\(\left(1:\dfrac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\dfrac{1}{49}\\ =7^2\left(2^6:2^5\right)\cdot\dfrac{1}{7^2}\\=\left(7^2\cdot\dfrac{1}{7^2}\right)\cdot2^{6-5}\\ =1\cdot2^1\\ =2\)

\(\left(1:\dfrac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\dfrac{1}{49}\)

\(=\dfrac{7^2}{49}\cdot\left(2^6:2^5\right)\)

\(=\dfrac{49}{49}\cdot2=2\)

\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}+\dfrac{x-4}{2008}=4\)

=>\(\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)+\left(\dfrac{x-3}{2009}-1\right)+\left(\dfrac{x-4}{2008}-1\right)=0\)

=>\(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}+\dfrac{x-2012}{2008}=0\)

=>\(\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\)

=>x-2012=0

=>x=2012

\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}+\dfrac{x-4}{2008}=4\\ \left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)+\left(\dfrac{x-3}{2009}-1\right)+\left(\dfrac{x-4}{2008}-1\right)=0\\ \dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}+\dfrac{x-2012}{2008}=0\\ \left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\\ x-2012=0\\ x=2012\)

\(\left[\left(\dfrac{4}{3}\right)^{-3}\cdot\left(\dfrac{3}{4}\right)^5\right]:\left(\dfrac{3}{8}\right)^7\\ =\left[\left(\dfrac{3}{4}\right)^3\cdot\left(\dfrac{3}{4}\right)^5\right]:\left(\dfrac{3}{8}\right)^7\\ =\left(\dfrac{3}{4}\right)^{3+5}:\dfrac{3^7}{8^7}\\ =\left(\dfrac{3}{4}\right)^8\cdot\dfrac{8^7}{3^7}\\ =\dfrac{3^8}{4^8}\cdot\dfrac{8^7}{3^7}\\ =\dfrac{3^8}{2^{16}}\cdot\dfrac{2^{21}}{3^7}=3\cdot2^5=3\cdot32=96\)

\(\left[\left(\dfrac{4}{3}\right)^{-3}\cdot\left(\dfrac{3}{4}\right)^5\right]:\left(\dfrac{3}{8}\right)^7\)

\(=\left[\left(\dfrac{3}{4}\right)^3\cdot\left(\dfrac{3}{4}\right)^5\right]:\dfrac{3^7}{8^7}\)

\(=\left(\dfrac{3}{4}\right)^8\cdot\dfrac{8^7}{3^7}=\dfrac{3^8}{4^8}\cdot\dfrac{8^7}{3^7}=\dfrac{3\cdot2^{21}}{2^{16}}=3\cdot2^5=3\cdot32=96\)

`-` Nhân vật là những nhân tố được tạo nên do tác giả nhằm tham gia cốt truyện để tạo thành một tác phẩm hoàn chỉnh

`-` Các yếu tố để bt đc nhân vật chính hay phụ:

`+` Số lần xuất hiện

`+` Sự ảnh hưởng đối vs cốt truyện đó

`+` mối quan hệ của nhân vật đó vs cái nhân vật kia

tk

Nhân vật văn học là con người được nhà văn miêu tả trong tác phẩm bằng phương tiện văn học

a: Các cặp góc đối đỉnh là: \(\widehat{xOt};\widehat{yOz}\) và \(\widehat{xOz};\widehat{yOt}\)

b: Các cặp góc kề bù là:

\(\widehat{xOt};\widehat{xOz}\)

\(\widehat{xOt};\widehat{tOy}\)

\(\widehat{zOy};\widehat{zOx}\)

\(\widehat{zOy};\widehat{tOy}\)

c: \(\widehat{xOt}+\widehat{xOz}=180^0\)(hai góc kề bù)

=>\(\widehat{xOz}+45^0=180^0\)

=>\(\widehat{xOz}=135^0\)

Ta có: \(\widehat{xOt}=\widehat{yOz}\)(hai góc đối đỉnh)

mà \(\widehat{xOt}=45^0\)

nên \(\widehat{yOz}=45^0\)

Ta có: \(\widehat{xOz}=\widehat{yOt}\)(hai góc đối đỉnh)

mà \(\widehat{xOz}=135^0\)

nên \(\widehat{yOt}=135^0\)

Ta có: xy\(\perp\)AB

x'y'\(\perp\)AB

Do đó: xy//x'y'

Bài 3:

Ta có: \(\dfrac{1}{5^2}< \dfrac{1}{4\cdot5};\dfrac{1}{6^2}< \dfrac{1}{5\cdot6};...;\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\\=> \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =>\dfrac{1}{5^2}+\dfrac{1}{6^2}+..+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\left(1\right)\)

Ta có: \(\dfrac{1}{5^2}>\dfrac{1}{5\cdot6};\dfrac{1}{6^2}>\dfrac{1}{6\cdot7};...;\dfrac{1}{100^2}>\dfrac{1}{100\cdot101}\)

\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+..+\dfrac{1}{100\cdot101}\\ =>\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ =>\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5}-\dfrac{1}{101}=\dfrac{96}{505}>\dfrac{96}{576}=\dfrac{1}{6}\left(2\right)\)

Từ (1) và (2) => \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)

He cycled carelessly and had an accident.

carelessly, mình đoán vậy vì đằng sau là had an accident 

a) Ta có: \(\widehat{cNb}+\widehat{MNb}=180^{\circ}\) (hai góc kề bù)

\(\Rightarrow\widehat{MNb}=180^{\circ}-\widehat{cNb}=180^{\circ}-55^{\circ}=125^{\circ}\)

b) Ta có: \(\widehat{MNb}=\widehat{aMN}\left(=125^{\circ}\right)\)

Mà hai góc này đều nằm ở vị trí so le trong

Nên \(Ma//Nb\)