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a) \(\frac{13}{26}-\frac{1}{3}-\frac{1}{2}+\frac{7}{21}\)
\(=\frac{1}{2}-\frac{1}{3}-\frac{1}{2}+\frac{1}{3}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)
\(=0+0\)
\(=0\)
b) \(\left(\frac{-5}{12}+\frac{6}{11}\right)+\left(\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\right)\)
\(=\frac{-5}{12}+\frac{6}{11}+\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\)
\(=\left(\frac{-5}{12}+\frac{5}{12}\right)+\left(\frac{7}{17}+\frac{5}{17}\right)+\frac{6}{11}\)
\(=0+\frac{12}{17}+\frac{6}{11}\)
\(=\frac{132}{187}+\frac{102}{187}\)
\(=\frac{234}{187}\)
c) \(\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{12}{10}\right)\)
\(=\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{6}{5}\right)\)
\(=\frac{13}{5}+\frac{7}{16}-\frac{11}{16}+\frac{6}{5}\)
\(=\left(\frac{13}{5}+\frac{6}{5}\right)+\left(\frac{7}{16}-\frac{11}{16}\right)\)
\(=\frac{19}{5}+\left(\frac{-4}{16}\right)\)
\(=\frac{19}{5}-\frac{1}{4}\)
\(=\frac{76}{20}-\frac{5}{20}\)
\(=\frac{71}{20}\)
d) \(-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{21}{30}-\frac{5}{11}\right)\)
\(=-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{7}{10}-\frac{5}{11}\right)\)
\(=-\frac{3}{10}+\frac{6}{11}-\frac{7}{10}+\frac{5}{11}\)
\(=
\left(-\frac{3}{10}-\frac{7}{10}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)\)
\(=\frac{-10}{10}+\frac{11}{11}\)
\(=-1+1\)
\(=0\)
1.
A=\(\frac{-5x+-5y+-5z}{21}=\frac{-5\left(x+y+z\right)}{21}=\frac{-5}{21}.x+y+z\)
A= -z+z=0
<p style="padding: 10000000000000000px;" class="alert success"></p>
\(A=\left(x-1\right)^2-3\)
a) Với x = -2, ta có:
\(A=\left(-2-1\right)^2-3=6\)
b) \(\left(x-1\right)^2-3\ge3\text{ vì }\left(x-1\right)^2\ge0\forall x\inℝ\)
\(\Rightarrow MIN_A=3\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy: \(MIN_A=3\Leftrightarrow x=1\)
Khong chac dau nhe .-.
A=(x-1)2-3
Với x=-2
Ta có:
A=(-2-1)2-3
A=(-3)2-3
A=9-6
A=3
Vậy A=3 với x=-2
b)Tính GTNN của biểu thức A
Để biểu thức A đạt GTNN <=>(x-1)2
<=>(x-1) đạt GTNN
<=>x=1
Vậy với x =1 thì biểu thức A đạt GTNN
a) Ta có:
\(A=\frac{3^{10}\cdot11+3^{10}\cdot5}{3^0\cdot2^4}\)
\(A=\frac{3^{10}\left(11+5\right)}{1\cdot16}\)
\(A=\frac{3^{10}\cdot16}{16}=3^{10}\)
b) Ta có:
\(B=\frac{2^{10}\cdot13+2^{10}\cdot65}{28\cdot104}\)
\(B=\frac{2^{10}\cdot13\cdot\left(1+5\right)}{2^2\cdot7\cdot2^3\cdot13}\)
\(B=\frac{2^{10}\cdot6\cdot13}{2^5\cdot7\cdot13}=\frac{2^{11}\cdot3\cdot13}{2^5\cdot7\cdot13}\)
\(B=\frac{2^6\cdot3}{7}=\frac{192}{7}\)