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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/ \(\left|x-3\right|=\left|4-x\right|\) \(\Rightarrow\left[\begin{matrix}x-3=4-x\\x-3=x-4\end{matrix}\right.\) => x = \(\frac{7}{2}\)
b/ \(\left|2x-1\right|+5=5\Rightarrow\left|2x-1\right|=0\Rightarrow x=\frac{1}{2}\)
c/ \(\left|x-1\right|.\left(-\frac{1}{2}\right)^3=\left(-\frac{1}{2}\right)^5\)
\(\Rightarrow\left|x-1\right|=\left(-\frac{1}{2}\right)^5:\left(-\frac{1}{2}\right)^3\)
\(\Rightarrow\left|x-1\right|=\frac{1}{4}\)
\(\Rightarrow x-1=\pm\frac{1}{4}\)
\(\Rightarrow\left[\begin{matrix}x=\frac{5}{4}\\x=\frac{3}{4}\end{matrix}\right.\)
d/ \(3x-\left|2x+1\right|=2\)
\(\Rightarrow\left[\begin{matrix}3x-\left(2x+1\right)=2\\3x-\left(-2x-1\right)=2\end{matrix}\right.\)
\(\Rightarrow x=3\) (loại x = 1/5)
e/ \(\left|x\right|+\left|-x\right|=3-x\)
Xét với x = 0 không thỏa mãn
Xét với x < 0 thì : (-x) + (-x) = 3 - x => x = -3
Xét với x > 0 thì : x + x = 3 - x => x = 1
Vậy x = 1 hoặc x = -3
a) \(\left(2x-3\right)-\left(x-5\right)=\left(x+2\right)-\left(x+1\right)\)
\(\Rightarrow2x-3-x+5=x+2-x-1\)
\(\Rightarrow2x-x-x+x=2-1-5+3\)
\(\Rightarrow x=-1\)
b) \(2x-3-x+5=x+2-x-1\)
\(\Rightarrow2x-x-x+x=2-1-5+3\)
\(\Rightarrow x=-1\)
b, 2x-3 - x+5 =x+2 -x-1 (2x − 3) − (x − 5) = (x + 2) − (x + 1) ⇒ 2x − 3 − x + 5 = x + 2 − x − 1 ⇒ 2x − x − x + x = 2 − 1 − 5 + 3 ⇒ x = −1 2x − 3 − x + 5 = x + 2 − x − 1 ⇒ 2x − x − x + x = 2 − 1 − 5 + 3