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a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)
b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))
\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)
\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)
d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)
\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)
\(\Leftrightarrow2x^2+2x=2x^2+1\)
\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).
(5x - 1)(2x - 1/3) = 0
<=> 5x - 1 = 0 hoặc 2x - 1/3 = 0
=> x = 1/5 hoặc x = 1/6
vậy x= 1/5 hoặc x= 1/6
a) (5x - 1) . ( 2x - 1/3 ) = 0
=> 5x - 1 = 0
2x - 1/3 = 0
=> 5x = 1
2x = 1/3
=> x = 1/5
x = 1/6
a) \(\left(5x-1\right)\left(\frac{2x-1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{2}\end{cases}}\)
b) \(6\left(x-1\right)+2x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\6+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
a) \(\left(5x-1\right)\cdot\frac{2x-1}{3}=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\\frac{2x-1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=1\\2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(x=\frac{1}{5};x=\frac{1}{2}\)
b) 6(x-1)+2x(x-1)=0
<=> (x-1)(6+2x)=0
<=> \(\orbr{\begin{cases}x-1=0\\6+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy x=1; x=-3
Trả lời :
*Nếu \(x\ge1\)
\(x+6+2x-1=3\) ( vô nghiệm )
* \(x\le6\)
\(6-x+2x-1=3\) ( vô nghiệm )
* \(6< x< 2x-1\)
\(x-6-2x-1=3\)
\(x-6+2x+1=3\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=\frac{12}{13}\) ( loại )
Lập bảng xét dấu :
Nếu x < - 6
=> |x + 6| = -(x + 6) = -x - 6
|2x - 1| = -(2x - 1) = -2x + 1
Khi đó : |x + 6| + |2x - 1| = 3 (1)
<=> - x - 6 - 2x + 1 = 3
=> -3x = 8
=> x = -8/3 (loại)
Nếu \(-6\le x\le0,5\)
=> |x + 6| = x + 6
|2x - 1| = -(2x - 1) = -2x + 1
Khi đó (1) <=> x + 6 - 2x + 1 = 3
=> x = 4 (loại)
Nếu x > 0,5
=> |x + 6| = x + 6
|2x - 1| = 2x - 1
Khi đó (1) <=> x + 6 + 2x - 1 = 3
=> 3x = -2
=> x = - 2/3 (loại)
Vậy \(x\in\varnothing\)