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\(120\left(\dfrac{1}{24.25}+\dfrac{1}{25.26}+...+\dfrac{1}{29.30}\right)+x=\dfrac{1}{3}\)
\(\Rightarrow120\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}\right)+x=\dfrac{1}{3}\)
\(\Rightarrow120\left(\dfrac{1}{24}-\dfrac{1}{30}\right)+x=\dfrac{1}{3}\)
\(\Rightarrow120.\dfrac{1}{120}+x=\dfrac{1}{3}\)
\(\Rightarrow1+x=\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{2}{3}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}=\frac{1}{5}-\frac{1}{25}=\frac{4}{25}\)
(1/24.25 + 1/25.26 + ... + 1/29.30) . 120 + x : 1/3 = -4
(1/24 - 1/25 + 1/25 - 1/26 + ... + 1/29 - 1/30) . 120 + x . 3 = -4
(1/24 - 1/30) . 120 + x . 3 = -4
(5/120 - 4/120) + x . 3 = -4
=> 1/120 . 120 + x . 3 = -4
=> 1 + x . 3 = -4
=> x . 3 = -4 - 1
=> x . 3 = -5
=> x = -5/3
Vậy x = -5/3
\((\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+y:\frac{1}{3}=-4\)
Ta có: \(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}\)
\(=\frac{1}{120}\)
Thay vào ta có: \(\frac{1}{120}.120+y:\frac{1}{3}=-4\)
\(\implies 1+y:\frac{1}{3}=-4\)
\(\implies y:\frac{1}{3}=-5\)
\(\implies y=-5.\frac{1}{3}\)
\(\implies y=\frac{-5}{3}\). Vậy \(y=\frac{-5}{3}\)
~ Học tốt a~
\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\frac{1}{120}.120+y:\frac{1}{3}=-4\)
\(\Rightarrow1+y:\frac{1}{3}=-4\)
\(\Rightarrow y:\frac{1}{3}=-4-1\)
\(\Rightarrow y:\frac{1}{3}=-5\)
\(\Rightarrow y=-5.\frac{1}{3}\)
\(\Rightarrow y=\frac{-5}{3}\)
Vậy \(y=\frac{-5}{3}\)
_Chúc bạn học tốt_
Ta có : \(\dfrac{x^2}{5.6}\text{=}\dfrac{x^2}{5}-\dfrac{x^2}{6}\)
\(\dfrac{x^2}{6.7}\text{=}\dfrac{x^2}{6}-\dfrac{x^2}{7}\)
\(...\)
\(\dfrac{x^2}{24.25}\text{=}\dfrac{x^2}{24}-\dfrac{x^2}{25}\)
\(\Rightarrow\) biểu thức chỉ còn :
\(\dfrac{x^2}{5}-\dfrac{x^2}{25}\text{=}\dfrac{5x^2-x^2}{25}\text{=}\dfrac{4x^2}{25}\)
Lời giải:
\(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}=\frac{25-24}{24.25}+\frac{26-25}{25.26}+...+\frac{30-29}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}=\frac{1}{120}\)
Vậy:
\(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(1+x:\frac{1}{3}=-4\)
\(x:\frac{1}{3}=-5\)
\(x=-15\)
\(\left(\dfrac{1}{24.25}+\dfrac{1}{25.26}+...+\dfrac{1}{29.30}\right).120+x:\dfrac{1}{3}=-4\)
\(\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}\right).120+x:\dfrac{1}{3}=-4\)
\(\left(\dfrac{1}{24}-\dfrac{1}{30}\right).120+x:\dfrac{1}{3}=-4\)
\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\)
\(1+x:\dfrac{1}{3}=-4\)
\(x:\dfrac{1}{3}=-4-1\)
\(x:\dfrac{1}{3}=-5\)
\(x=-5.\dfrac{1}{3}\)
\(x=\dfrac{-5}{3}\)
= 25.(124-24) + 212
= 25 . 100 +212
= 2500 +212
= 2712
25.124 - 24.25 + 212
= 25(124 - 24) + 212
= 25. 100 + 212
= 2500 + 212
= 2712