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1) \(A=23+\left|2x-\frac{1}{3}\right|\)
Ta có: \(\left|2x-\frac{1}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|2x-\frac{1}{3}\right|+23\ge23\forall x\)
\(A=23\Leftrightarrow\left|2x-\frac{1}{3}\right|=0\Leftrightarrow2x-\frac{1}{3}=0\Leftrightarrow2x=\frac{1}{3}\Leftrightarrow x=\frac{1}{6}\)
Vậy Amin=23 \(\Leftrightarrow x=\frac{1}{6}\)
Câu b, câu c tương tự
2) \(\left|x-3,5\right|+\left|y-1,3\right|=0\)
Ta có: \(\orbr{\begin{cases}\left|x-3,5\right|\ge0\forall x\\\left|y-1,3\right|\ge0\forall y\end{cases}}\Rightarrow\left|x-3,5\right|+\left|y-1,3\right|\ge0\forall x\)
Mà \(\left|x-3,5\right|+\left|y-1,3\right|=0\)
\(\Rightarrow\orbr{\begin{cases}\left|x-3,5\right|=0\\\left|y-1,3\right|=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-3,5=0\\y-1,3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3,5\\y=1,3\end{cases}}}\)
Vậy x=3,5 ; y=1,3
a) 5x.(x+3/4) = 0
=> x = 0
x+3/4 = 0 => x = -3/4
b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)
\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)
\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)
\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)
\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)
\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)
=> x + 2017 = 0
x = -2017
a) để 2x - 3 > 0
=> 2x > 3
x > 3/2
b) 13-5x < 0
=> 5x < 13
x < 13/5
c) \(\frac{x+3}{2x-1}>0\)
=> x + 3 > 0
x > -3
d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)
Để x+7/x+3 < 1
=> 1 + 4/x+3 < 1
=> 4/x+3 < 0
=> không tìm được x thỏa mãn điều kiện
\(\left|x\right|=2\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
Thay x=-2 vào B ta có:
\(B=4x^3+x-2022=4.\left(-2\right)^3+\left(-2\right)-2022=-32-2-2022=-2056\)
Thay x=2 vào B ta có:
\(B=4x^3+x-2022=4.2^3+2-2022=32+2-2022=-1988\)