\(\dfrac{80}{1\cdot6}+\dfrac{80}{6\cdot11}+...+\dfrac{80}{251\cdot256}\)

\(=16\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{251\cdot256}\right)\)

\(=16\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{251}-\dfrac{1}{256}\right)\)

\(=16\left(1-\dfrac{1}{256}\right)=16\cdot\dfrac{255}{256}=\dfrac{255}{16}\)

\(5^{n+2}+3^{n+2}-5^n-3^n\)

\(=5^n\left(5^2-1\right)+3^n\left(3^2-1\right)\)

\(=5^n\cdot24+3^{n-1}\cdot3\cdot8=24\left(5^n+3^{n-1}\right)⋮24\)

\(x^2-7x+2=x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{41}{4}\)

\(=\left(x-\dfrac{7}{2}\right)^2-\dfrac{41}{4}>=-\dfrac{41}{4}\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{7}{2}=0\)

=>\(x=\dfrac{7}{2}\)

\(x^2-12x+5\)

\(=x^2-12x+36-31=\left(x-6\right)^2-31>=-31\forall x\)

Dấu '=' xảy ra khi x-6=0

=>x=6

`x^2 - 7x + 2`

`= x^2 - 2.x . 7/2 + (7/2)^2 - 41/4`

`= (x - 7/2)^2 - 41/4`

Do `(x - 7/2)^2 >= 0=>  (x - 7/2)^2 - 41/4 >= - 41/4`

Dấu = xảy ra khi: 

`x - 7/2 = 0`

`<=> x = 7/2`

Vậy ...

-----------------------

`x^2 - 12x + 5`

`= x^2 - 2.x.6 + 6^2 - 31`

`= (x-6)^2 - 31`

Do `(x-6)^2 >= 0 =>  (x-6)^2 - 31>= -31`

Dấu = có khi: 

`x - 6 = 0`

`<=> x = 6`

Vậy .... (không có max )

Tổng số phần bằng nhau là 4+3=7(phần)

Tuổi mẹ là 56:7x4=8x4=32(tuổi)

Tuổi con là 56-32=24(tuổi)

Mẹ hơn con 32-24=8(tuổi)

Tuổi của con :

\(56:4x1=14\left(tuổi\right)\)

Tuổi của mẹ :

\(56-14=42\left(tuổi\right)\)

Mẹ hơn con :

\(42-14=28\left(tuổi\right)\)

Đáp số...

1:

6,52:2

72:15

2,52:2,1=25,2:21

42:1,4=420:14

2:

396:8,8=3960:88

42:2,4=420:24

9:0.16=900:16

\(1\cdot4+2\cdot5+...+100\cdot103\)

\(=1\left(1+3\right)+2\left(2+3\right)+...+100\left(100+3\right)\)

\(=3\left(1+2+...+100\right)+\left(1^2+2^2+...+100^2\right)\)

\(=\dfrac{3\cdot100\cdot101}{2}+\dfrac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}\)

\(=3\cdot50\cdot101+338350=353500\)

\(c,C=4^{100}-4^{98}-4^{96}-...-1\\ C=4^{100}-\left(1+4^2+...+4^{98}\right)\)

Đặt: `D=1+4^2+...+4^98`

`4^2D=4^2+4^4+...+4^100`

`16D-D=(4^2+4^4+...+4^100)-(1+4^2+...+4^98)`

`15D=4^100-1`

`D=(4^100-1)/15`

\(C=4^{100}-\dfrac{4^{100}-1}{15}=\dfrac{15\cdot4^{100}-4^{100}+1}{15}=\dfrac{14\cdot4^{100}+1}{15}\)

\(d,D=3^{100}-3^{99}+3^{98}-...-3+1\\ D=3^{100}-\left(3^{99}-3^{98}+...+3-1\right)\)

Đặt: `S=3^99-3^98+....+3-1`

`3S=3^100-3^99+...+3^2-3`

`3S+S=(3^100+3^99+...+3^2-3)+(3^99-3^98+3-1)`

`4S=3^100-1`

`S=(3^100-1)/4`

\(D=3^{100}-\dfrac{3^{100}-1}{4}=\dfrac{4\cdot3^{100}-3^{100}+1}{4}=\dfrac{3^{101}+1}{4}\)

ai giúp mik giải bài này vs

\(1)5\left(x-7\right)=0\\ x-7=0\\ x=7\\ 2)25\left(x-4\right)=0\\ x-4=0\\ x=4\\ 3)34\left(2x-6\right)=0\\ 2x-6=0\\ 2x=6\\ x=\dfrac{6}{2}\\ x=3\\ 4)2007\left(3x-12\right)=0\\ 3x-12=0\\ 3x=12\\ x=\dfrac{12}{3}\\ x=4\\ 5)47\left(5x-15\right)=0\\5x=15\\ x=\dfrac{15}{5}\\ x=3\\ 6)13\left(4x-24\right)=0\\ 4x-24=0\\ 4x=24\\ x=\dfrac{24}{4}\\ x=6\\ 7)49\left(6x-12\right)=0\\ 6x-12=0\\ 6x=12\\ x=\dfrac{12}{6}\\ x=2\)

\(8)17\left(5x-45\right)=0\\ 5x-45=0\\ 5x=45\\ x=\dfrac{45}{5}\\ x=9\\ 9)105\left(17x-34\right)=0\\ 17x-34=0\\ 17x=34\\ x=\dfrac{34}{17}\\ x=2\\ 10)57\left(9x-27\right)=0\\ 9x-27=0\\ 9x=27\\ x=\dfrac{27}{9}\\ x=3\\ 11)25+\left(15-x\right)=30\\ 15-x=30-25\\ 15-x=5\\ x=15-5\\ x=10\\ 12)43-\left(24-x\right)=20\\ 24-x=43-20\\ 24-x=23\\ x=24-23\\ x=1\\ 13)2\left(x-5\right)-17=25\\ 2\left(x-5\right)=17+25\\ 2\left(x-5\right)=42\\ x-5=\dfrac{42}{2}=21\\ x=21+5\\ x=26\\ 14)3\left(x+7\right)-15=27\\ 3\left(x+7\right)=27+15\\ 3\left(x+7\right)=42\\ x+7=\dfrac{42}{3}\\ x+7=14\\ x=14-7\\ x=7\)