)
=>\(x^2-4-x^2+2x-1=2x+2\)
=>2x-5=2x+2
=>-7=0(vô lý)
=>\(x\in\varnothing\)
c: \(\dfrac{9x+4}{6}=1-\dfrac{3x-5}{9}\)
=>\(\dfrac{3\left(9x+4\right)}{18}=\dfrac{18}{18}-\dfrac{2\left(3x-5\right)}{18}\)
=>3(9x+4)=18-2(3x-5)
=>27x+12=18-6x+10
=>27x+12=-6x+28
=>33x=16
=>\(x=\dfrac{16}{33}\left(nhận\right)\)
d: ĐKXĐ: \(x\notin\left\{2;5\right\}\)
\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(\dfrac{6x+1}{\left(x-2\right)\left(x-5\right)}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(6x+1+5\left(x-5\right)=3\left(x-2\right)\)
=>6x+1+5x-25=3x-6
=>11x-24=3x-6
=>8x=18
=>\(x=\dfrac{9}{4}\left(nhận\right)\)
V
26 tháng 1
a: x−3(2x−6)=21−(5x+3)
=>x−6x+18=21−5x−3
=>18=18(luôn đúng)
=>x∈R
b: (x−2)(x+2)−(x−1)2=2(x+1)
=>x2−4−x2+2x−1=2x+2
=>2x-5=2x+2
=>-7=0(vô lý)
=>x∈∅
c: 9x+46=1−3x−59
=>3(9x+4)18=1818−2(3x−5)18
=>3(9x+4)=18-2(3x-5)
=>27x+12=18-6x+10
=>27x+12=-6x+28
=>33x=16
=>x=1633(nhận)
d: ĐKXĐ: x∉{2;5}
6x+1x2−7x+10+5x−2=3x−5
=>6x+1(x−2)(x−5)+5x−2=3x−5
=>6x+1+5(x−5)=3(x−2)6
=>6x+1+5x-25=3x-6
=>11x-24=3x-6
=>8x=18
=>x=94(nhận)
20 tháng 7 2022
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
19 tháng 12 2021
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
24 tháng 8 2021
\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)
\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)
\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)
24 tháng 8 2021
\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)
Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:
\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)
Tick plzz
5 tháng 3 2021
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
AH
Akai Haruma
Giáo viên
5 tháng 3 2021
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
17 tháng 7 2022
b: \(\Leftrightarrow\dfrac{2}{\left(x+7\right)\left(x-3\right)}=\dfrac{3x+21}{\left(x-3\right)\left(x+7\right)}\)
=>3x+21=2
=>x=-19/3
d: \(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=8\)
\(\Leftrightarrow4x^2+4x+1-4x^2+4x-1=8\)
=>8x=8
hay x=1
15 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
`=> (x-3)5 = (2x+1)3`
`=> 5x-15 = 6x+3`
`=> 5x-6x = 15+3`
`=> -x=18`
`=> x=-18`
\(\dfrac{x+1}{22}=\dfrac{6}{x}\)
`=> (x+1)x = 22*6`
`=> (x+1)x = 132`
`=> x^2 + x = 132`
`=> x^2+x-132=0`
`=> (x-11)(x+12)=0`
`=>`\(\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.\)
\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)
`=> (2x-1)x = 2*5`
`=> 2x^2 - x =10`
`=> 2x^2 - x - 10 =0`
`=> 2x^2 + 4x - 5x - 10 =0`
`=> (2x^2 + 4x) - (5x+10)=0`
`=> 2x(x+2) - 5(x+2)=0`
`=> (2x-5)(x+2)=0`
`=>`\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)
`=> (2x-1)(2x+1)=21*3`
`=> 4x^2 + 2x - 2x - 1 = 63`
`=> 4x^2 - 1=63`
`=> 4x^2 - 1 - 63=0`
`=> 4x^2 - 64 = 0`
`=> 4(x^2 - 16)=0`
`=> 4(x^2 + 4x - 4x - 16)=0`
`=> 4[(x^2+4x)-(4x+16)]=0`
`=> 4[x(x+4)-4(x+4)]=0`
`=> 4(x-4)(x+4)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)
`=> (2x+1)(x+1) = 9*5`
`=> (2x+1)(x+1)=45`
`=> 2x^2 + 2x + x + 1 = 45`
`=> 2x^2 + 3x + 1 =45`
`=> 2x^2 + 3x + 1 - 45 =0`
`=> 2x^2+3x-44=0`
`=> 2x^2 + 11x - 8x - 44=0`
`=> (2x^2 +11x) - (8x+44)=0`
`=> x(2x+11) - 4(2x+11)=0`
`=> (x-4)(2x+11)=0`
`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)
BD
15 tháng 6 2023
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)
20 tháng 1 2021
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
9 tháng 5 2021
Giải:
a) \(\dfrac{12}{16}=\dfrac{-x}{4}=\dfrac{21}{y}=\dfrac{z}{80}\)
\(\Rightarrow x=\dfrac{12.-4}{16}=-3\)
\(\Rightarrow y=\dfrac{16.21}{12}=28\)
\(\Rightarrow z=\dfrac{12.80}{16}=60\)
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)\) =0
\(\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)
\(x.\left(\dfrac{1}{3}+\dfrac{2}{5}\right)\) \(=0+\dfrac{2}{5}\)
\(x.\dfrac{11}{15}\) \(=\dfrac{2}{5}\)
x \(=\dfrac{2}{5}:\dfrac{11}{15}\)
x \(=\dfrac{6}{11}\)
c) (2x-3)(6-2x)=0
⇒2x-3=0 hoặc 6-2x=0
x=3/2 hoặc x=3
d) \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-2}{3}-\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}\)
\(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}\)
\(2x-5=\dfrac{-13}{2}\)
\(2x=\dfrac{-13}{2}+5\)
\(2x=\dfrac{-3}{2}\)
\(x=\dfrac{-3}{2}:2\)
\(x=\dfrac{-3}{4}\)
e) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}:2\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{8}\) hoặc \(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-1}{8}\)
\(x=\dfrac{11}{12}\) hoặc \(x=\dfrac{5}{12}\)
a) -x/8 = -9/(2x)
x.2x = -9.(-8)
2x² = 72
x² = 72: 2
x² = 36
x = -6 hoặc x = 6
b) x/3 = 10/(x + 1)
x.(x + 1) = 3.10
x.(x + 1) = 30
x² + x - 30 = 0
x² - 5x + 6x - 30 = 0
(x² - 5x) + (6x - 30) = 0
x(x - 5) + 6(x - 5) = 0
(x - 5)(x + 6) = 0
x - 5 = 0 hoặc x + 6 = 0
*) x - 5 = 0
x = 0 + 5
x = 5
*) x + 6 = 0
x = 0 - 6
x = -6
c) 2 5/6 x - 1 2/3 + 2 3/4 = 1 1/3
17/6 x - 5/3 + 11/4 = 4/3
17/6 x = 4/3 + 5/3 - 11/4
17/6 x = 1/4
x = 1/4 : 17/6
x = 3/34
d) (2x - 1)/21 = 3/(2x + 1)
(2x - 1)(2x + 1) = 3.21
4x² + 2x - 2x - 1 = 63
4x² = 63 + 1
4x² = 64
x² = 64 : 4
x² = 16
x = -4 hoặc x = 4