a) \(\dfrac{1}{3}+\dfrac{2}{5}(x+1)=1\)
b) \(\dfrac{1}{4}+\dfrac{1}{3}:3x=-5\)
K
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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!
22 tháng 7 2023
a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
3 tháng 2 2022
a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)
\(\Rightarrow-x^2=-36\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{6;-6\right\}\)
b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{5}{9}+\dfrac{-9x}{9}=\dfrac{-3}{9}\)
\(\Rightarrow5-9x=-3\)
\(\Rightarrow-9x=-8\)
\(\Rightarrow x=\dfrac{8}{9}\)
Vậy: \(x=\dfrac{8}{9}\)
c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)
\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)
\(\Rightarrow x=\dfrac{24}{5}\)
Vậy: \(x=\dfrac{24}{5}\)
d) \(\dfrac{3x-1}{-5}=\dfrac{-5}{3x-1}\)
\(\Rightarrow\left(3x-1\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)
22 tháng 3 2021
a) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
Suy ra: x+2=0
hay x=-2(thỏa ĐK)
Vậy: S={-2}
22 tháng 3 2021
d)
ĐKXĐ: \(x\notin\left\{1;3\right\}\)
Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)
Suy ra: \(x^2-3x+5x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+9=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3(loại)
Vậy: \(S=\varnothing\)
22 tháng 3 2021
a) Ta có: \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{6\left(x-2\right)}\)
Suy ra: \(2x+5-3x+6=6x-9\)
\(\Leftrightarrow-x+11-6x+9=0\)
\(\Leftrightarrow20-7x=0\)
\(\Leftrightarrow7x=20\)
hay \(x=\dfrac{20}{7}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{20}{7}\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
a)
\(\dfrac{1}{3}+\dfrac{2}{5}\left(x+1\right)=1\\ \Leftrightarrow\dfrac{2}{5}\left(x+1\right)=\dfrac{2}{3}\\ \Leftrightarrow x+1=\dfrac{5}{3}\\ \Rightarrow x=\dfrac{2}{3}\)
a) `1/3 + 2/5 (x+1)=1`
`2/5 (x+1) = 2/3`
`x+1=5/3`
`x=2/3`
b) `1/4 + 1/3 : 3x =-5`
`1/3 : 3x = -21/4`
`3x = -4/63`
`x=-4/189`