1/5 . (-5/5 - 10 ) + 5x = x-2/3
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cm A = 3n + 3 - 2n+3 + 3n+1 - 2n+1 ⋮ 10
ta có : A = 3n+ 1( 32 + 1) - 2n. ( 23 + 2)
⇔A = 3n+1.10 - 2n.10
⇔ A = 10.( 3n+1- 2n)
10 ⋮ 10 ⇔ 10. ( 3n+1 - 2n)⋮ 10
⇔ A = 10 . (3n+1 - 2n) ⋮10 (đpcm)
\(\dfrac{9^5.8^2}{27^3.16}\) = \(\dfrac{3^{10}.2^6}{3^9.2^4}\) = 12
Cách giải:
A=1/2.4+1/4.6+........+1/100.102
A=1/2-1/4+1/4-1/6+.......+1/100-1/102
A=1/2-1/102
A=51/102-1/102
A=50/102
A=25/51
Đặt A = \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{100\cdot102}\)
2A = \(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{100\cdot102}\)
2A = \(\dfrac{2}{2}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{6}+...+\dfrac{2}{100}-\dfrac{2}{102}\)
2A = \(1-\dfrac{1}{51}=\dfrac{50}{51}\)
A = \(\dfrac{50}{51}:2=\dfrac{25}{51}\)
a, (x- \(\dfrac{2}{7}\))(x+\(\dfrac{3}{4}\))
⇔ \(\left[{}\begin{matrix}x-\dfrac{2}{7}=0\\x+\dfrac{3}{4}=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
b, \(\dfrac{2}{3}\) x - \(\dfrac{2}{5}\) = \(\dfrac{1}{2}\) x - \(\dfrac{1}{3}\)
⇔ \(\dfrac{2}{3}\) x - \(\dfrac{1}{2}\) x = \(\dfrac{2}{5}\) - \(\dfrac{1}{3}\)
⇔ \(\dfrac{x}{6}\) = \(\dfrac{1}{15}\)
⇔ x = \(\dfrac{1}{15}\) x 6
x = \(\dfrac{2}{5}\)
c, \(\dfrac{1}{3}\) x + \(\dfrac{2}{5}\) (x+1)= 0
⇔ \(\dfrac{5x+6x+6}{15}\) = 0
⇔ 11x + 6 = 0
⇔ x = - \(\dfrac{6}{11}\)
tìm x
a, (x - 2/7 ) ( x + 3/4 )=0
<=> x - 2/7= 0 hoặc x+3/4 = 0
=> X = 2/7 hoặc x = -3/4
c,1/3x +2/5 (x+1 )=0
=> 1/3x + 2/5x + 2/5.1 = 0
=> x.(1/3 + 2/5 ) +2/5 = 0
=> x.11/15 + 2/5 = 0
=> x.11/15 = 0 - 2/5
=> x.11/15 = -2/5
=> x = -2/5 : 11/15
=> x = -6/11
Vậy x = -6/11
\(\dfrac{1}{6}\) - 0,4 . \(\dfrac{5}{8}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{2}\)
= \(\dfrac{2}{12}\) - \(\dfrac{3}{12}\) + \(\dfrac{6}{12}\)
= \(\dfrac{5}{12}\)
Đề vẫn giải được nhưng số khá xấu, không phù hợp với lớp 7. Bạn xem lại xem đã đúng đề chưa vậy>
`1/5.(-5/5-10)+5x=x-2/3`
`1/5.(-1-10)+5x-x=-2/3`
`1/5 .(-11)+4x=-2/3`
`4x=-2/3+11/5=23/15`
`x=23/15:4=23/60`