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\(T=1\cdot2\cdot3+2\cdot3\cdot4+...+20\cdot21\cdot22\)
\(4T=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+20\cdot21\cdot22\cdot4\)
\(4T=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\cdot\left(5-1\right)+..+20\cdot21\cdot22\cdot\left(23-19\right)\)
\(4T=0\cdot1\cdot2\cdot3-1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+..+20\cdot21\cdot22\cdot23-19\cdot20\cdot21\cdot22\)
\(4T=21\cdot22\cdot23\cdot24\)
\(T=\frac{21\cdot22\cdot23\cdot24}{4}=21\cdot22\cdot23\cdot6=63756\\ k.cho.mk.nha\)
1.2.3+2.3.4+3.4.5+...+20.21.22
=1/4.(1.2.3.4+2.3.4.4+3.4.5.4+...+20.21.22.4)
=1/4.[1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+...+20.21.22.(23-19)]
=1/4.(1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+....-19.20.21.22+20.21.22.23)
=1/4.20.21.22.23
=53130
a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)
b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)
=9/17+8/17=1
c: =>x-3/10=7/15*1/5=7/75
=>x=7/75+3/10=59/150
\(x-\frac{10}{3}=\frac{7}{15}.\frac{3}{5}\)
\(x-\frac{10}{3}=\frac{7}{25}\)
\(x=\frac{7}{25}+\frac{10}{3}=\frac{271}{75}\)
Vậy x = 271/75
\(x-\frac{10}{3}\)\(=\)\(\frac{7}{15}\)\(.\)\(\frac{3}{5}\)
\(x-\frac{10}{3}\)\(=\) \(\frac{7}{25}\)
\(x\) \(=\)\(\frac{7}{25}\)\(+\)\(\frac{10}{3}\)
\(x\) \(=\) \(\frac{271}{75}\)
a/ 2/3.x+1/2=1/10
nên:2/3x=1/10-1/2
nên:2/3x=-2/5
suy ra :x=-2/5:2/3
vậy x=-3/5
a. 2/3x+1/2=1/10
2/3x=1/10-1/2
2/3x=-2/5
x=-2/5:2/3
x=-3/5
b.(7/2-2x).4/3=22/3
(7/2-2x).=22/3:4/3
7/2-2x=11/2
2x=7/2-11/2
2x=-2
x=-1
nho k cho minh voi nhe
Bài làm
a) x² - 3 = 22
=> x² = 25
=> x = + 5
Vậy x = + 5
b) 2x³ + 5 = -11
2x³ = -16
x³ = -8
x = -2
Vậy x = -2
c) ( x + 2 )² = 81
=> x + 2 = 9
=> x = 7
Vậy x = 7
d) ( 2x + 1 )² = 25
=> 2x + 1 = 5
=> 2x = 4
=> x = 2
Vậy x = 2
e) 5x + 2 = 625
5x = 623 ( vô lí )
g) ( 2x - 3 )² = 36.
=> 2x - 3 = 6
=> 2x = 9
=> x = 4,5
Vậy x = 4,5
h) ( 2x - 1 )³ = -8
=> 2x - 1 = -2
=> 2x = -1
=> x = -1/2
Vậy x = -1/2
i) ( x - 1 )x + 2 = ( x - 1 )x + 6
=> [ (x - 1 )x - ( x - 1 )x ] = 6 - 2
=> 0 = 4 ( vô lí )
Vậy x thuộc rỗng.
k) x² + x = 0
=> x( x + 1 ) = 0
=> x = 0 hoặc x + 1 = 0
=> x = 0 hoặc x = -1
Vậy x = 0 hoặc x = -1
\(2\left[x-\frac{1}{2}\right]+3\left[x-\frac{1}{3}\right]=-22\)
\(\Rightarrow2x-1+3x-1=-22\)
\(\Rightarrow2x+3x-1-1=-22\)
\(\Rightarrow5x=-22+1+1\)
\(\Rightarrow5x=-20\Leftrightarrow x=-4\)