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 \(222\)

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

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

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

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

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

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

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

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

\(\dfrac{-3}{100}=\dfrac{-3\cdot3}{100\cdot3}=\dfrac{-9}{100};\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot100}{3\cdot100}=\dfrac{-200}{300}\)

mà -9>-200

nên \(\dfrac{-3}{100}>\dfrac{2}{-3}\)

Ta có : 

+) \(\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-6}{9}\)

+) \(\dfrac{-3}{100}=\dfrac{-6}{200}\)

Vì \(9< 200\Rightarrow\dfrac{6}{9}>\dfrac{6}{200}\Rightarrow\dfrac{-6}{9}< \dfrac{-6}{200}\)

\(\Rightarrow\dfrac{2}{-3}< \dfrac{-3}{100}\)

Bài 3:

a) 

\(\dfrac{2}{3}x+4=-12\\ \Rightarrow\dfrac{2}{3}x=-12-4=-16\\ \Rightarrow x=-16:\dfrac{2}{3}\Rightarrow x=-24\)

b) \(\dfrac{5}{6}-\dfrac{1}{6}:x=-2\dfrac{1}{2}\)

\(\Rightarrow\dfrac{5}{6}-\dfrac{1}{6}:x=-\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{6}:x=\dfrac{5}{6}+\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{6}:x=\dfrac{14}{6}\\ \Rightarrow x=\dfrac{1}{6}:\dfrac{14}{6}=\dfrac{1}{14}\) 

c) \(x:20-25\%x=-1\dfrac{1}{5}\)

\(\Rightarrow\dfrac{x}{20}-\dfrac{x}{4}=-\dfrac{4}{5}\)

\(\Rightarrow\dfrac{x}{20}-\dfrac{5x}{20}=-\dfrac{4}{5}\)

\(\Rightarrow-\dfrac{4x}{20}=\dfrac{-4}{5}\)

\(\Rightarrow-\dfrac{x}{5}=-\dfrac{4}{5}\)

\(\Rightarrow x=-4\)

d) \(\dfrac{35}{x-1}=\dfrac{15}{-6}\left(x\ne1\right)\)

\(\Rightarrow-6\cdot35=15\left(x-1\right)\\ \Rightarrow-210=15x-15\\ \Rightarrow15x=-210+15=-195\\ \Rightarrow x=\dfrac{-195}{15}\\ \Rightarrow x=-13\)

e) 

\(\left(\dfrac{9}{25}\right)^x=\left(\dfrac{5}{3}\right)^{-1}\cdot\left(\dfrac{3}{5}\right)^5\\ \Rightarrow\left[\left(\dfrac{3}{5}\right)^2\right]^x=\dfrac{3}{5}\cdot\left(\dfrac{3}{5}\right)^5\\ \Rightarrow\left(\dfrac{3}{5}\right)^{2x}=\left(\dfrac{3}{5}\right)^6\\ \Rightarrow2x=6\\ \Rightarrow x=3\)

f) 

\(0,5^{x+1}+0,5^x=1,5\\ \Rightarrow0,5^x\cdot\left(0,5+1\right)=1,5\\ \Rightarrow0,5^x\cdot1,5=1,5\\ \Rightarrow0,5^x=1,5:1,5=1\\ \Rightarrow0,5^x=0,5^0\\ \Rightarrow x=0\)

Bài 3:

a: \(\dfrac{2}{3}x+4=-12\)

=>\(\dfrac{2}{3}x=-12-4=-16\)

=>\(x=-16:\dfrac{2}{3}=-16\cdot\dfrac{3}{2}=-24\)

b: \(\dfrac{5}{6}-\dfrac{1}{6}:x=-2\dfrac{1}{2}\)

=>\(\dfrac{5}{6}-\dfrac{1}{6}:x=-\dfrac{5}{2}\)

=>\(\dfrac{1}{6}:x=\dfrac{5}{6}+\dfrac{5}{2}=\dfrac{5}{6}+\dfrac{15}{6}=\dfrac{20}{6}=\dfrac{10}{3}\)

=>\(x=\dfrac{1}{6}:\dfrac{10}{3}=\dfrac{1}{6}\cdot\dfrac{3}{10}=\dfrac{1}{20}\)

c: \(x:20-25\%\cdot x=-1\dfrac{1}{5}\)

=>\(0,05x-0,25x=-1,2\)

=>-0,2x=-1,2

=>x=1,2:0,2=6

d: \(\dfrac{35}{x-1}=\dfrac{15}{-6}\)(ĐKXĐ: \(x\ne1\))

=>\(x-1=\dfrac{35\cdot\left(-6\right)}{15}=\dfrac{-210}{15}=-14\)

=>x=-14+1=-13(nhận)

e: \(\left(\dfrac{9}{25}\right)^x=\left(\dfrac{5}{3}\right)^{-1}\cdot\left(\dfrac{3}{5}\right)^5\)

=>\(\left(\dfrac{3}{5}\right)^{2x}=\left(\dfrac{3}{5}\right)\cdot\left(\dfrac{3}{5}\right)^5=\left(\dfrac{3}{5}\right)^6\)

