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\(\dfrac{2}{3}-\left|x-2,4\right|=\dfrac{1}{2}\)
\(\left|x-2,4\right|=\dfrac{2}{3}-\dfrac{1}{2}\)
\(\left|x-2,4\right|=\dfrac{1}{6}\)
*) Với \(x\ge2,4\) ta có:
\(x-2,4=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+2,4\)
\(x=\dfrac{77}{30}\) (nhận)
*) Với \(x< 2,4\) ta có:
\(x-2,4=-\dfrac{1}{6}\)
\(x=-\dfrac{1}{6}+2,4\)
\(x=\dfrac{67}{30}\) (nhận)
Vậy \(x=\dfrac{67}{30};x=\dfrac{77}{30}\)
a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{17}{35}\) + 0,4
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{3}{35}\)
\(x\) = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)
\(x\) = - \(\dfrac{1}{5}\)
b, 0,2.(\(x\) - 3) +2,4 = 10
0,2.(\(x\) - 3) = 10 - 2,4
0,2.(\(x\) - 3) = 7,6
\(x\) - 3 = 7,6:0,2
\(x\) - 3 = 38
\(x\) = 38 + 3
\(x\) = 41
a: Sửa đề: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)
=>\(4⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7\right\}\)
=>\(x\in\left\{16;4;25;1;49\right\}\)
b:
a) \(\left|x-1\right|=5\)
\(\Rightarrow\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
b) \(\left|x-0,5\right|=2,4\)
\(\Rightarrow\orbr{\begin{cases}x-0,5=2,4\\x-0,5=-2,4\end{cases}}\Rightarrow\orbr{\begin{cases}x=2,9\\x=-1,9\end{cases}}\)
c) \(2\left|3x-1\right|-1=6\)
\(\Leftrightarrow2\left|3x-1\right|=7\)
\(\Leftrightarrow\left|3x-1\right|=3,5\)
\(\Rightarrow\orbr{\begin{cases}3x-1=3,5\\3x-1=-3,5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=4,5\\2x=-2,5\end{cases}\Rightarrow\orbr{\begin{cases}x=1,5\\-1,25\end{cases}}}\)
a) |x-1|=5
1) x-1 = 5
x= 6
2) x-1 = -5
x= -4
Vậy x= 5 hoặc x= -4
b) tương tự câu a
c) ) 2|3x-1|-1=6
2|3x-1| = 7
|3x-1| = 7/2
1) 3x - 1 = 7/2
3x = 9/2
x= 9/2.1/3
x= 3/2
2) 3x - 1 = -7/2
3x = -5/2
x= -5/2 . 1/3
x= -5/6
Vậy x = 3/2 hoặc x= -5/6
a)vì |x-1|=5 nên x-1=5 hoặc x-1=-5
Th1:x-1=5 suy ra x=5+1=6
Th2:x-1=-5 suy ra x=-5+1=-4
Vậy x=6 hoặc x=-4
Phần b+c tương tự
3/4 - (x - 2/3) = 1 1/3
3/4 - x + 2/3 = 4/3
-x = 4/3 - 3/4 - 2/3
-x = -1/12
x = 1/12
3/4 - (x - 2/3) = 1 1/3
3/4 - x + 2/3 = 4/3
-x = 4/3 - 3/4 - 2/3
-x = -1/12
x = 1/12
2/3 - |x - 1/2| = 2/3
|x - 1/2| = 2/3 - 2/3
|x - 1/2| = 0
x - 1/2 = 0
x = 0 + 1/2
x = 1/2
a, \(\dfrac{5}{2}\)\(x\) - \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)
\(\dfrac{5}{2}\)\(x\) = \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)
\(\dfrac{5}{2}\)\(x\) = 1
\(x\) = 1: \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{2}{5}\)
b, \(\dfrac{x+4}{20}\) = \(\dfrac{5}{x+4}\) (đk \(x\) ≠ -4)
(\(x\)+4).(\(x\) + 4) = 20.5
(\(x\)+ 4)2 = 100
(\(x\) + 4)2 = 102
\(\left[{}\begin{matrix}x+4=-10\\x+4=10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-10-4\\x=10-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-14\\x=6\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-14; 6}
| x - 2,4| = \(\dfrac{1}{2}\)
\(\left[{}\begin{matrix}x-2,4=\dfrac{1}{2}(đk:x>2,4)\\x-2,4=-\dfrac{1}{2}(đk:x< 2,4)\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{2}+2,4\\x=-\dfrac{1}{2}+2,4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2,9(tm)\\x=1,9(tm)\end{matrix}\right.\)
vậy \(x\in\) { 1,9 ; 2,9}
x=2/9