Đừng chê chữ của tui nữa nha

Thử xem -1 đi đc k ?

Bài 3:

a, Ta có:

\(\dfrac{4n+2}{2n+6}=\dfrac{4n+12-10}{2n+6}=\dfrac{4n+12}{2n+6}-\dfrac{10}{2n+6}\)

Để \(4n+2⋮2n+6\) thì \(10⋮2n+6\)

\(\Rightarrow2n+6\inƯ\left(10\right)\)

\(\Rightarrow2n+6\in\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

mà \(2n+6\) là số chăn nên

\(2n+6\in\left\{-10;-2;2;10\right\}\)

\(\Rightarrow2n\in\left\{-16;-8;-4;4\right\}\)

\(\Rightarrow n\in\left\{-8;-4;-2;2\right\}\)

Vậy \(n\in\left\{-8;-4;-2;2\right\}\) thì \(4n+2⋮2n+6\)

Chúc bạn học tốt!!!

b, Ta có:

\(2^{15}+16^5=2^{15}+\left(2^4\right)^5=2^{15}+2^{4.5}\)

\(=2^{15}+2^{20}=2^{15}.\left(2^5+1\right)=2^{15}.33\)

\(=2^{15}.3.11\)

Vì \(3⋮3\);\(11⋮11\) nên \(2^{15}.3.11⋮3\) và \(⋮11\)

\(\Rightarrow2^{15}+16^5\) chia hết cho 3 và 11 (đpcm)

Chúc bạn học tốt!!!

a, Ta có: \(\left\{{}\begin{matrix}xy=\dfrac{1}{3}\\yz=\dfrac{-2}{5}\\xz=\dfrac{-3}{10}\end{matrix}\right.\Rightarrow x^2y^2z^2=\dfrac{1}{25}\Rightarrow xyz=\pm\dfrac{1}{5}\)

+) Xét \(xyz=\dfrac{1}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}xyz:xy=\dfrac{1}{5}:\dfrac{1}{3}\\xyz:yz=\dfrac{1}{5}:\dfrac{-2}{5}\\xyz:xz=\dfrac{1}{5}:\dfrac{-3}{10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=\dfrac{3}{5}\\x=\dfrac{-1}{2}\\y=\dfrac{-2}{3}\end{matrix}\right.\)

+) Xét \(xyz=\dfrac{-1}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}z=\dfrac{-3}{5}\\x=\dfrac{1}{2}\\y=\dfrac{-2}{3}\end{matrix}\right.\)

Vậy....

b, Ta có: \(\left\{{}\begin{matrix}x+y=\dfrac{-7}{6}\\y+z=\dfrac{1}{4}\\x+z=\dfrac{1}{12}\end{matrix}\right.\Rightarrow2\left(x+y+z\right)=-\dfrac{5}{6}\)

\(\Rightarrow x+y+z=\dfrac{-5}{12}\)

\(\Rightarrow\left\{{}\begin{matrix}z=\dfrac{-5}{12}+\dfrac{7}{6}\\x=\dfrac{-5}{12}-\dfrac{1}{4}\\y=-\dfrac{5}{12}-\dfrac{1}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=\dfrac{3}{4}\\x=\dfrac{-2}{3}\\y=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy...

Câu a

a, \(A=1+3+3^2+...+3^{2012}\)

\(\Rightarrow3A=3+3^2+...+3^{2013}\)

\(\Rightarrow3A-A=\left(3+3^2+...+3^{2013}\right)-\left(1+3+...+3^{2012}\right)\)

\(\Rightarrow2A=3^{2013}-1\)

\(\Rightarrow A=\dfrac{3^{2013}-1}{2}< 3^{2013}\)

\(\Rightarrow A< B\)

b, \(A=1+5+5^2+...+5^{100}\)

\(\Rightarrow5A=5+5^2+...+5^{101}\)

\(\Rightarrow5A-A=\left(5+5^2+...+5^{101}\right)-\left(1+5+...+5^{100}\right)\)

\(\Rightarrow4A=5^{101}-1\)

\(\Rightarrow A=\dfrac{5^{101}-1}{4}< \dfrac{5^{101}}{4}\)

\(\Rightarrow A< B\)

Dễ òm

Bài 2.6:

\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{2.1}\)

\(C=-\left(\dfrac{-1}{100}+\dfrac{1}{100.99}+......+\dfrac{1}{2.1}\right)\)

\(C=-\left(\dfrac{-1}{100}+\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+....+\dfrac{1}{2}-\dfrac{1}{1}\right)\)

\(C=-1\)

Chúc bạn học tốt!!!

2.4:

\(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\left(\dfrac{1}{3}-\dfrac{2}{9}\right)-\dfrac{3}{4}+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{1}{72}-\dfrac{1}{36}\right)\)

\(=\left(\dfrac{3}{9}-\dfrac{2}{9}\right)-\dfrac{3}{4}+\left(\dfrac{9}{15}+\dfrac{1}{15}\right)+\left(\dfrac{1}{72}-\dfrac{2}{72}\right)\)

\(=\dfrac{1}{9}-\dfrac{3}{4}+\dfrac{10}{15}+\left(-\dfrac{1}{72}\right)\)

\(=\dfrac{1}{9}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{72}\)

\(=\dfrac{1}{72}\)

Câu hỏi của Monkey D.Luffy - Toán lớp 7 - Học toán với OnlineMath

mk thấy chưa chi tiết lắm

O X Y

Thưa! Cách vẽ đâu và hình cũng sai luôn

Cái này là kiến thức cơ bản nhất của lớp 7 rồi bạn, tự làm đi sẽ tốt hơn vì nó rất dễ.

Hoàng Ngọc Anh thanks