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1 x 3 = ? 1 x 3 = 1 + 1 + 1 = 3 1 x 3 = 3 | 1 x 4 = ? 1 x 4 = 1 + 1 + 1 + 1 = 4 1 x 4 = 4
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1 x 6 = ? 1 x 6 = 1 + 1 + 1 + 1 + 1 + 1 = 6 1 x 6 = 6 | 1 x 5 = ? 1 x 5 = 1 + 1 + 1 + 1 + 1= 5 1 x 5 = 5 |
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\(a,\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13=\frac{3}{7}\)
\(b,\frac{2}{3}\cdot x+\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{2}=-\frac{2}{5}\)
\(x=-\frac{2}{5}:\frac{2}{3}=-\frac{3}{5}\)
\(c,x:\frac{8}{11}=\frac{11}{3}\)
\(x=\frac{11}{3}\cdot\frac{8}{11}=\frac{8}{3}\)
\(d,\frac{2}{9}-\frac{7}{8}\cdot x=\frac{1}{3}\)
\(\frac{7}{8}.x=\frac{2}{9}-\frac{1}{3}=-\frac{1}{9}\)
\(x=-\frac{1}{9}:\frac{7}{8}=-\frac{8}{63}\)
cảm ơn Ngyễn Thị Ngọc Ánh nha bạn lm đúng mà mình k nhầm xl bạn nhìu
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chứng minh \(\frac{3}{2}\ge\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\)
ta có \(\left(x-1\right)^2\ge0\Leftrightarrow x^2+1\ge2x\Leftrightarrow\frac{2x}{1+x^2}\le1\)
\(\left(y-1\right)^2\ge0\Leftrightarrow y^2+1\ge2y\Leftrightarrow\frac{2y}{1+y^2}\le1\)
\(\left(z-1\right)^2\ge0\Leftrightarrow z^2+1\ge2z\Leftrightarrow\frac{2z}{1+z^2}\le1\)
\(\Rightarrow\frac{2x}{1+x^2}+\frac{2y}{1+y^2}+\frac{2x}{1+z^2}\le3\Leftrightarrow\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\le\frac{3}{2}\)
chứng minh \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{3}{2}\)
áp dụng bất đẳng thức Cauchy ta có:
\(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge3\sqrt[3]{\frac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}=\frac{3}{\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\)
ta lại có \(\frac{\left(1+x\right)\left(1+y\right)\left(1+z\right)}{3}\ge\sqrt[3]{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
vậy \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{3}{\frac{\left(1+x\right)+\left(1+y\right)+\left(1+z\right)}{3}}=\frac{3}{2}\)
kết hợp ta có \(\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\le\frac{3}{2}\le\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\)
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a: \(=60\cdot\dfrac{17}{20}=51\)
b: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}=\dfrac{1}{5}\)
TL
1 + 1 X 1 + 2 - 2 +3 = .....
Đáp án 5
HT
TL:
= 5
_HT_