1. 25.124-24.25+212
2. 80-4.5^2+3.2^3
3.35.48+65.68+20.35
4.2^7:2^3+2^3.2^0-1^10
a/ 53 : 52 + 96 c/ 17.75 + 17.25 – 125 e/ 21.16 + 37.21 + 21.63
b/ 25.7.5.4.2 d/ 25.124 – 24.25 + 212 f/ 35.48 + 65.68 + 20.35
h/ 711: 79 - 62 + 23.2 g/ 117:{[79 – 3(33 – 17 )]:7 + 2} m/ 27:23 + 23.20 - 110
n/ 52 + [250 – 150 :(35.22 – 90)] p/ 3978 : [359 – (27:24 + 39.8]
q/ 12 + 15 + 18 +…+ 90 k/ 8 + 12 + 16 +…+ 100 l/ 7 + 11 +15 +…+ 43 + 47
r/ 1 + 6 +11 + …+ 46 +51 t/ 20 – [30 – (5 – 1)2] v/80 –[130 – (12 -4)2]
s/ 80 – 4.52 + 3.23 o/ 140:{80 – [130 – (12 -4)2]}
đỉ mẹ, đỉ má, cái lồn, con cặc.
(1/24.25+1/25.26+....+1/29.30).120+x=1/3
\(120\left(\dfrac{1}{24.25}+\dfrac{1}{25.26}+...+\dfrac{1}{29.30}\right)+x=\dfrac{1}{3}\)
\(\Rightarrow120\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}+\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}\right)+x=\dfrac{1}{3}\)
\(\Rightarrow120\left(\dfrac{1}{24}-\dfrac{1}{30}\right)+x=\dfrac{1}{3}\)
\(\Rightarrow120.\dfrac{1}{120}+x=\dfrac{1}{3}\)
\(\Rightarrow1+x=\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{2}{3}\)
Tính tổng 1/5.6+1/6.7+1/7.8+...+1/24.25
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}=\frac{1}{5}-\frac{1}{25}=\frac{4}{25}\)
1/5.6+1/6.7+1/7.8+...+1/24.25
= 1/5-1/6+1/6+1/7-1/8+...+1/24-1/25
= 1/5-1/25
= 5/25-1/25
= 4/25
k mik nhé mọi người
{1\24.25 + 1\25.26 +...+ 1\29.30} .120+x:1\3 = -4
(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
{1\24.25 + 1\25.26 +...+ 1\29.30} .120+x:1\3 = -4
tìm y :(1/24.25 + 1/25.26 + ... + 1/29.30) . 120 + y : 1/3 = -4
\((\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+y:\frac{1}{3}=-4\)
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_
\(\dfrac{x^2}{5.6}\)+\(\dfrac{x^2}{6.7}\)+\(\dfrac{x^2}{7.8}\)+...+\(\dfrac{x^2}{24.25}\)
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 :
\(\dfrac{x^2}{5}-\dfrac{x^2}{25}\text{=}\dfrac{5x^2-x^2}{25}\text{=}\dfrac{4x^2}{25}\)
Tìm y: (1/24.25 + 1/25.26 + ...+ 1/29.30 ) x 120 + y: 1/3 = - 4
Tìm x biết:
(\(\dfrac{1}{24.25}\)+\(\dfrac{1}{25.26}\)+...+\(\dfrac{1}{29.30}\)).120+x:\(\dfrac{1}{3}\)=\(-4\)
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:
\(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(1+x:\frac{1}{3}=-4\)
\(x:\frac{1}{3}=-5\)
\(x=-15\)
\(\left(\dfrac{1}{24.25}+\dfrac{1}{25.26}+...+\dfrac{1}{29.30}\right).120+x:\dfrac{1}{3}=-4\)
\(\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}\right).120+x:\dfrac{1}{3}=-4\)
\(\left(\dfrac{1}{24}-\dfrac{1}{30}\right).120+x:\dfrac{1}{3}=-4\)
\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\)
\(1+x:\dfrac{1}{3}=-4\)
\(x:\dfrac{1}{3}=-4-1\)
\(x:\dfrac{1}{3}=-5\)
\(x=-5.\dfrac{1}{3}\)
\(x=\dfrac{-5}{3}\)