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Vì \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{3}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0\) với mọi x
=>\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|\ge0\) với mọi x
=>\(4x\ge0=>x\ge0\), do đó PT ban đầu trở thành:
\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x< =>3x+1=4x< =>x=1\)
Vậy x=1
Ta có:
(x+1/3)^2 >=0 với mọi x
|y+5| >=0 với mọi y
=>GTNN A=(x+1/3)^2+|y+5| -2/5 >= -2/5
dấu = xảy ra khi và chỉ khi:
x+1/3=0 =>x=-1/3
y+5=0 => y=-5
KL:
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)
vì \(\left|\frac{3}{2}x+\frac{1}{9}\right|\ge0;\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0=>\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\) (với mọi x,y)
Mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\) (theo đề)
Nên \(\left|\frac{3}{2}x+\frac{1}{9}\right|=0=>\frac{3}{2}x=-\frac{1}{9}=>x=-\frac{2}{27}\)
\(\left|\frac{1}{5}y-\frac{1}{2}\right|=0=>\frac{1}{5}y=\frac{1}{2}=>y=\frac{5}{2}\)
Vậy...........
\(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\)
\(\left|\frac{1}{2}-x\right|=\frac{2}{5}-6\)
\(\left|\frac{1}{2}-x\right|=\frac{-28}{5}\)
vì | 1/2 - x | \(\ge\)0 \(\forall\)x nên x không tồn tại
\(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\)
\(\left|\frac{1}{2}-x\right|=\frac{2}{5}-6\)
\(\left|\frac{1}{2}-x\right|=\frac{-28}{5}\)
Vì \(\left|\frac{1}{2}-x\right|=\frac{-28}{5}\)
=> x ko tồn tại