Tìm x biết |x-1|+|x-3|+|x-5|+|x-7|=8
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a) 3/8 . x = 9/8 - 1
3/8 . x = 1/8
x = 1/8 : 3/8
x = 1/3
b) 4/5 . x = 7/5 - 1/5
4/5 . x = 6/5
x = 6/5 : 4/5
x = 3/2
c) 12/7 : x + 2/3 = 7/5
12/7 : x = 7/5 - 2/3
12/7 : x = 11/15
x = 12/7 : 11/15
x = 180/77
d) 3.(x + 7) - 15 = 27
3.(x + 7) = 27 + 15
3.(x + 7) = 42
x + 7 = 42 : 3
x + 7 = 14
x = 14 - 7
x = 7
a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)
\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)
b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)
c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)
d) \(3\left(x+7\right)-15=27\)
\(\Leftrightarrow3\left(x+7\right)=42\)
\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)
mình đang cần gấp câu trả lời ,bạn nào giải được nhanh k luôn ,hứa đấy
Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
Ta có: |x-1|+|x-3|+|x-5|+|x-7| = (|x-1|+|7-x|)+(|x-3|+|5-x|) \(\ge\) |x-1+7-x| + |x-3+5-x| = 6+2 = 8 (1)
Mà |x-1|+|x-3|+|x-5|+|x-7|=8 suy ra (1) xảy ra dấu "=" khi:
\(\hept{\begin{cases}\left(x-1\right)\left(7-x\right)\ge0\\\left(x-3\right)\left(5-x\right)\ge0\end{cases}\Rightarrow\hept{\begin{cases}1\le x\le7\\3\le x\le5\end{cases}\Rightarrow}3\le x\le5}\)
Do x nguyên nên \(x\in\left\{3;4;5\right\}\)
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
Xét 5 trường hợp
- TH1: x < 1 thì |x - 1| = 1 - x; |x - 3| = 3 - x; |x - 5| = 5 - x; |x - 7| = 7 - x
Ta có: (1 - x) + (3 - x) + (5 - x) + (7 - x) = 8
=> 16 - 4x = 8
=> 4x = 16 - 8 = 8
=> x = 8 : 4 = 2, không thỏa mãn x < 1
- TH2: \(1\le x< 3\) thì |x - 1| = x - 1; |x - 3| = 3 - x; |x - 5| = 5 - x; |x - 7| = 7 - x
Ta có: (x - 1) + (3 - x) + (5 - x) + (7 - x) = 8
=> 14 - 2x = 8
=> 2x = 14 - 8 = 6
=> x = 6 : 2 = 3, không thỏa mãn \(1\le x< 3\)
- TH3: \(3\le x< 5\) thì |x - 1| = x - 1; |x - 3| = x - 3; |x - 5| = 5 - x; |x - 7| = 7 - x
Ta có: (x - 1) + (x - 3) + (5 - x) + (7 - x) = 8
=> 8 = 8, luôn đúng
- TH4: \(5\le x< 7\) thì |x - 1| = x - 1; |x - 3| = x - 3; |x - 5| = x - 5; |x - 7| = 7 - x
Ta có: (x - 1) + (x - 3) + (x - 5) + (7 - x) = 8
=> 2x - 2 = 8
=> 2x = 8 + 2 = 10
=> x = 10 : 2 = 5, thỏa mãn \(5\le x< 7\)
- TH5: \(7\le x\) thì |x - 1| = x - 1; |x - 3| = x - 3; |x - 5| = x - 5; |x - 7| = x - 7
Ta có: (x - 1) + (x - 3) + (x - 5) + (x - 7) = 8
=> 4x - 16 = 8
=> 4x = 8 + 16 = 24
=> x = 24 : 4 = 6, không thỏa mãn \(7\le x\)
Vậy \(3\le x\le5\) thỏa mãn đề bài