\(\frac14-2x=5\)

\(2x=\frac14-5\)

\(2x=\frac{-19}{4}\)

\(x=-\frac{19}{4}:2\)

\(x=\frac{-19}{8}\)

\(\frac12x-\frac13=25\%\)

\(\frac12x-\frac13=\frac14\)

\(\frac12x=\frac14+\frac13\)

\(\frac12x=\frac{7}{12}\)

\(x=\frac{7}{12}:\frac12\)

\(x=\frac76\)

mik sẽ tick cho 1 người làm nhanh và đúng nhất

Ta có:

\(\left(\right. a - \frac{1}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\) (1)

Và: \(a + 1 = b + 2 = c + 3\)

\(\Rightarrow a = b + 2 - 1 = b + 1\)

Thay vào (1) ta có:
\(\left(\right. b + 1 - \frac{1}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\)

\(\Rightarrow \left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\) (2)

Mà: \(b + 2 = c + 3\)

\(\Rightarrow c = b + 2 - 3 = b - 1\) 

Thay vào (2) ta có:
\(\left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. b - 1 - 3 \left.\right) = 0\)

\(\Rightarrow \left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. b - 4 \left.\right) = 0\)

\(\Rightarrow \left[\right. b = - \frac{2}{3} \\ b = - \frac{1}{2} \\ b = 4\)

TH1 khi b=\(- \frac{2}{3}\)

\(\Rightarrow a = b + 1 = - \frac{2}{3} + 1 = \frac{1}{3}\)

\(\Rightarrow c = b - 1 = - \frac{2}{3} - 1 = - \frac{5}{3}\)

TH2 khi \(b = - \frac{1}{2}\)

\(\Rightarrow a = b + 1 = - \frac{1}{2} + 1 = \frac{1}{2}\)

\(\Rightarrow c = b - 1 = - \frac{1}{2} - 1 = - \frac{3}{2}\)

TH3 khi \(b = 4\)

\(\Rightarrow a = b + 1 = 4 + 1 = 5\)

\(\Rightarrow c = b - 1 = 4 - 1 = 3\)

sai mình xin lỗi

Câu 1:

c: \(\frac19+\frac28+\frac37+\cdots+\frac91\)

\(=\left(\frac19+1\right)+\left(\frac28+1\right)+\cdots+\left(\frac82+1\right)+1\)

\(=\frac{10}{2}+\frac{10}{3}+\cdots+\frac{10}{10}=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

Ta có: \(\left(\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\right)\cdot x=\frac19+\frac28+\frac37+\cdots+\frac91\)

=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

=>x=10

Câu 2:

d: \(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023\cdot2024}\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

Hẹ hẹ

a) \(x+2x+3x+...+100x=-213\)

\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)

\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)

\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)

d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)

                                                                    \(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)

c) \(3\left(x-2\right)+2\left(x-1\right)=10\)

\(\Rightarrow3x-6+2x-2=10\)

\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)

a) \(x+2x+3x+4x+...+100x=-213\)

\(x.\left(1+2+3+4+...+100\right)=-213\)

\(x.5050=-213\)

\(x=-\frac{213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)

\(\frac{1}{2}x=-\frac{43}{12}\)

\(x=\frac{-43}{6}\)

\(\left(25-11^2\right)\left(25-10^2\right)\left(25-9^2\right)...\left(25-5^2\right)\left(25-4^2\right)\left(25-3^2\right)\left(25-2^2\right)\left(25-1^2\right)\)

\(=\left(25-11^2\right)\left(25-10^2\right)\left(25-9^2\right)...\left(25-25\right)\left(25-4^2\right)\left(25-3^2\right)\left(25-2^2\right)\left(25-1^2\right)\)

\(=\left(25-11^2\right)\left(25-10^2\right)\left(25-9^2\right)...0.\left(25-4^2\right)\left(25-3^2\right)\left(25-2^2\right)\left(25-1^2\right)\)

\(=0\)

\(\left(25-11^2\right)\left(25-10^2\right)\left(25-9^2\right)...\left(25-5^2\right)\left(25-4^2\right)\left(25-3^2\right)\left(25-2^2\right)\left(25-1^2\right)\)

\(=\left(25-11^2\right)\left(25-10^2\right)\left(25-9^2\right)...\left(25-25\right)\left(25-4^2\right)\left(25-3^2\right)\left(25-2^2\right)\left(25-1^2\right)\)

\(=\left(25-11^2\right)\left(25-10^2\right)\left(25-9^2\right)....0.\left(25-4^2\right)\left(25-3^2\right)\left(25-2^2\right)\left(25-1^2\right)\)

\(=0\)

Cậu thậc thú zị :v

một câu hỏi rất đáng khen ,.. very good!

Đặt A=(25-1^2).(25-2^2).....(25-100^2)

   Vì trong tích A trên có chứa thừa số 25-5^2

Mà 25-5^2=25-25=0

     một số nào nhân với 0 cũng bằng 0

nên A=0

Vậy.....

Chúc bạn hok tốt!!!

Bạn nhớ k đúng cho mik nha!!!

C