Gọi biểu thức là A

A=\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}\)

A=\(\frac{3.8.15...899}{2^2.3^2.4^2...30^2}\)

A=\(\frac{3.8.15...899}{4.9.16...900}\)

A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{899}{900}\)

Cần làm tiếp thì

a) \(x-\dfrac{3}{5}=\dfrac{4}{-10}\)

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

\(x=\dfrac{-4}{10}+\dfrac{6}{10}\)

\(x=\dfrac{1}{5}\)

b) \(\dfrac{3}{x}-2=\dfrac{4}{x}+4\)

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

\(\dfrac{3}{x}=\dfrac{4}{x}+4\)

\(\dfrac{3}{x}=\dfrac{4x+4}{x}\)

\(3x=\left(4x+4\right)x\)

\(3x=5x\cdot x+4x\)

\(3x=x\left(5x+4\right)\)

\(3=5x+4\)

\(5x=-1\)

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

Toán này lớp 6 á???

\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)

b: Ta có: \(3^x+3^{x+2}=20\)

\(\Leftrightarrow3^x\cdot10=20\)

\(\Leftrightarrow3^x=2\left(loại\right)\)

a.\(\frac{1}{4}+\frac{1}{8}\)

b.\(\frac{1}{2}+\frac{1}{4}\)

Giải ht ra nhá r mới tk .

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Sai rồi, để tôi sửa lại:

\(\frac{3}{2^2}.\frac{8}{3^2}.....\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{29.31}{30.30}\)

\(=\frac{1.2.3.....31}{2.3.4.....30}.\frac{3.4.5....29}{2.3.4....30}=31.\frac{1}{60}=\frac{31}{60}\)