Nếu a/-7=b/5 thì:
A.a/-7=b/5=b-a/-7-5
B.a/-7=b/5=a-b/-7=5
C.a/-7=b/5=b-a/5-7
D.a/-7=b/5=b-a/5+7
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Ta có đánh giá: \(\frac{a^7+b^7}{a^5+b^5}\ge\frac{a^2+b^2}{2}\)
\(\Leftrightarrow2a^7+2b^7\ge a^7+b^7+a^5b^2+a^2b^5\)
\(\Leftrightarrow a^5\left(a^2-b^2\right)-b^5\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\ge0\) (luôn đúng)
Tương tự \(\frac{b^7+c^7}{b^5+c^5}\ge\frac{b^2+c^2}{2}\) ; \(\frac{c^7+a^7}{c^5+a^5}\ge\frac{a^2+c^2}{2}\)
\(\Rightarrow VT\ge a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\dfrac{a}{b}=5-\dfrac{3}{5}=\dfrac{25}{5}-\dfrac{3}{5}=\dfrac{22}{5}\)
b)\(\dfrac{a}{b}=\dfrac{5}{6}+\dfrac{4}{7}=\dfrac{35}{42}+\dfrac{24}{42}=\dfrac{59}{42}\)
c)\(\dfrac{a}{b}=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{9}{10}\)
Nếu `a/-7=b/5` thì:
`A.a/-7=b/5=`\(\dfrac{b-a}{-7-5}\)
`B.a/-7=b/5=`\(\dfrac{a-b}{-7=5}\)
`C.a/-7=b/5=`\(\dfrac{b-a}{5-7}\)
D.`a/-7=b/5=`\(\dfrac{b-a}{5+7}\)
`---`
Vì `a/-7=b/5`
Theo tính chất tỉ lệ thức `-> a/-7=b/5=`\(\dfrac{b-a}{5-\left(-7\right)}=\dfrac{b-a}{5+7}\)
`->` Đáp án `D.`