BCNN là j v ạ!?

\(9=3^2;24=3\cdot2^3\)

=>\(BCNN\left(9;24\right)=3^2\cdot2^3=72\)

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

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

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

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

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

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

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

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

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

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

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

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

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

\(x-\left(11-x\right)=-48+\left(-12+x\right)\\ =>x-11+x=-48-12+x\\ =>2x-11=-60+x\\ =>2x-x=-60+11\\ =>x=-49\)

Vậy: ... 

Vì 480 ⋮ a và 600 ⋮ a nên a ∈ ƯC(480; 600)

Vì a là số tự nhiên lớn nhất nên a là ƯCLN của 480 và 600

Ta có: 480 = 25.3.5

600 = 23.3.52

ƯCLN(480;600) = 23.3.5 = 120

Vậy a = 120

\(A=1\times2+2\times3+3\times4+98\times99\\ =2+6+12+9702\\ =8+12+9702\\ =12+9710\\ =9722\)

`A = 1.2 + 2.3 + 3.4 + ... + 98.99`

`3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ... + 98 . 99 . (100 - 97) `

`3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 98.99.100 - 97.98.99`

`3A = 98 . 99 . 100`

`A = 98.33.100 `

`A = 323400`

\(\left(x-40\right)^2-48=-12\\ =>\left(x-40\right)^2=-12+48\\ =>\left(x-40\right)^2=36\\ =>\left(x-40\right)^2=6^2\\ TH1:x-40=6\\ =>x=6+40\\ =>x=46\\ TH2:x-40=-6\\ =>x=40-6\\ =>x=34\)

Vậy: ... 

\(\left(x-40\right)^2-48=-12\)

\(\left(x-40\right)^2=\left(-12\right)+48\)

\(\left(x-40\right)^2=36\)

\(\left(x-40\right)^2=6^2\)

\(x-40=6\)

\(x=6+40\)

\(x=46\)

Số lượng số hạng:

`(2024-1):1+1=2024` (số hạng)

Tổng của dãy số:

`(2024+1)*2024:2=2049300` 

ĐS: ...