Tính nhanh:

\(A=\frac{2}{1+2}+2+\frac{3}{12+3}+...+2+3+\frac{20}{1+2+3+...+20}\)

Đặt \(A=\frac{2}{1+2}+2+\frac{3}{12+3}+...+2+3+\frac{20}{1+2+3+...+20}\)

\(=2-1+2+\frac{3}{12+3}+...+2+3+\frac{20}{1+2+3+...+20}\)

\(=\) Không biết! Nhờ Doraeiga  với At the speed of light - Trang của At the speed of light - Học toán với OnlineMath giải nhé! Tui mới lớp 6 thôi! Chưa học tới bài này

\(A=\frac{2}{1+2}+\frac{2+3}{1+2+3}+....+\frac{2+3+...+20}{1+2+3+...+20}\)

\(A=\frac{2}{3}+\frac{5}{6}+...+\frac{209}{210}\)

\(A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{210}\right)\)

\(A=\left(1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{6}+....+\frac{1}{210}\right)\)

\(A=19-\left(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{420}\right)\)

\(A=19-\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\right)\)

\(A=19-\left[2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\right]\)

\(A=19-\left[2\cdot\left(\frac{1}{2}-\frac{1}{21}\right)\right]\)

\(A=19-\left[2\cdot\frac{19}{42}\right]=19-\frac{19}{21}=\frac{380}{21}\)

Vậy A = .....

bài toán này ko phải như vậy bạn nhé! phần tổng ko có dấu ngoặc chính là tử số của phần có dấu ngoặc đằng sau nó nhé!!

a, (3 - \(x\))(4y + 1) = 20

   Ư(20) = { -20; -10; -5; -4; -2; -1; 1; 2; 4; 5; 10; 20}

Lập bảng ta có:

Vậy các cặp \(x;y\) nguyên thỏa mãn đề bài là:

(\(x;y\)) =(-1; 1); (-17; 0)

b, \(x\left(y+2\right)\)+ 2\(y\) = 6

    \(x\) = \(\dfrac{6-2y}{y+2}\)

\(x\in\) Z ⇔ 6 - \(2y⋮\) \(y\) + 2 ⇒-(2y + 4) +10 ⋮ \(y\) + 2 ⇒ -2(\(y\)+2) +10 ⋮ \(y\)+2

⇒ 10 ⋮ \(y\) + 2

Ư(10) = { -10; -5; -2; -1; 1; 2; 5; 10}

Lập bảng ta có:

 Theo bảng trên ta có các cặp \(x;y\)

 nguyên thỏa mãn đề bài lần lượt là:

(\(x;y\)    ) =(-3; -12); (-4; -7); (-12; -3); (8; -1); (3; 0); (0;3 (-1; 8)                           

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

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

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)

b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)

\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)

c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)

\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)

\(\Rightarrow x=-2\)

d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)

\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{25}{9}\)

e) \(\dfrac{1}{2}x+650\%x-x=-6\)

\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=\dfrac{-6}{6}=-1\)

g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)

\(\Rightarrow2x-1-3+x=2-x\)

\(\Rightarrow3x-4=2-x\)

\(\Rightarrow3x+x=2+4\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)

h) \(x-\dfrac{2}{20}=-\dfrac{5}{2}-x\)

\(\Rightarrow x+x=-\dfrac{5}{2}+\dfrac{2}{20}\)

\(\Rightarrow2x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}:2=-\dfrac{6}{5}\)

i) \(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\)

\(\Rightarrow\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)

\(\Rightarrow\dfrac{x}{2}-1=\sqrt[3]{-\dfrac{27}{8}}\)

\(\Rightarrow\dfrac{x}{2}-1=-\dfrac{3}{2}\)

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

\(\Rightarrow x=-\dfrac{1}{2}.2=-1\)

k) \(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{3}{4}-1\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.3=-\dfrac{3}{2}\\x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\end{matrix}\right.\)

TÌM X À

Đunga rồi >< mình nhầm. Đây toán lớp 7 ><

BEAR CHỊU

2 + 7 =18

~hok tốt~

#Trang#

2+7=18 hoặc 2+7=9

hok tốt

Cái này là cố ý đăng linh tinh đấy chứ,bớt xàm,bớt zả dối

1) \(\frac{45^{20}\cdot20^{10}}{3^{15}\cdot6^3}=\frac{3^{40}\cdot5^{20}\cdot5^{10}\cdot2^{20}}{3^{15}\cdot2^3\cdot3^3}\)

\(=\frac{2^{20}\cdot3^{40}\cdot5^{30}}{2^3\cdot3^{18}}=2^{17}\cdot3^{22}\cdot5^{30}\)

2) Ta có: \(2^{2009}+2^{2008}+...+2^1+2^0\)

\(=2^{2010}-1\) đã CM ở rất nhiều bài rồi

=> \(2^{2010}-2^{2010}+1=1\)