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\(+,x< -2\Rightarrow\left\{{}\begin{matrix}x+2< 0\\2x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|=-2-x\\\left|2x-3\right|=3-2x\end{matrix}\right.\Rightarrow1-3x=5\Rightarrow x=-\frac{4}{3}\left(\text{loại}\right)\)
\(+,x\ge\frac{3}{2}\Rightarrow\left\{{}\begin{matrix}2x-3\ge0\\x+2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|2x-3\right|=2x-3\\\left|x+2\right|=x+2\end{matrix}\right.\Rightarrow3x-1=5\Rightarrow x=2\left(\text{thoa man}\right)\)
\(+,-2\le x< \frac{3}{2}\Rightarrow\left\{{}\begin{matrix}x+2\ge0\\2x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|=x+2\\\left|2x-3\right|=3-2x\end{matrix}\right.\Rightarrow5-x=0\Rightarrow x=0\left(\text{thoa man}\right)\)
\(2.\text{ Ta co:}\left\{{}\begin{matrix}\left|x-102\right|\ge102-x\\\left|2-x\right|\ge x-2\end{matrix}\right.\Rightarrow A\ge102-x+x-2=100.\Rightarrow A_{min}=100.\text{dâu "=" xay ra}\Leftrightarrow\left\{{}\begin{matrix}102-x\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow2\le x\le102\)
a) |5/3 - x| - |-5/6| = |-5/9|
=> |5/3 - x| - 5/6 = 5/9
=> |5/3 - x| = 5/9 + 5/6
=> |5/3 - x| = 25/18
=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)
a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)
\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)
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Từ \(\frac{x-2016}{100}\rightarrow\frac{x-2}{2114}\) có tất cả \(1008\) số.
Ta có: \(\frac{x-2016}{100}+\frac{x-2014}{102}+...+\frac{x-2}{2114}=1008\)
\(\Leftrightarrow\frac{x-2016}{100}+\frac{x-2014}{102}+...+\frac{x-2}{2114}-1008=1008-1008\)
\(\Leftrightarrow\left(\frac{x-2016}{100}-1\right)+\left(\frac{x-2014}{102}-1\right)+...+\left(\frac{x-2}{2114}-1\right)=0\) (*)
\(\Leftrightarrow\frac{x-2116}{100}+\frac{x-2116}{102}+...+\frac{x-2116}{2114}=0\)
\(\Leftrightarrow\left(x-2116\right)\left(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\right)=0\)
Vì \(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\ne0\)
Nên từ pt (*) \(\Leftrightarrow x-2116=0\Leftrightarrow x=2116\)
Vậy...
\(A=\left|x-102\right|+\left|2-x\right|\)
Áp dụng bất đẳng thức:
\(\left|A+B\right|\le\left|A\right|+\left|B\right|\)
\(\Rightarrow A=\left|x-102\right|+\left|2-x\right|\le\left|x-102+2-x\right|\)
\(A\le\left|-100\right|\Rightarrow A\le100\)
Dấu "=" xảy ra khi:
\(-2\le x\le102\)
a, Vì /x-2/ ≥ 0 (với mọi x ∈ R )
=> /x-2/ +5 ≥ 5
Dấu " = " xảy ra khi và chỉ khi /x-2/ = 0 => x-2 = 0 => x=2
Vậy Amin = 5 khi x =2
a,Nhận xét:
\(\left|x-2\right|\ge0\)
\(\rightarrow\left|x-2\right|+5\ge5\)
Vậy Min A=5 khi \(\left|x-2\right|=0\)
\(x-2=0\)
\(x=2\)
b,Nhận xét:
\(\left|x+4\right|\ge0\)
\(12-\left|x+4\right|\)\(\ge12\)
Vậy Max B=12 khi x+4=0
x=4
ojk
Ta co:
A=/x-102/+/2-x/
=>minA=/2-x/
<=> /x-102/=0=>x-102=0=>x=102
khi do
minA=/2-102/=100
Vay minA=100 khi x=102
nho tich tui nha