f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)( ĐKXĐ : \(x\ne-\frac{1}{2}\))
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x^2=\left(\pm4\right)^2\)
\(\Leftrightarrow x=\pm4\)(tmđk)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)( ĐKXĐ : \(x\ne-1\))
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=6\cdot5\)
\(\Leftrightarrow10x^2+15x+5=30\)
\(\Leftrightarrow10x^2+15x+5-30=0\)
\(\Leftrightarrow10x^2+15x-25=0\)
\(\Leftrightarrow5\left(2x^2+3x-5\right)=0\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)(tmđk)
f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21.3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)\(\Leftrightarrow x^2=4^2\)\(\Leftrightarrow x=4\)
Vậy \(x=4\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=5.6\)
\(\Leftrightarrow5\left(2x+1\right)\left(x+1\right)=30\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=6\)
\(\Leftrightarrow2x^2+3x+1=6\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{-5}{2};1\right\}\)
\(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63 (1)
Đặt 2x = t
Khi đó (1) <=> (t - 1)(t + 1) = 63
=> t2 + t - t - 1 = 63
=> t2 - 1 = 63
=> t2 = 64
=> t = \(\pm\)8
Khi t = 8
=> 2x = 8
=> x = 4
Khi t = -8
=> 2x = -8
=> x = -4
Vậy \(x\in\left\{4;-4\right\}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)
=> (10x + 5)(x + 1) = 6.5
=> 5(2x + 1)(x + 1) = 30
=> (2x + 1)(x + 1) = 6
=> 2x2 + 2x + x + 1 = 6
=> 2x2 + 3x + 1 = 6
=>2x2 + 3x - 5 = 0
=> 2x2 - 2x + 5x - 5 = 0
=> 2x(x - 1) + 5(x - 1) = 0
=> (2x + 5)(x - 1) = 0
=> \(\orbr{\begin{cases}2x+5=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2,5\\x=1\end{cases}}\)
Vậy \(x\in\left\{-2,5;1\right\}\)
f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\) \(\left(x\ne-\frac{1}{2}\right)\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)
\(\Rightarrow x=4\)
g) giống f
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\) \(\left(x\ne-1\right)\)
\(\Leftrightarrow10x^2+15x+5=30\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
