Đề yc gì em?

chỗ \(\dfrac{3}{5}\) là \(\left|\dfrac{-3}{5}\right|\)

1: \(=5.7\left(-6.5-3.5\right)=-10\cdot5.7=-57\)

a: \(=-15,5\left(20,8+9,2\right)+3,5\left(9,2+20,8\right)\)

\(=30\cdot\left(-12\right)=-360\)

b: \(=\dfrac{21}{20}:\dfrac{27}{10}+2+\left(\dfrac{2}{5}:\dfrac{5}{2}\right)\cdot\left(\dfrac{21}{5}-\dfrac{13}{10}\right)\)

\(=\dfrac{21}{10}\cdot\dfrac{10}{27}+2+\dfrac{4}{25}\cdot\dfrac{29}{10}\)

\(=\dfrac{7}{9}+2+\dfrac{58}{125}=\dfrac{3647}{1125}\)

\(a,-\dfrac{13}{20}+x=\dfrac{-11}{15}\\ \Rightarrow x=\dfrac{-11}{15}+\dfrac{13}{20}\\ \Rightarrow x=-\dfrac{1}{12}\\ b,\left(x-3,5\right):3\dfrac{1}{2}-2,5=-1\dfrac{3}{4}\\ \Rightarrow\left(x-\dfrac{7}{2}\right):\dfrac{7}{2}-\dfrac{5}{2}=\dfrac{-7}{4}\\ \Rightarrow\left(x-\dfrac{7}{2}\right):\dfrac{7}{2}=\dfrac{3}{4}\\ \Rightarrow x-\dfrac{7}{2}=\dfrac{21}{8}\\ \Rightarrow x=\dfrac{49}{8}\)

=1/2(2/3*5+2/5*7+...+2/97*99)

=1/2(1/3-1/5+1/5-1/7+...+1/97-1/99)

=1/2*32/99

=16/99

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

a)Ta có:

 \(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)

\(\Rightarrow x-3,5=y-\dfrac{1}{10}=0\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}=0,1\end{matrix}\right.\)

b) Ta có:

\(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)

b: ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)

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

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

= \(\dfrac{1}{16}-3,5+\)√7

= \(\dfrac{-55+16\sqrt{7}}{16}\)