Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=-5-\frac{1}{4}\)
\(\frac{1}{3}:2x=-\frac{21}{3}\)
\(2x=\frac{1}{3}:\left(\frac{-21}{3}\right)\)
\(2x=-\frac{1}{21}\)
\(x=\frac{-1}{42}\)
b)\(\left(3x-\frac{1}{4}\right).\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{12}\\x=-\frac{1}{2}\end{array}\right.\)
c)\(\left(2x-5\right).\left(\frac{3}{2}x+9\right).\left(0,3x-12\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-5=0\\\frac{3}{2}x+9=0\\0,3x-12=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=5\\\frac{3}{2}x=-9\\0,3x=12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-6\\x=40\end{array}\right.\)
a) 1/4 + 1/3 : 2x = -5
=> 1/3 : 2x = -5 - 1/4
=> 1/3 : 2x = -21/4
=> 2x = 1/3 : (-21/4) = -4/63
=> x = -4/63 : 2 = -2/63
a, 11/12 - ( 2/5 + x ) = 2/3
<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)
=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)
b, 2x . ( x - 1/7 ) = 0
<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
vậy x={\(0;\frac{1}{7}\)}
c, 3/4 + 1/4 : x = 2/5
<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)
vậy x=-5/7
a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)
\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)
\(\Leftrightarrow x=-\frac{3}{20}\)
b) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
\(\Leftrightarrow4x=\frac{-20}{7}\)
\(\Leftrightarrow x=-\frac{5}{7}\)
\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=-5-\frac{1}{4}\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{1}{3}:\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{-4}{63}:2\)
\(x=\frac{-2}{63}\)
\(\)
\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\Rightarrow\frac{1}{3}:2x=-\frac{21}{4}\)
\(\Rightarrow2x=\frac{-4}{63}\)
\(\Rightarrow x=\frac{-2}{63}\)
\(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}}\)
\(\left(2x-5\right)\left(\frac{3}{2}x+9\right)\left(0,3x-12\right)=0\)
Th1 : \(2x-5=0\Rightarrow x=\frac{5}{2}\)
Th2 : \(\frac{3}{2}x+9=0\Rightarrow x=-6\)
Th3 : \(0,3x-12=0\Rightarrow x=\frac{12}{0,3}\)
a) -4/5 + 5/2x = -3/10
5/2x = -3/10 + 4/5
5/2x = 1/5
5/2x = 1/2
x = 1/2 : 5/2
x = 1/5
b) 4/3 + 5/8 : x = 1/12
5/8x = 1/12 - 4/3
5/8x = -5/4
5 = -5/4.8x
5 = -10x
5/-10 = x
-1/2 = x
x = -1/2
c) (x - 1/3)(x - 2/5) = 0
x - 1/3 = 0 hoặc x - 2/5 = 0
x = 0 + 1/3 x = 0 + 2/5
x = 1/3 x = 2/5
\(a,\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=-\dfrac{3}{20}\\ b,\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\\ c,\Rightarrow\dfrac{1}{4}:x=-\dfrac{2}{5}-\dfrac{3}{4}=-\dfrac{23}{20}\\ \Rightarrow x=\dfrac{1}{4}:\left(-\dfrac{23}{20}\right)=-\dfrac{5}{23}\)
a)
\(x-\dfrac{2}{-4}=\dfrac{-9}{5}-x\\ x+x=\dfrac{-9}{5}+\dfrac{2}{-4}\\ 2x=\dfrac{-9}{5}-\dfrac{1}{2}\\ 2x=\dfrac{-23}{10}\\ x=\dfrac{-23}{20}\)
b)
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\\ \dfrac{x-5}{3}=\dfrac{12}{x-5}\\ \left(x-5\right)^2=3\cdot12\\ \left(x-5\right)^2=36\\ \left(x-5\right)^2=6^2\)
TH1: x - 5 = 6 => x = 6 + 5 = 11
TH2: x - 5 = -6 => x = -6 + 5 = -1
c)
\(\left(2x-1\right)^2-\left(2x-1\right)=0\\ \Rightarrow\left(2x-1\right)\left(2x-1-1\right)=0\\ \Rightarrow\left(2x-1\right)\left(2x-2\right)=0\)
TH1: 2x - 1=0 => 2x = 1 => x =1/2
TH2: 2x - 2=0 => 2x = 2 => x = 1
a:
ĐKXĐ: x<>5
\(\dfrac{x-2}{-4}=\dfrac{-9}{5-x}\)
=>\(\dfrac{\left(x-2\right)}{-4}=\dfrac{9}{x-5}\)
=>\(\left(x-2\right)\left(x-5\right)=-4\cdot9=-36\)
=>\(x^2-7x+10+36=0\)
=>\(x^2-7x+46=0\)
\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot46=49-184=-135< 0\)
=>Phương trình vô nghiệm
b: ĐKXĐ: x<>5
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\)
=>\(\dfrac{x-5}{3}=\dfrac{12}{x-5}\)
=>\(\left(x-5\right)\left(x-5\right)=3\cdot12=36\)
=>\(\left(x-5\right)^2=36\)
=>\(\left[{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
c: \(\left(2x-1\right)^2-\left(2x-1\right)=0\)
=>(2x-1)(2x-1-1)=0
=>(2x-1)(2x-2)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)