Người thợ săn trộn số lít sơn xanh và trắng là :
        3,8 + 0,7 = 4,5 ( l )
Người thợ sơn còn lại số lít sơn xanh nhạt là :
       4,5 - 1,25 = 3,25 ( l )
            Đáp số : 3,25l sơn xanh nhạt

a) \(\dfrac{-3}{5}+\dfrac{28}{5}.\left(\dfrac{13}{56}-\dfrac{5}{24}+\dfrac{1}{7}\right)\\ =-\dfrac{3}{5}+\dfrac{28}{5}.\dfrac{13}{56}-\dfrac{28}{5}.\dfrac{5}{24}+\dfrac{28}{5}.\dfrac{1}{7}\\ =-\dfrac{3}{5}+\dfrac{28.13}{5.2.28}-\dfrac{4.7.5}{5.4.6}+\dfrac{4.7.1}{5.7}\\ =-\dfrac{3}{5}+\dfrac{13}{10}-\dfrac{7}{6}+\dfrac{4}{5}\\ =\dfrac{1}{5}+\dfrac{13}{10}-\dfrac{7}{6}\\ =\dfrac{6}{30}+\dfrac{39}{30}-\dfrac{35}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)

b) \(\dfrac{5}{7}.\dfrac{11}{18}+\dfrac{3}{7}.\dfrac{5}{8}+\dfrac{4}{9}\\ =\dfrac{5}{7}.\dfrac{11}{18}+\dfrac{5}{7}.\dfrac{3}{8}+\dfrac{4}{9}\\ =\dfrac{5}{7}.\left(\dfrac{11}{18}+\dfrac{3}{8}\right)+\dfrac{4}{9}\\ =\dfrac{5}{7}.\dfrac{7}{4}+\dfrac{4}{9}\\ =\dfrac{5}{4}+\dfrac{4}{9}=\dfrac{61}{36}\)

c) \(4\dfrac{5}{9}:\left(\dfrac{-5}{7}\right)+\dfrac{49}{9}:\left(\dfrac{-5}{7}\right)\\=\left(4\dfrac{5}{9}+\dfrac{49}{9}\right):\dfrac{-5}{7}\\ =\left(\dfrac{41}{9}+\dfrac{49}{9}\right).\dfrac{-7}{5}\\ =10.\dfrac{-7}{5}=-14\)

Mình sửa lại câu b nhé:

\(=\dfrac{5}{7}.\dfrac{11}{18}+\dfrac{5}{7}.\dfrac{3}{8}+\dfrac{4}{9}\\ =\dfrac{5}{7}.\left(\dfrac{11}{18}+\dfrac{3}{8}\right)+\dfrac{4}{9}\\ =\dfrac{5}{7}.\dfrac{71}{72}+\dfrac{4}{9}=\dfrac{355}{504}+\dfrac{4}{9}\\ =\dfrac{193}{168}\)

a)

b)Ta có: \(x-y=7\Rightarrow x=y+7\)

\(\dfrac{2x+7}{3x-y}=\dfrac{2\left(y+7\right)+7}{3\left(y+7\right)-y}=\dfrac{2y+21}{2y+21}=1\)

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

\(\Rightarrow C=\dfrac{2x+7}{3x-y}+\dfrac{3y-7}{3y-x}=1+1=2\)

\(\left(\dfrac{1}{4}-\dfrac{2}{3}\right):\dfrac{3x}{5}=\dfrac{5}{2}\Leftrightarrow\dfrac{-5}{12}:\dfrac{3x}{5}=\dfrac{5}{2}\Leftrightarrow\dfrac{3x}{5}=-\dfrac{1}{6}\Leftrightarrow x=-\dfrac{5}{18}\)

\(\left(\dfrac{1}{4}-\dfrac{2}{3}\right):\dfrac{3x}{5}=\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{-5}{12}:\dfrac{3x}{5}=\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{3x}{5}=\dfrac{-5}{12}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{3x}{5}=\dfrac{-1}{6}\)

\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{5}\)

\(\Leftrightarrow x=-\dfrac{5}{18}\)

Vậy \(x=-\dfrac{5}{18}\)

5A:

\(x-128=\left(4\dfrac{20}{21}-5\right):\left(\dfrac{4141}{4242}-1\right):\left(\dfrac{636364}{646464}-1\right)\)

\(\Rightarrow x-128=\left(\dfrac{104}{21}-5\right):\left(\dfrac{41}{42}-1\right):\left(\dfrac{63}{64}-1\right)\)

\(\Rightarrow x-128=-\dfrac{1}{21}:\left(-\dfrac{1}{42}\right):\left(-\dfrac{1}{64}\right)\)

\(\Rightarrow x-128=-\dfrac{1}{21}.\left(-42\right).\left(-64\right)\)

\(\Rightarrow x-128=-128\)

\(\Rightarrow x=-128+128=0\)

6A:

Vì \(15558000< 21960000< 70560000< 85133000< 168723000\)

nên thứ tự tăng dần của các đại dương đó là:

