tim gia tri :(1-1/1+2).(1-1/1+2+3).(1-1/1/2/3/4)....(1-1/1+2+3+...+2006)
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Ta có
(1-1/(1+2))=(0/(1+2))=0
(1-1/(1+2+3))=(0/(1+2+3))=0
..........................................
.(1-1/(1+2+3+...+2006))=.(0/(1+2+3+...+2006))=0
=>0.0.0.0.....0=0
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\(A=\left(2x+\frac{1}{3}\right)^4-1\) . Có: \(\left(2x+\frac{1}{3}\right)\ge0\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^4-1\ge-1\)
Dấu = xảy ra khi: \(2x+\frac{1}{3}=0\)
\(\Rightarrow2x=-\frac{1}{3}\)
\(\Rightarrow x=-\frac{1}{3}:2=-\frac{1}{6}\)
Vậy: \(Min_A=-1\) tại \(x=-\frac{1}{6}\)
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A
\(\frac{3}{5}\div\frac{4}{5}+\frac{1}{2}\times\frac{2}{3}\)
\(=\frac{3}{4}+\frac{1}{3}\)
\(=\frac{13}{12}\)
B
\(\frac{5}{4}\times x=\frac{3}{8}+\frac{1}{4}\)
\(\frac{5}{4}\times x =\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{5}{4}\)
\(x=\frac{4}{8}=\frac{1}{2}\)
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\(4\frac{3}{5}-1\frac{2}{7}-1\frac{1}{7}=\frac{23}{5}-\frac{1}{7}=\frac{156}{35}\)
\(\frac{27}{12}-\frac{5}{4}-\frac{1}{2}=\frac{27}{12}-\frac{15}{12}-\frac{6}{12}=\frac{6}{12}=\frac{1}{2}\)
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a: \(\left(2x+1\right)^4-1\ge-1\)
Dấu '=' xảy ra khi x=-1/2
b: \(\left(x^2-16\right)^2+\left|y-3\right|-2\ge-2\)
Dấu '=' xảy ra khi \(\left(x,y\right)\in\left\{\left(4;3\right);\left(-4;3\right)\right\}\)
chiu bo
cha mi