Question

[Ye Chi-Lien]

\(\frac{1}{4}\)\(+\frac{1}{28}\)\(+\frac{1}{70}\)\(+\frac{1}{130}\)\(+\frac{1}{x\cdot\left(x+3\right)}\)

( ai onl giúp vs ạ, tick cho 5 bn đầu, giải ra hộ mik nhoa, thanks !!!)

Answer 1

[ミ★ṡѧṅєṃı{ღṭєѧṃღғѧṅღҡṅʏღ}★彡]

CÁCH LÀM NHƯ SAU : 

(7/28 + 1/28) + 1/70 + 1/130 + 1/x.(x+3)

8/28 + 1/70 +1/130 +1/x.(x+3)

2/7+1/70+1/130+1/x.(x+3)

(20/70 +1/70)+1/130+1/x.(x+3)

3/10+1/130+1/x.(x+3)

39/130+1/130+1/x.(x+3)

4/13+1/x.(x+3)

Đến đây bn tự làm hộ mình vớ. chúc hok tốt k cho mình nhé

Answer 2

[༺༒༻²ᵏ⁸]

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{x\left(x+3\right)}\)

\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{x\left(x+3\right)}\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(=\frac{1}{3}\left(\frac{12}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(=\frac{1}{3}.\frac{12}{13}+\frac{1}{3}.\frac{1}{x}-\frac{1}{3}.\frac{1}{x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3}.\frac{1}{x+3}\)

\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3x}\)

\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3}.\frac{1}{x}\)

\(=\frac{4}{13}=\frac{1}{3}\left(\frac{1}{x+3}-\frac{1}{x}\right)\)

\(=\frac{4}{13}:\frac{1}{3}=\frac{1}{x+1}-\frac{1}{x}\)

\(=\frac{12}{13}=\frac{1}{x+1}-\frac{1}{x}\)

\(=\frac{12}{13}=\frac{x-\left(x+1\right)}{\left(x+1\right)x}\)

\(=\frac{12}{13}=-\frac{1}{x^2+x}\)

\(\Leftrightarrow=12\left(x^2+x\right)=13.\left(-1\right)\)

\(=12\left(x^2+x\right)=-13\)

\(=x^2+x=-\frac{13}{12}\)

\(=x\left(x+1\right)=-\frac{13}{12}\)

.... Chiụ