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Nhận thấy \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
=> \(\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\forall x\)
Dấu "=" xảy ra <=> \(2x+\frac{1}{3}=0\Rightarrow x=-\frac{1}{6}\)
Vậy Min A = -1 <=> X = -1/6
a, \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\)
Dấu "=" xảy ra <=> 2x+1/3=0 <=> x= -1/6
a) \(\left(2x+\frac{1}{3}\right)^4\ge0\Rightarrow A\ge-1\)
Dấu \(=\)xảy ra khi \(2x+\frac{1}{3}=0\Leftrightarrow x=-\frac{1}{6}\).
b) \(\left(\frac{4}{9}x-\frac{2}{15}\right)^6\ge0\Rightarrow B\le3\)
Dấu \(=\)xảy ra khi \(\frac{4}{9}x-\frac{2}{15}=0\Leftrightarrow x=\frac{3}{10}\).
a) \(\left|2x-1\right|=5\)
\(2x-1=5\)
\(2x=6\)
\(x=3\)
b) \(x:4\frac{1}{3}=-2,5\)
\(x:\frac{4}{3}=\frac{-5}{2}\)
\(x=\frac{-5}{2}.\frac{4}{3}=\frac{-10}{3}\)
c) \(x^3-\frac{9}{10}.x=0\)
\(xx^2-\frac{9}{10}.x=0\)
\(x\left(x^2-\frac{9}{10}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x^2-\frac{9}{10}=0\Rightarrow x^2=\frac{9}{10}\Rightarrow x=\sqrt{\frac{9}{10}}\end{cases}}\)
Câu c có 2 kết quả là \(\hept{\begin{cases}x=0\\x=\sqrt{\frac{9}{10}}\end{cases}}\)
Câu b nhầm
\(x:4\frac{1}{3}=-2,5\)
\(x:\frac{13}{3}=\frac{-5}{2}\)
\(x=\frac{-5}{2}.\frac{13}{3}=\frac{-65}{6}\)
a: \(\Leftrightarrow\left(x-\dfrac{2}{5}\right):\dfrac{3}{2}=-\dfrac{5}{4}+\dfrac{5}{2}=\dfrac{5}{4}\)
\(\Leftrightarrow x-\dfrac{2}{5}=\dfrac{5}{4}\cdot\dfrac{3}{2}=\dfrac{15}{8}\)
hay x=91/40
b: \(\Leftrightarrow\left(2.5x-3.6\right)=-1\cdot\dfrac{12}{7}=\dfrac{-12}{7}\)
=>2,5x=66/35
hay x=132/175
c: \(\Leftrightarrow\left(\dfrac{15}{4}-2x\right)=\dfrac{19}{9}:\dfrac{4}{3}=\dfrac{19}{9}\cdot\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x=15/4-19/12=45/12-19/12=26/12
=>x=13/12
\(A=\left|x+3\right|+\dfrac{5}{2}\ge\dfrac{5}{2}\forall x\)
Dấu '=' xảy ra khi x=-3
\(B=\left|2x+1\right|-\dfrac{1}{4}\ge-\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi x=-1/2