Mik lười quá bạn tham khảo câu 3 tại đây nhé:

Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath

\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)

\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)

\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)

\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)

1) \(\frac{2}{3}+x=-\frac{4}{5}\)

\(x=\left(-\frac{4}{5}\right)-\frac{2}{3}\)

\(x=-1\frac{7}{15}\)

Vậy \(x=-1\frac{7}{15}\)

2) \(\frac{2}{5}-x=-\frac{1}{3}\)

\(x=\frac{2}{5}-\left(-\frac{1}{3}\right)\)

\(x=\frac{11}{15}\)

Vậy \(x=\frac{11}{15}\)

3) \(1-\frac{x}{3}=1\frac{1}{2}\)

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

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

\(\Rightarrow x=\frac{\left(-1\right)\cdot3}{2}\)

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

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

\(\frac{2x}{3}+2=1-\left(-1\right)\)

\(\frac{2x}{3}+2=2\)

\(\frac{2x}{3}=2-2\)

\(\frac{2x}{3}=0\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)

\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)

\(=\frac{4}{3}.\frac{1}{3}+4\)

\(=4+4\)

\(=8\)

đọc cách lm trrong sbt nha bạn lớp 8

mk học lớp 6 lên 7

a) \(\frac{x}{3}-\frac{1}{4}=\frac{-5}{6}\)

\(\Rightarrow\frac{x}{3}=\frac{-5}{6}+\frac{1}{4}\)

\(\Rightarrow\frac{x}{3}=\frac{-7}{12}\)

\(\Rightarrow x=\frac{\left(-7\right).3}{12}\)

\(\Rightarrow x=\frac{-7}{4}\)

Vậy x = \(\frac{-7}{4}\)

b) \(\frac{x+3}{15}=\frac{1}{3}\)

\(\Rightarrow\left(x+3\right).3=15\)

\(\Rightarrow x+3=15:3\)

\(\Rightarrow x+3=5\)

\(\Rightarrow x=5-3\)

\(\Rightarrow x=2\)

Vậy x = 2

c) \(\frac{x-12}{4}=\frac{1}{2}\)

\(\Rightarrow\left(x-12\right).2=4\)

\(\Rightarrow x-12=4:2\)

\(\Rightarrow x-12=2\)

\(\Rightarrow x=2+12\)

\(\Rightarrow x=14\)

Vậy x = 14

d) \(\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\)

\(\Rightarrow\frac{12}{13}x=\frac{25}{9}-\frac{7}{9}\)

\(\Rightarrow\frac{12}{13}x=2\)

\(\Rightarrow x=2:\frac{12}{13}\)

\(\Rightarrow x=\frac{13}{6}\)

Vậy x = \(\frac{13}{6}\)

_Chúc bạn học tốt_

a)\(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)

\(\frac{x}{3}=-\frac{5}{6}+\frac{1}{4}\)

\(\frac{x}{3}=-\frac{7}{12}\)

\(x=-\frac{7}{12}\times3\)

\(\Rightarrow x=-\frac{7}{4}\)

b) \(\frac{x+3}{15}=\frac{1}{3}\)

\(\frac{x+3}{15\div5}=\frac{1}{3}\)

\(\Rightarrow x=5-3\)

\(\Rightarrow x=2\)

c) \(\frac{x-12}{4}=\frac{1}{2}\)

\(\frac{x-12}{4\div2}=\frac{1}{2}\)

\(x=12+2\)

\(\Rightarrow x=14\)

d) tự làm nhé cũng dễ mà

giúp mik nha chiều này 6:00 mik nộp rồi

ai nhanh mik sẽ k cho 3 k

\(2\frac{3}{5}x-\frac{1}{7}=1\frac{9}{35}\)

\(\frac{13}{5}x=\frac{44}{35}+\frac{1}{7}\)

\(\frac{13}{5}x=\frac{7}{5}\)

\(x=\frac{7}{5}:\frac{13}{5}\\ x=\frac{7}{13}\)

\(A>\left(\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}\right)+\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)\) (mỗi ngoặc có 50 số hạng)

\(;A>\left(\frac{1}{150}.50\right)+\left(\frac{1}{200}.50\right)=50.\left(\frac{1}{150}+\frac{1}{200}\right)=50.\frac{7}{600}=\frac{7}{12}\)

Mấy câu trên dễ , bạn có thể tự làm được 

Chứng minh \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< 1\)

Đặt  \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}\)

Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)

...

\(\frac{1}{10^2}=\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)

=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1}-\frac{1}{10}\)

=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{9}{10}\)

Lại có : \(\frac{9}{10}< 1\)

=> \(A< \frac{9}{10}< 1\)

=> \(A< 1\left(đpcm\right)\)