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a) \(\frac{x}{27}=-\frac{2}{3}.6\)
\(\frac{x}{27}=-4\)
\(x:27=-4\)
\(x=-4.27\)
\(x=-108\)
Rồi tương tự nha
\(-0.5x:x=-9.36:16.38\)
\(\Leftrightarrow-0.5x:x=-\frac{4}{7}\)
\(\Leftrightarrow-\frac{1}{2}x:x=-\frac{4}{7}\Leftrightarrow\left(-\frac{1}{2}:1\right)x=-\frac{4}{7}\)
\(\Leftrightarrow-\frac{1}{2}x=-\frac{4}{7}\Leftrightarrow x=\left(-\frac{4}{7}\right):\left(-\frac{1}{2}\right)\)
\(x=1\frac{1}{7}\Leftrightarrow x=\frac{8}{7}\)
xin lỗi tai mik lười nên mik viết nha pạn ...
- bạn tính phép tính về bên phải trước
- rồi lấy -0,5 chia cho kết qur vừa tìm là ra x nha
~ hok tốt ~
\(C=\dfrac{4^9.36+64^4}{16^4.100}=\dfrac{4^9.36+4^{12}}{4^9.25}=\dfrac{4^9\left(36+4^3\right)}{4^9.25}=\dfrac{100}{25}=4\)
a, \(B=\dfrac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(=\dfrac{2^{10}.\left(13+65\right)}{2^8.2^3.13}\)
\(=\dfrac{2^{10}.78}{2^{11}.13}\)\(=\dfrac{1.6}{2.1}=\dfrac{1.3}{1.1}=3\)
b: \(=\dfrac{2^{20}\cdot3^2+2^{54}}{2^{18}\cdot5^2}=\dfrac{2^{20}\left(3^2+2^{32}\right)}{2^{18}\cdot5^2}=\dfrac{2^2\left(3^2+2^{32}\right)}{25}\)
c: \(=\dfrac{2^9\cdot3^6\cdot3^6\cdot2^2}{2^8\cdot3^{12}}=\dfrac{2^{11}}{2^8}=8\)
d: \(=\dfrac{2^{12}\cdot3^4\cdot3^{10}}{2^{12}\cdot3^{12}}=9\)
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 \(\frac{1}{4}\le x\le\frac{3}{4}\)
X x X x X x X x X = X5 = 210.105 = 405
=> X = 40
=> Y = 40 - 30 = 10
=> Z = 20 : 10 = 2
Cách 1 : -0,52 : x = -9,36 : 16,83
-0,52 : x = -\(\frac{104}{187}\)
x = -0,25 : -\(\frac{104}{187}\)
x = \(\frac{187}{416}\)
Cách 2 : -0,52 : x = -9,36 : 16,83
=> \(\frac{-0.25}{x}\) = \(\frac{-9.36}{16.83}\)
=>x. (-9,36) = -0,25. 16,83
x.(-9,36) = -4,2075
x = -4,2075 : (-9,36)
x = \(\frac{187}{416}\)
\(\Rightarrow x=\frac{-0,52.16,83}{-9,86}=\frac{187}{200}\)
#Walker