Bắc Băng Dương; Nam Đại Dương; Ấn Độ Dương; Đại Tây Dương; Thái Bình Dương

đây là 5B mà cậu, lm 5A í

4A:

\(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\left(\dfrac{2}{193}.\dfrac{193}{17}-\dfrac{3}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right):\left(\dfrac{7}{1931}.\dfrac{1931}{25}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right)\)

\(=\left(\dfrac{2}{17}-\dfrac{3}{34}+\dfrac{33}{34}\right):\left(\dfrac{7}{25}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)

\(=\left(\dfrac{4}{34}+\dfrac{30}{34}\right):\left(\dfrac{14}{50}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)

\(=\dfrac{34}{34}:\left(\dfrac{1}{2}+\dfrac{9}{2}\right)\)

\(=1:\dfrac{10}{2}=\dfrac{1}{5}\)

4B:

a, \(A=\dfrac{1,11+0,19-13.2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\)

\(=\dfrac{1,3-26}{2,6}-\dfrac{3}{4}:2\)

\(=\dfrac{-24,7}{2,6}-\dfrac{3}{8}=\dfrac{-19}{2}-\dfrac{3}{8}=-\dfrac{79}{8}\)

\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\)

\(=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)

\(=\left(\dfrac{29}{8}-\dfrac{1}{2}\right).\dfrac{26}{75}\)

\(=\dfrac{25}{8}.\dfrac{26}{75}=\dfrac{13}{12}\)

b, Để \(A< x< B\) thì \(-\dfrac{79}{8}< x< \dfrac{13}{12}\)

\(\Rightarrow x\in\left\{-9;-8;-7;...;1\right\}\) (vì \(x\in\mathbb{Z}\))

$Toru$

a) Để \(\dfrac{2}{x-2}\) là số hữu tỉ dương thì: 

\(\left\{{}\begin{matrix}\dfrac{2}{x-2}\in\mathbb{Q}\\\dfrac{2}{x-2}>0\\x-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\in\mathbb{Q}\\x-2>0\left(\text{vì }2>0\right)\\x\ne2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\in\mathbb{Q}\\x>2\end{matrix}\right.\)

b) Để \(\dfrac{-3}{x+5}\) là số hữu tỉ dương thì:

\(\left\{{}\begin{matrix}\dfrac{-3}{x+5}\in\mathbb{Q}\\\dfrac{-3}{x+5}>0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\in\mathbb{Q}\\x+5< 0\left(\text{vì }-3< 0\right)\\x\ne-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\in\mathbb{Q}\\x< -5\end{matrix}\right.\)

c) Để \(\dfrac{12}{2x+4}\) là số hữu tỉ dương thì:

\(\left\{{}\begin{matrix}\dfrac{12}{2x+4}\in\mathbb{Q}\\\dfrac{12}{2x+4}>0\\2x+4\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\in\mathbb{Q}\\2x+4>0\left(\text{vì }12>0\right)\\x\ne-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\in\mathbb{Q}\\x>-2\end{matrix}\right.\)

tớ gửi chước câu mà sao mà sao chép được

Để \(x=\dfrac{m+3}{m-2}\) là số hữu tỉ thì: \(\left\{{}\begin{matrix}m+3\in\mathbb{Q}\\m-2\in\mathbb{Q}\\m-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m\in\mathbb{Q}\\m\ne2\end{matrix}\right.\)

\(D=10^9+10^8+10^7\)

\(=10^7\left(10^2+10+1\right)\)

\(=10^7\cdot101=10^6\cdot1010=10^6\cdot555\cdot2=10^6\cdot222\cdot5\)

=>D chia hết cho 555 và D chia hết cho 222

Ta có :

\(D=10^9+10^8+10^7\)

\(=10^7.\left(10^2+10+1\right)\)

\(=10^7.111\)

\(=10^6.5.2.111\)

\(=10^6.555.2=10^6.5.222\)

\(\Rightarrow D\) chia hết cho \(555\) và \(222\)

\(0,275+\left(\dfrac{-8}{17}\right)+\dfrac{29}{40}+\left(\dfrac{-9}{17}\right)-1\dfrac{1}{3}\)

\(=\dfrac{11}{40}-\dfrac{8}{17}+\dfrac{29}{40}-\dfrac{9}{17}-\dfrac{4}{3}\)

\(=\left(\dfrac{11}{40}+\dfrac{29}{40}\right)-\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{4}{3}\)

\(=1-1-\dfrac{4}{3}\)

\(=-\dfrac{4}{3}\)

\(0,275+\left(-\dfrac{8}{17}\right)+\dfrac{29}{40}+\left(-\dfrac{9}{17}\right)-1\dfrac{1}{3}\)

\(=\left(\dfrac{11}{40}+\dfrac{29}{40}\right)+\left(-\dfrac{8}{17}-\dfrac{9}{17}\right)-\dfrac{4}{3}\)

\(=1-1-\dfrac{4}{3}=-\dfrac{4}{3}\)