có kết quả r mà

4/7 *1/3+5/7:2/3

\(\frac{4}{21}\)+ \(\frac{15}{14}\)= \(\frac{53}{42}\)

=4/21+15/14

=53/42

\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7-2\sqrt{21}+3}+\sqrt{7+2\sqrt{21}+3}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)

\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=\sqrt{7}+\sqrt{7}=2\sqrt{7}\)

Ta có: \(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

\(=\sqrt{7-2\sqrt{21}+3}+\sqrt{7+2\sqrt{21}+3}\)

\(=\sqrt{\sqrt{7}^2-2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}+\sqrt{\sqrt{7}^2+2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}\)

\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)

\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

\(=2\sqrt{7}\)

\(\dfrac{\left(-2\right).2}{15.2}-\dfrac{3.3}{10.3}=\dfrac{-4}{30}-\dfrac{9}{30}=\dfrac{-13}{30}\)

\(=\dfrac{-2}{15}+\dfrac{3}{10}=\dfrac{-4}{30}+\dfrac{9}{30}=\dfrac{5}{30}=\dfrac{1}{6}\)

\(=\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{15}=\dfrac{6}{60}+\dfrac{5}{60}+\dfrac{4}{60}=\dfrac{15}{60}=\dfrac{1}{4}\)

\(=\dfrac{1.6}{10.6}-\dfrac{\left(-1\right).5}{12.5}+\dfrac{1.4}{15.4}=\dfrac{6}{60}+\dfrac{-5}{60}+\dfrac{4}{60}=\dfrac{5}{60}=\dfrac{1}{12}\)

Sửa đề: Biểu thức luôn có giá trị dương

Ta có: \(3x^2+2x-5\)

\(=3\left(x^2+\dfrac{2}{3}x-\dfrac{5}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{16}{9}\right)\)

\(=3\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}\ge-\dfrac{16}{3}\forall x\)

\(\Leftrightarrow\dfrac{1}{3\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}}\le\dfrac{1}{\dfrac{-16}{3}}=\dfrac{-3}{16}\forall x\)

\(\Leftrightarrow\dfrac{-1}{3\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}}\ge\dfrac{3}{16}>0\forall x\)(đpcm)

\(=\dfrac{7}{20}-\dfrac{1}{4}=\dfrac{7}{20}-\dfrac{5}{20}=\dfrac{1}{10}\)

\(\dfrac{7}{20}+\dfrac{-1}{4}=\dfrac{7}{20}+\dfrac{-5}{20}=\dfrac{2}{20}=\dfrac{1}{10}\)

bài 2 là dương nhé

Bài 2: 

a: Để \(\dfrac{4}{x+2}>0\) thì x+2>0

hay x>-2

b: Để \(\dfrac{3x+2}{-4}>0\) thì 3x+2<0

hay x<-2/3