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a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a: \(\dfrac{x}{0.9}=\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
b: \(\dfrac{-6}{x}=\dfrac{9}{-15}\)
\(\Leftrightarrow x=10\)
c: \(\dfrac{\dfrac{14}{15}}{\dfrac{9}{10}}=\dfrac{x}{\dfrac{3}{7}}\)
\(\Leftrightarrow x=\dfrac{3}{7}\cdot\dfrac{14}{15}:\dfrac{9}{10}=\dfrac{2}{5}\cdot\dfrac{10}{9}=\dfrac{20}{45}=\dfrac{4}{9}\)
4, Q = |x+\(\frac{1}{5}\) | -x +\(\frac{4}{7}\)
xét x \(\ge\) \(-\frac{1}{5}\)
Ta Có Q = |x+\(\frac{1}{5}\) | -x + \(\frac{4}{7}\) = x+\(\frac{1}{5}\) - x +\(\frac{4}{7}\) = \(\frac{27}{35}\) (1)
xét x \(< -\frac{1}{5}\)
Ta có Q = | x +\(\frac{1}{5}\) | - x + \(\frac{4}{7}\) = -x - \(\frac{1}{5}\) - x + \(\frac{4}{7}\) = -2x + \(\frac{13}{35}\)
với x \(< -\frac{1}{5}\)
=> -2x \(>\) \(\frac{2}{5}\)
=> -2x + \(\frac{13}{35}\) \(>\frac{27}{35}\) (2)
Từ (1) và (2) => MinQ = \(\frac{27}{35}\) khi \(x\ge-\frac{1}{5}\)
5 , D = |x| + |8-x|
D = |x| + |8-x| \(\ge\) |x+8-x| = |8| = 8
Dấu ''='' xảy ra khi x(8-x) \(\ge\) 0 <=> 0\(\le\)x\(\le\) 8
Vậy MinD = 8 khi \(0\le x\le8\)
6,L= |x - 2012| + |2011 - x|
L = |x-2012| + |2011-x| \(\ge\) | x-2012 + 2011 - x | = |-1| = 1
Dấu ''= '' xảy ra khi ( x-2012)(2011-x) \(\ge\) 0
làm nốt câu 6 nãy ấn nhầm
<=> 2011\(\le\) x \(\le\) 2012
Vậy MinL = 1 khi \(2011\le x\le2012\)
7 , E = | x- \(\frac{2006}{2007}\) | + |x-1|
Ta có :
E = |x-\(\frac{2006}{2007}\) | + |1-x|
E = | x - \(\frac{2006}{2007}\) | + |1-x| \(\ge\) | x - \(\frac{2006}{2007}\) + 1 - x | = \(\frac{1}{2007}\)
Dấu ''='' xảy ra khi (x- \(\frac{2006}{2007}\) ) ( 1-x ) \(\ge0\) <=> \(\frac{2006}{2007}\le x\le1\)
Vậy MinE = \(\frac{1}{2007}\) khi \(\frac{2006}{2007}\le x\le1\)
8 ,F = | x -\(\frac{1}{4}\) | + | \(x-\frac{3}{4}\) |
Ta có :
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) - x |
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) -x | \(\ge\) | x - \(\frac{1}{4}\) + \(\frac{3}{4}\) -x | = \(\frac{1}{2}\)
Dấu ''='' xảy ra khi ( x-\(\frac{1}{4}\) ) ( \(\frac{3}{4}-x\) ) \(\ge\) 0 <=> \(\frac{1}{4}\le x\le\frac{3}{4}\)
Vậy MinF = \(\frac{1}{2}\) khi \(\frac{1}{4}\le x\le\frac{3}{4}\)
a, x + 1/9 - 3/5 = 3/6
x + 1/9 = 3/6 - 3/5
x + 1/9 = -1/10
x = -1/10 - 1/9
x = -19/90
19/90 nha