Bài 2: Tìm x, biết:
a) 2x + 12 = 434
b) 25 + 27(x – 8) = 106
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a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.
b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.
Lời giải:
a.
$|4x-1|-|3x-\frac{1}{2}|=0$
$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)
b. Nếu $x\geq 1$ thì:
$|x-1|-2x=\frac{1}{2}$
$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$
$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)
Nếu $x< 1$ thì:
$1-x-2x=\frac{1}{2}$
$\Leftrightarrow x=\frac{1}{6}$ (tm)
\(2x-5+x=2^0\)
\(2x+x-5=2^0\)
\(3x-5=2^0=1\)
\(3x=1+5=6\)
\(x=6:3\)
\(x=2\)
Vậy \(x=3\)
2x - 5 + x = 2^0
2x - 5 + x = 1
2x + x - 5 = 1
3x - 5 =1
3x = 6
x = 2
\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)
\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)
\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)
\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)
\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)
\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)
\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)
\(a,\left|2x-5\right|=1\)
\(\Rightarrow\orbr{\begin{cases}2x-5=1\\2x-5=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\2x=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
b, đề thiếu
a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Tham khảo:Tìm x thuộc N , biết:a) 2x + 2x+3 =144b) (4x -1)2 =25 x 9 - Hoc24
a, 2x + 12 = 434
2x = 434 - 12 = 422
x = 422 : 2 = 211
b. 25 + 27 ( x - 8 ) = 106
27 ( x - 8 ) = 106 - 25 = 81
x - 8 = 81 : 27 = 3
x = 3 + 8 = 11