=>2x=6

=>x=3

f: \(0,5^{x+1}+0,5^x=1,5\)

=>\(0,5^x\cdot\left(0,5+1\right)=1,5\)

=>\(0,5^x=1\)

=>x=0

Bài 2:

a: \(\dfrac{-23}{32}+\dfrac{14}{21}+\dfrac{-9}{32}+\dfrac{28}{21}\)

\(=\left(-\dfrac{23}{32}-\dfrac{9}{32}\right)+\left(\dfrac{14}{21}+\dfrac{28}{21}\right)\)

\(=-\dfrac{32}{32}+\dfrac{42}{21}=-1+2=1\)

b: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)\)

\(=\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{3}{12}\right)\cdot\left(\dfrac{16}{20}-\dfrac{15}{20}\right)\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{20}=\dfrac{17}{240}\)

c: \(\dfrac{-5}{7}\cdot16\dfrac{1}{3}+\dfrac{5}{7}\cdot\left(-23\dfrac{2}{3}\right)\)

\(=\dfrac{-5}{7}\cdot\left(16+\dfrac{1}{3}+23+\dfrac{2}{3}\right)\)

\(=-\dfrac{5}{7}\cdot40=-\dfrac{200}{7}\)

d: \(6\dfrac{4}{9}:\dfrac{7}{2}+7\dfrac{5}{9}:\left(\dfrac{2}{7}\right)^{-1}-\dfrac{1}{2}\)

\(=\left(6+\dfrac{4}{9}\right)\cdot\dfrac{2}{7}+\left(7+\dfrac{5}{9}\right)\cdot\dfrac{2}{7}-\dfrac{1}{2}\)

\(=\dfrac{2}{7}\left(6+\dfrac{4}{9}+7+\dfrac{5}{9}\right)-\dfrac{1}{2}=\dfrac{2}{7}\cdot14-\dfrac{1}{2}=4-\dfrac{1}{2}=\dfrac{7}{2}\)

e: \(\left(2^3:\dfrac{1}{2}\right)\cdot\dfrac{1}{8}+\dfrac{1}{9}\cdot\left(-3\right)^2-\left(-\dfrac{1}{2015}\right)^0\)

\(=\left(8\cdot2\right)\cdot\dfrac{1}{8}+\dfrac{1}{9}\cdot9-1=2+1-1=2\)

a) Ta có: `3/4=0,75` và `3/5=0,6` 

Ta cần viết các số hữu tỉ nhỏ hơn 0,75 và lớn hớn 0,6 

\(0,65=\dfrac{13}{20}\)

\(0,7=\dfrac{7}{10}\)

\(0,72=\dfrac{18}{25}\) 

b) Ta có: `2/3=0,66...` và `3/4=0,75`

Ta cần viết các số hữu tỉ nhỏ hơn 0,75 và lớn hơn 0,66... 

\(0,68=\dfrac{17}{25}\)

\(0,7=\dfrac{7}{10}\)

\(0,72=\dfrac{18}{25}\)

a) 

\(\dfrac{267}{268}< 1\Rightarrow-\dfrac{267}{268}>-1\)

\(\dfrac{1347}{1343}>1\Rightarrow-\dfrac{1347}{1343}< -1\)

\(\Rightarrow-\dfrac{1347}{1343}< -\dfrac{267}{268}\)

b) \(\dfrac{2022\cdot2023-1}{2022\cdot2023}=\dfrac{2022\cdot2023}{2022\cdot2023}-\dfrac{1}{2022\cdot2023}=1-\dfrac{1}{2022\cdot2023}\)

\(\dfrac{2023\cdot2024-1}{2023\cdot2024}=\dfrac{2023\cdot2024}{2023\cdot2024}-\dfrac{1}{2023\cdot2024}=1-\dfrac{1}{2023\cdot2024}\)

Vì: \(2022\cdot2023< 2023\cdot2024\)

\(\Rightarrow\dfrac{1}{2022\cdot2023}>\dfrac{1}{2023\cdot2024}\)

\(\Rightarrow1-\dfrac{1}{2022\cdot2023}< 1-\dfrac{1}{2023\cdot2024}\)

Hay: `(2022*2023-1)/(2022*2023) < (2023*2024 - 1)/(2023*2024)` 

c) \(\dfrac{2022\cdot2023}{2022\cdot2023+1}=\dfrac{2023\cdot2023+1-1}{2022\cdot2023+1}=1-\dfrac{1}{2022\cdot2023+1}\)

\(\dfrac{2023\cdot2024}{2023\cdot2024+1}=\dfrac{2023\cdot2024+1-1}{2023\cdot2024+1}=1-\dfrac{1}{2023\cdot2024+1}\)
Vì: \(2022\cdot2023+1< 2023\cdot2024+1\)

\(\Rightarrow\dfrac{1}{2022\cdot2023+1}>\dfrac{1}{2023\cdot2024+1}\)

\(\Rightarrow1-\dfrac{1}{2022\cdot2023+1}< 1-\dfrac{1}{2023\cdot2024+1}\)

Hay: `(2022*2023)/(2022*2023+1)<(2023*2024)/(2023*2024+1)`