bài 1: Tìm x biết
b, [(1.2 + 2.3 + 3.4 + ... + 98.99) . x] / 26950= 90/7 : -3/2
tÌM X:
(1.2+2.3+3.4+...+98.99).x/26950=90/7:3/2
Tìm x biết:
a) (1.2+2.3+3.4+...+98.99).x/26950 = 90/7:3/2
b) |2x-1|/|x| = 3/5
d) 12√x+1 -x-1=0
(1.2+2.3+3.4+...+98.99).x/26950=90/7:1,5
(1.2 + 2.3 + 3.4 +...+ 98.99) . \(\dfrac{x}{26950}\) = \(\dfrac{90}{7}\):1,5
Áp dụng ct: 1.2 + 2.3+ 3.4 +....+n.(n+1) = n(n+1)(n+2):3
Ta có:
\(\dfrac{98.99.100}{3}\).\(\dfrac{x}{26950}\) = \(\dfrac{90}{7}\):\(\dfrac{3}{2}\)
12\(x\) = \(\dfrac{60}{7}\)
\(x\) = \(\dfrac{60}{7}\): 12
\(x\) = \(\dfrac{5}{7}\)
( 1.2 + 2.3 + 3.4 + ...... + 98.99 ) x / 26950 = 12/6/7 : 3/2
Khó quá mk chịu thôi bạn giải giùm mk nha
k mình đi
kb luôn nhé
Tìm x biết \(\frac{1.2+2.3+3.4+...+98.99:49}{26950}.x=12+\frac{\frac{6}{7}}{\frac{-3}{2}}\)
(1.2+2.3+3.4 +...+98.99).x/26950 =12và 6/7 :1,5
Đặt (1.2+2.3+3.4+..+98.99)=a
=>3a= 1.2.(3-0)+2.3.(4-1).....98.99.(100-97)
=>3a=1.2.3-0+2.3.4-1.2.3+...+98.99.100-97.98.99
=>3a=98.99.100
=>a-98.99.100/3=323400
Thay vào,ta được
323400.x/26950=12
=>323400.x=12.26950=323400
=>x=1
6/7:1.5=4/7
Tìm giá trị của y:
(1.2+2.3+3.4+...+98.99)y/26950= 12.6/7:3/2
Tìm x:
\(\dfrac{\left(1.2+2.3+3.4+...+98.99\right).x}{26950}=12\dfrac{6}{7}:\dfrac{-3}{2}\)
\(\dfrac{\left(1.2+2.3+3.4+...+98.99\right).x}{26950}=12\dfrac{6}{7}:\dfrac{-3}{2}\\ \Rightarrow\left(1.2+2.3+3.4+...+98.99\right).x:26950=\dfrac{90}{7}:\dfrac{-3}{2}\\ \left(1.2+2.3+3.4+...+98.99\right).x:26950=\dfrac{-60}{7}\\ \left(1.2+2.3+3.4+...+98.99\right).x=\dfrac{-60}{7}.26950\\ \left(1.2+2.3+3.4+...+98.99\right).x=-231000\\ \left\{\left[99.98.\left(98+2\right)\right]:3\right\}.x=-231000\\ 323400x=-231000\\ x=-231000:323400\\ x=\dfrac{-5}{7}\)
Đặt A=1.2+2.3+...+98.99
=>3A=1.2.3+2.3.(4-1)+...+98.99.(100-97)
=1.2.3-1.2.3+2.3.4-...-97.98.99+98.99.100
=98.99.100
=>A=98.99.100:3=323400
=>\(\dfrac{323400x}{26950}=\dfrac{90}{7}\cdot\dfrac{2}{-3}\)
<=>12x=\(-\dfrac{60}{7}\)
<=>x=\(-\dfrac{60}{12.7}\)
<=>x=\(-\dfrac{5}{7}\)
Vậy...
\(\frac{\left(1.2+2.3+3.4+....+98.99\right).x}{26950}=12\frac{6}{7}:\frac{-3}{2}\)