cho C= 1x2+2x3+3x4+...+X x(x-1)
giúp mình với nhé. hôm nay nhiều bài quá. cảm ơn các bạn nhé. ai đúng mình tích cho nhé
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\(E=\dfrac{3}{1x5}+\dfrac{3}{5x9}+...+\dfrac{3}{121x125}\)
\(\dfrac{4}{3}xE=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{121}-\dfrac{1}{125}\)
\(\dfrac{4}{3}xE=1-\dfrac{1}{125}\)
\(E=\dfrac{124}{125}x\dfrac{3}{4}=\dfrac{93}{125}\)
219 - 7(x + 1) = 10²
7(x + 1) = 219 - 10²
7(x + 1) = 219 - 100
7(x + 1) = 119
x + 1 = 119 : 7
x + 1 = 17
x = 17 - 1
x = 16
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(3x - 2⁴) . 7³ = 2.7⁴
(3x - 16).7³ = 2.7⁴
3x - 16 = 2.7⁴:7³
3x - 16 = 2.7
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 30 : 3
x = 10
D = \(\dfrac{1}{1.4}\) + \(\dfrac{1}{4.7}\) + \(\dfrac{1}{7.10}\)+...+ \(\dfrac{1}{91.94}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\)+...+ \(\dfrac{1}{91}\) - \(\dfrac{1}{94}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{94}\)
D = \(\dfrac{93}{94}\)
\(10+2\cdot x=4^5:4^3\)
\(\Rightarrow10+2\cdot x=4^{5-3}\)
\(\Rightarrow10+2\cdot x=4^2\)
\(\Rightarrow10+2\cdot x=16\)
\(\Rightarrow2\cdot x=16-10\)
\(\Rightarrow2\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{2}=3\)
__________________
\(2\cdot x-6^2:18=3\cdot2^2\)
\(\Rightarrow2\cdot x-36:18=12\)
\(\Rightarrow2\cdot x-2=12\)
\(\Rightarrow2\cdot x=12+2\)
\(\Rightarrow2\cdot x=14\)
\(\Rightarrow x=\dfrac{14}{2}=7\)
_________________
\(70-5\cdot\left(x-3\right)=3\cdot2\)
\(\Rightarrow70-5\cdot\left(x-3\right)=6\)
\(\Rightarrow70-5x+15=6\)
\(\Rightarrow-5x+15=6-70\)
\(\Rightarrow-5x+15=64\)
\(\Rightarrow-5x=64-15\)
\(\Rightarrow-5x=49\)
\(\Rightarrow x=-\dfrac{49}{5}\)
\(5\cdot4^2-18:3^2\)
\(=5\cdot16-18:9\)
\(=80-2\)
\(=78\)
________________
\(6^2:4\cdot3+2\cdot5^3\)
\(=36:12+2\cdot125\)
\(=3+250\)
\(=253\)
_______________
\(80-\left(4\cdot5^2-3\cdot2^3\right)\)
\(=80-\left(4\cdot25-3\cdot8\right)\)
\(=80-\left(100-24\right)\)
\(=80-76\)
\(=4\)
6² : 4 . 3 + 2 . 5³
= 36 : 4 . 3 + 2 . 125
= 9 . 3 + 250
= 27 + 250
= 277
Hai bài còn lại xem của bạn Phong nhé. Bài này Phong giải chưa chính xác nên Thầy giải lại cho em!
\(C=1.2+2.3+3.4+...+x.\left(x-1\right)\)
\(\Rightarrow3C=1.2.3+2.3.3+3.4.3+...+x.\left(x-1\right).3\)
\(\Rightarrow3C=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+x.\left(x-1\right).\left[\left(x+1\right)-\left(x-2\right)\right]\)
\(\Rightarrow3C=\left(1.2.3-0.12\right)+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.4\right)+...+\left[x.\left(x-1\right)\left(x+1\right)-x.\left(x-1\right)\left(x-2\right)\right]\)
\(\Rightarrow3C=-0.1.2+x.\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow3C=x.\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow C=\dfrac{x.\left(x-1\right)\left(x+1\right)}{3}\)
3C=1x2x3+2x3x3+3x4x3+...+Xx(X+1)=
=1x2x3+2x3x(4-1)+3x4x(5-2)+...+Xx(X+1)[(X+2)-(X-1)]=
=1x2x3-1x2x3+2x3x4-2x3x4+3x4x5-...-(X-1)xXx(X+1)+Xx(X+1)x(X+2)=
=Xx(X+1)(X+2)