TH1: x<2020
Q=2020-x+2021-x+2022-x=6063-3x
TH2: 2020<=x<2021
Q=x-2020+2021-x+2022-x=2023-x
TH3: 2021<=x<2022
Q=x-2020+x-2021+2022-x=x-2019
TH4: x>=2022
Q=x-2020+x-2021+x-2022=3x-6063
tìm x,y,z thuộc z biết ;
\(\left|x-4\right|+\left|x-10\right|+\left|x-2020\right|+\left|y-2015\right|+\left|z-2016\right|=2016\)
Tìm x, biết:
a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
b) \(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)
c) \(2020^{\left(x-2\right).\left(2x+3\right)}=1\)
d) \(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\) với \(x\in Z\)
e) \(\dfrac{3}{4}\sqrt{x}-\dfrac{1}{2}=\dfrac{1}{3}\)
Tìm tất cả các cặp số \(\left(x,y\right)\) thoả mãn: \(\left(2x-y+7\right)^{2022}+\left|x-3\right|^{2023}\le0\)
Tìm GTNN của biểu thức:
a) \(\left|x-23\right|+\left|x-10\right|\)
b) \(\left|x-2\right|+\left|x-5\right|+\left|x+3\right|\)
c) \(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|+\left|x-5\right|\)
Tìm giá trị nhỏ nhất của biểu thức :
a) A=\(\left|x+2\right|+\left|2x-3\right|+\left|x-5\right|\)
b) B=\(\left|x+2\right|+\left|3x-1\right|+\left|x-7\right|+5\)
c) C=\(\left|x+1\right|+4\left|2x-7\right|+\left|x-5\right|\)
d) D=\(\left|x+4\right|+5\left|x+1\right|+\left|x-2\right|+5\)
Bài 4.1: Tìm x, biết
a) \(4\left|3x-1\right|+\left|x\right|-2\left|x-5\right|+7\left|x-3\right|=12\)
b) \(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9\right|=5\)
c) \(\left|2\frac{1}{5}-x\right|+\left|x-\frac{1}{5}\right|+8\frac{1}{5}=1,2\)
d) \(2\left|x+3\frac{1}{2}\right|+\left|x\right|-3\frac{1}{2}=\left|2\frac{1}{5}-x\right|\)
Tìm x biết :
\(\dfrac{3}{\left(x+2\right)\cdot\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\cdot\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\cdot\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\cdot\left(x+17\right)}\) Vs x \(\notin\){-2;-5;-17;-10}
1.\(\left|x+1\right|\)+\(\left|x+5\right|\)+\(\left|x+10\right|\)=4x
2.\(\left|x-1\right|\)+\(\left|x-2\right|\)=5
3.\(\left|x-3\right|\)+\(\left|x+1\right|\)=10
4.\(\left|2x-1\right|\)+\(\left|3x-5\right|\)+\(\left|4-5x\right|\)=10-3x
\(\dfrac{2}{\left(x-2\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}\)+\(^{\dfrac{12}{\left(x-8\right)\left(x-20\right)}}\)-\(\dfrac{1}{x-20}\)= \(\dfrac{-3}{4}\)
