\(a)\)\(\left[\left(\frac{1}{9}\div\frac{8}{27}\right)\div-\frac{1}{3}\right]\div\frac{81}{128}\)

\(=\)\(\left[\left(\frac{1}{9}.\frac{27}{8}\right)\div-\frac{1}{3}\right]\div\frac{81}{128}\)

\(=\)\(\left[\frac{3}{8}\div-\frac{1}{3}\right]\div\frac{81}{128}\)

\(=\)\(\left[\frac{3}{8}.-3\right]\div\frac{81}{128}\)

\(=\)\(-\frac{9}{8}\div\frac{81}{128}\)

\(=\)\(-\frac{9}{8}.\frac{128}{81}\)

\(=\)\(-\frac{16}{9}\)

\(b)\)\(\left(-\frac{7}{15}\right).\frac{5}{8}.\left(\frac{15}{-7}\right).\left(-32\right)\)

\(=\)\(\frac{-7.5.15}{15.8.-7}.\left(-32\right)\)

\(=\)\(\frac{5}{8}.\left(-32\right)\)

\(=\)\(-20\)

 điền dấu <

Ta có : \(\frac{2020}{2021}=\frac{2020.2022}{2021.2022}=\frac{\left(2021-1\right)\left(2021+1\right)}{2021.2022}=\frac{2021^2-1}{2021.2022}\)

\(\frac{2021}{2022}=\frac{2021^2}{2021.2022}\)

Vì 2021² > 2021² - 1 nên \(\frac{2021^2-1}{2021.2022}< \frac{2021^2}{2021.2022}\)

Hay \(\frac{2020}{2021}< \frac{2021}{2022}\)

\(VT=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{1018081}=\)

\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}=\)

\(=\frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{1009-1008}{1008.1009}=\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1008}-\frac{1}{1009}=\frac{3}{4}-\frac{1}{1009}< \frac{3}{4}\)

bài này âm bạn ạ

Ta có: 5x=2y⇒2x=5y5x=2y⇒2x=5y(1)

3y=5z⇒5y=3z3y=5z⇒5y=3z (2)

Từ (1) và (2) ,đặt: 2x=5y=3z=k⇒⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x=2k=2288y=5k=5288z=3k=32882x=5y=3z=k⇒{x=2k=2288y=5k=5288z=3k=3288 (3)

Từ (1) và (2) theo tính chất tỉ dãy số bằng nhau ,ta có:

2x=5y=3z=2−5+3x−y+z=02882x=5y=3z=2−5+3x−y+z=0288(4)

Suy ra k = 288. Dựa và (3) ta có: ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x=2k=2288y=5k=5288z=3k=3288{x=2k=2288y=5k=5288z=3k=3288

Vậy .....

này áp dụng tính chất dãy tỉ số bằng nhau

\(M=\left|x-2021\right|+\left|x-2020\right|=\left|2021-x\right|+\left|x-2020\right|\)

Ta có: \(\hept{\begin{cases}\left|2021-x\right|\ge2021-x\\\left|x-2020\right|\ge x-2020\end{cases}}\Rightarrow M\ge2021-x+x-2020=1\)

Dấu '' = '' xảy ra khi: \(\hept{\begin{cases}2021-x\ge0\\x-2020\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\le2021\\x\ge2020\end{cases}}\Rightarrow2020\le x\le2021\)