Tìm x biết: |x-30|+|x-4|+|x-1975|+|x-44|+|x-2019|=3960
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Ta có: \(\left|x-1\right|+\left|x-2020\right|=\left|x-1\right|+\left|2020-x\right|\ge\left|x-1+2020-x\right|=2019\)
Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(2020-x\right)\ge0\)\(\Leftrightarrow1\le x\le2020\)
Vì \(\hept{\begin{cases}\left|x-30\right|\ge0\\\left|y-4\right|\ge0\\\left|z-1975\right|\ge0\end{cases}}\forall x,y,z\)\(\Rightarrow\left|x-1\right|+\left|x-30\right|+\left|y-4\right|+\left|z-1975\right|+\left|x-2020\right|\ge2019\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-30=0\\y-4=0\\z-1975=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=30\\y=4\\z=1975\end{cases}}\)
So sánh \(x=30\)với điều kiện \(1\le x\le2020\)ta được x thoả mãn
Vậy \(x=30\); \(y=4\); \(z=1975\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2^x+2^x+1+2^x+2+......+2^x+2015=2^2019-8
=>2^x(1+2+2^2+2^3+...+2^2015)=2^2019-2^3
=>2^x(2^2016-1)=2^3.(2^2016-1)
=>x=3
nhớ link nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
x + 30%x = -1,3
x + \(\frac{30}{100}\)x = -1,3
x + 0,3x = -1,3
x.( 1 + 0,3 ) =-1,3
x.1,3 = -1,3
x = -1,3 : 1,3
x = -1
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Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|x-2007\right|=\left|x-1\right|+\left|2007-x\right|\ge\left|x-1+2007-x\right|=2006\)Dấu "=" xảy ra khi \(\left(x-1\right)\left(2007-x\right)\ge0\Rightarrow1\le x\le2007\)
Lại có
\(\left\{\begin{matrix}\left|x-30\right|\ge0\\\left|y-4\right|\ge0\\\left|z-1975\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-1\right|+\left|x-30\right|+\left|y-4\right|+\left|z-1975\right|+\left|x-2007\right|\ge2006\)Dấu "="xảy ra khi \(\left\{\begin{matrix}\left|x-30\right|=0\\\left|y-4\right|=0\\\left|z-1975\right|=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left\{\begin{matrix}x-30=0\\y-4=0\\z-1975=0\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=30\\y=4\\z=1975\end{matrix}\right.\)
Vậy x=30,y=4,z=1975
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\\ 5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\\ \Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a-7b+5c}{63-98+50}=\dfrac{-30}{15}=-2\\ \Rightarrow\left\{{}\begin{matrix}a=-42\\b=-28\\c=-20\end{matrix}\right.\)
\(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow x=3k;y=4k;z=5k\)
\(2x^2+2y^2-3z^2=-100\\ \Rightarrow18k^2+32k^2-75k^2=-100\\ \Rightarrow-25k^2=-100\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=8;z=10\\x=-6;y=-8;z=-10\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
| x + 2 | = | 2 - 3x |
xét 2 trường hợp :
+) TH1 :
2 - 3x = x + 2
-3x + x = 2 + 2
2x = 4
x = 4 : 2 = 2
+) TH2 :
2 - 3x = - ( x + 2 )
2 - 3x = -x - 2
-3x - x = 2 - 2
-4x = 0
x = 0 : ( -4 )
x = 0
bài còn lại tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
30 : 5 / 35 : 5 = 40/x
5/6 = 40/x
5 x 8 / 6x 8 = 40/x
40/42 = 42/x
=> x = 42
Nhận thấy:\(2019+1975-30-4=3960\)
Qua đó,ta biến đổi như sau.
Do \(\hept{\begin{cases}\left|x-2019\right|=\left|2019-x\right|\\\left|x-1975\right|=\left|1975-x\right|\end{cases}}\Rightarrow\hept{\begin{cases}\left|2019-x\right|+\left|1975-x\right|\ge\left|2019-x+1975-x\right|=\left|3994-2x\right|\\\left|x-30\right|+\left|x-4\right|\ge\left|x-30+x-4\right|=\left|2x-34\right|\end{cases}}\)
Dấu "=" xảy ra lần lượt là:\(\hept{\begin{cases}x< 1975;x< 2019\\x>30;x>4\end{cases}}\)
\(\Rightarrow\left|2019-x\right|+\left|1975-x\right|+\left|x-30\right|+\left|x-4\right|\ge\left|3994-2x+2x-34\right|=3960\)
Dấu "=" xảy ra khi và chỉ khi:\(30< x< 1975\)
\(\Rightarrow\left|2019-x\right|+\left|1975-x\right|+\left|x-30\right|+\left|x-4\right|+\left|x-44\right|\ge3960\)
Dấu "=" xảy ra khi và chỉ khi:\(\left|x-44\right|=0\Leftrightarrow x=44\)
Thử vào thấy thỏa mãn.
Vậy \(x=44\)