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Bài 1:
a)Ta thấy:\(-\left|x+\frac{4}{7}\right|\le0\)
\(\Rightarrow-\left|x+\frac{4}{7}\right|+\frac{12}{19}\le0+\frac{12}{19}=\frac{12}{19}\)
\(\Rightarrow A\le\frac{12}{19}\)
Dấu "=" xảy ra <=>x=-4/7
Vậy...
b)Ta thấy:\(-\left|x-5,3\right|\le0\)
\(\Rightarrow19,18-\left|x-5,3\right|\le19,18-0=19,18\)
\(\Rightarrow B\le19,18\)
Dấu "=" xảy ra <=>x=5,3
Vậy...
Bài 2:
a)Áp dụng BĐT |a|+|b|>=|a+b| ta có:
\(\left|x-20\right|+\left|x-2016\right|\ge\left|x-20+2016-x\right|=1996\)
\(\Rightarrow A\ge1996\)
Dấu "=" xảy ra <=>x=20 hoặc 2016
b)bạn xét từng trường hợp rồi tìm ra Min xét dấu "=" là ok
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a) \(\frac{x}{4}=\frac{9}{12}\)
\(\Rightarrow x=\frac{9}{12}\cdot4\)
\(\Rightarrow x=3\)
b) \(-\frac{8}{3}:x=\frac{5}{6}\)
\(\Rightarrow x=-\frac{8}{3}:\frac{5}{6}\)
\(\Rightarrow x=-\frac{16}{5}=-3,2\)
c) \(\left(\frac{3}{4}-x\right)\cdot\frac{10}{3}=\frac{2}{5}\)
\(\Rightarrow\frac{3}{4}-x=\frac{2}{5}:\frac{10}{3}\)
\(\Rightarrow x=\frac{3}{4}-\frac{3}{25}=\frac{63}{100}=0,63\)
Nhớ k mk đó nkoa!@^_^@! Love you
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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
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a, Xét : x-4 = 0 => x= 4
2x+1 = 0 => x= \(\frac{1}{2}\)
x+3 = 0 => x = -3
x + 9 = 0 => x = -9
Khi đó ta có bảng xét dấu :
x | -9 | -3 | \(\frac{1}{2}\) | 4 |
x-4 | -13 | -7 | \(\frac{-7}{2}\) | 0 |
2x+1 | -17 | -5 | 2 | 9 |
x+3 | -6 | 0 | \(\frac{7}{2}\) | 7 |
x+9 | 0 | 6 | \(\frac{19}{2}\) | 13 |
=> có 5 trường hợp:
TH1 : \(x\le-9\)
TH2 : \(-9\le x< -3\)
TH3 : \(-3\le x< \frac{1}{2}\)
TH4 : \(\frac{1}{2}\le x< 4\)
Do đó :
TH1 : \(x\le-9\)
Ta có : /x-4/ = -(x-4) = 4 - x
/2x+1/ = -(2x+1) = -2x -1
/x+3/ = -(x + 3 ) = -x - 3
/x-9/ = -(x-9) = -x + 9 Thay vào đề bài ta có:
3.(4-x) + 2x-1 +5(-x - 3) -x-9 = 5
=> 12 - 3x + 2x - 1 + -5x - 15 - x - 9 = 5
=>(12 - 1 - 15 -9 ) +(-3x +2x -5x -x) = 5
=> -13 - 7x = 5
7x = -13 - 5
7x = -18
x = \(\frac{-18}{7}\)( Ko TM)
Tương tự với 4 trường hợp còn lại.
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\(\text{a, 3(x+1)+4x=10}\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow7x+3=10\)
\(\Rightarrow7x=10-3=7\)
\(\Rightarrow x=1\)
c, x+1/10+x+2/9=x+3/8+x+4/7
=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)
=> x+11/10 + x+11/9 = x+11/8 + x+11/7
...............
a) \(3\left(x+1\right)+4x=10\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow3x+4x=10-3\)
\(\Rightarrow7x=7\)
\(\Rightarrow x=7\)
a)
\(\left|x-5\right|=4\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-5=4\\x-5=-4\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=1\end{array}\right.\)
Vậy x = 9 ; x = 1
b)
\(\left|x+9\right|=12\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+9=12\\x+9=-12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-21\end{array}\right.\)
Vậy x = 3 ; x = - 21
\(\left|x-5\right|=4\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-5=4\\x-5=-4\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=4+5\\x=-4+5\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=1\end{array}\right.\)
Vậy: \(x\in\left\{1;9\right\}\)
\(\left|x+9\right|=12\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+9=12\\x+9=-12\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=12-9\\x=-12-9\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-21\end{array}\right.\)
Vậy: \(x\in\left\{-21;3\right\}\)