\(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)-\dfrac{7}{10}\\ =\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{7}{10}\\ =\dfrac{56}{70}+\dfrac{20}{70}-\dfrac{49}{70}\\ =\dfrac{27}{70}\\ 3,5-\left(-\dfrac{2}{7}\right)\\ =\dfrac{7}{2}+\dfrac{2}{7}\\ =\dfrac{49}{14}+\dfrac{4}{14}\\ =\dfrac{53}{14}\\ \left(-3\right).\left(-\dfrac{7}{12}\right)\\ =\dfrac{21}{12}\\ =\dfrac{7}{4}\)

Ta có: \(\left(\frac87-\frac43\right)\cdot1\frac13-\frac27\cdot\left(-3,5\right)^2\)

\(=\left(\frac{24}{21}-\frac{28}{21}\right)\cdot\frac43-\frac27\cdot\left(-\frac72\right)^2\)

\(=\frac{-4}{21}\cdot\frac43-\frac27\cdot\frac{49}{4}=\frac{-16}{63}-\frac72=\frac{-32-441}{126}=\frac{-473}{126}\)

\(=\dfrac{1}{16}-\dfrac{7}{2}+\sqrt{7}=\dfrac{-55}{16}+\sqrt{7}\)

\(\left(-\dfrac{1}{4}\right)^2-3,5+\sqrt{7}\\ =\dfrac{1}{16}-3,5+\sqrt{7}\\ =-\dfrac{55}{16}+\sqrt{7}\\ =\dfrac{-55+16\sqrt{7}}{16}\)

a) \(...-\dfrac{22}{7}:-\dfrac{15}{9}=\dfrac{22}{7}.\dfrac{9}{15}=\dfrac{66}{35}\)

b) \(...\dfrac{8}{3}:\left(-\dfrac{42}{7}\right)=\dfrac{8}{3}.\left(-\dfrac{7}{42}\right)=-\dfrac{8}{9}\)

c) ...\(\left(-\dfrac{7}{2}\right):\left(-\dfrac{13}{5}\right)=\dfrac{7}{2}.\dfrac{5}{13}=\dfrac{35}{36}\)

\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot\dfrac{1}{3}-\dfrac{2}{7}\cdot\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right)\cdot\dfrac{1}{3}-\dfrac{2}{7}\cdot\dfrac{49}{4}\)
\(=\dfrac{1}{8}\cdot\dfrac{1}{3}-\dfrac{7}{2}\)
\(=\dfrac{1}{24}-\dfrac{7}{2}\)
\(=\dfrac{1}{24}-\dfrac{84}{24}\)
\(=-\dfrac{83}{24}\)
#データネ

`(7/8-3/4)xx1/3-2/7xx(3,5)^2`

`=(7/8-6/8)xx1/3-2/7xx12,25`

`=1/8xx1/3-2/7xx12,25`

`=1/24-7/2`

`=1/24-84/24`

`=-83/24`

\(\frac{3a-b}{2a+7}+\frac{3b-a}{2b-7}=\frac{3a-b}{2a+a-b}+\frac{3b-a}{2b-\left(a-b\right)}=\frac{3a-b}{3a-b}+\frac{3b-a}{3b-a}=2\)

Bài làm:

\(3,5\times6\times7-\left(\frac{1}{2}+\frac{10}{4}\right)\)

\(=21\times7-3\)

\(=147-3\)

\(=144\)

Học tốt 

Bài làm

3,5 . 6 . 7 - ( 1/2 + 10/4 )

= 21 . 7 - ( 2/4 + 10/4 )

= 147 - 12/4

= 147 - 3

= 144 

\(P=-\left(2x+1\right)^2-7\left(y-3,5\right)^2+\dfrac{2}{3}\)

 vì \(\left\{{}\begin{matrix}-\left(2x+1\right)^2\le0,\forall x\\-7\left(y-3,5\right)^2\le0,\forall y\end{matrix}\right.\)

\(\Rightarrow P=-\left(2x+1\right)^2-7\left(y-3,5\right)^2+\dfrac{2}{3}\le\dfrac{2}{3}\)

Dấu "=" xảy ra khi

\(\left\{{}\begin{matrix}2x+1=0\\y-3,5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3,5=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(GTLN\left(P\right)=\dfrac{2}{3}\left(tạix=-\dfrac{1}{2};y=\dfrac{7}{2}\right)\)

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