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C1: \(\left(\frac{3}{5}+\frac{4}{9}\right)\cdot\frac{3}{8}=\frac{47}{45}\cdot\frac{3}{8}=\frac{141}{360}=\frac{47}{120}\)
C2: \(\left(\frac{3}{5}+\frac{4}{9}\right)\cdot\frac{3}{8}=\frac{3}{5}\cdot\frac{3}{8}+\frac{4}{9}\cdot\frac{3}{8}=\frac{9}{40}+\frac{12}{72}=\frac{47}{120}\)
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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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\(-\frac{2}{3}\cdot x+\frac{13}{15}=\frac{7}{10}\)
\(-\frac{2}{3}\cdot x=\frac{7}{10}-\frac{13}{15}=-\frac{1}{6}\)
\(x=-\frac{1}{6}:\left(-\frac{2}{3}\right)=\frac{1}{4}\)
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hình như cái đề saisai sao ấy bạn ak ??????
tk tui nha
mơn mọi người nhiều lắm !!!!!!!!
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Đề có sai không bạn, mình thấy đề là \(\frac{2}{3}\times\frac{4}{5}+\frac{2}{3}\times\frac{1}{6}\)như vậy đúng hơn
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x + 3/4 = 1/2 + 3/5 = 5/10 + 6/10
x + 3/4 = 11/10
x = 11/10 : 3/4
x = 11/10 : 3/4 = 11/10 x 4/3
x = 44/30 = 22/15
Mình không biết làm đúng không
3695 - x = 120 : 6
3695 - x = 20
x = 3695 - 20
x = 3675
\(x+\frac{3}{4}=\frac{1}{2}+\frac{3}{5}\)
\(x+\frac{3}{4}=\frac{11}{10}\)
\(x=\frac{11}{10}-\frac{3}{4}\)
\(x=\frac{7}{20}\)
3695 - x = 120 : 6
3695 - x = 20
x = 3695 - 20
x = 3675
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\(\frac{2}{3}\)\(\div\)\(\frac{5}{7}\)
= \(\frac{2}{3}\)\(\times\)\(\frac{7}{5}\)
= \(\frac{2\times7}{3\times5}\)
= \(\frac{14}{35}\)
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:V toán lp 3 cơ ak
A = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+...+\frac{1}{4347}\)
\(A\cdot3=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1449}\)
\(A\cdot3-A=\left(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1449}\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+...+\frac{1}{4347}\right)\)
\(A\cdot2=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1449}-\frac{1}{2}-\frac{1}{6}-\frac{1}{18}-...-\frac{1}{4347}\)
\(A\cdot2=\frac{3}{2}-\frac{1}{4347}\)
\(A\cdot2=\frac{13039}{8694}\)
\(A=\frac{13039}{8694}:2\)
\(A=\frac{13039}{17388}\)
Kết quả hơi lớn nên kiểm tra lại đề :))
\(\frac{2}{3}x3=\frac{2}{3}x\frac{3}{1}=\frac{6}{3}=2\)
ủng hộ mk giùm nha %@#Q#
Tích đó là :
2/3 x 3 = 2
Đáp số : 2