Toán

Ta có:

`C=3+3^2 +..+3^100`

`=>3C=3^2 +3^3 +...+3^101`

`=> 3C-C=3^2 +3^3 +...+3^101 - 3-3^2 -...-3^100`

`=>2C= 3^101 -3`

Ta có:

`2C+3=3^{2n-1}`

`<=>3^101 -3+3=3^{2n-1}`

`<=>3^101=3^{2n-1}`

`<=>2n-1=101`

<=>2n=102`

`<=>n=51`

Vậy `n=51`

gúp với

\(15.2.6 = 5.3.12 = 15.3.4\)

\(4.4.9 = 8.18 = \)\(8 . 2 . 9\)

15.2.6 = 5.3.2.2.3

5.3.12 = 5.3.2.2.3

15.3.4 = 5.3.3.2.2

=> 15.2.6 = 5.3.12 = 15.3.4

4.4.9 = 2.2.2.2.3.3

8.19 = 2.2.2.2.3.3

8.2.9 = 2.2.2.2.3.3

=> 4.4.9 = 8.19 = 8.2.9

ấn đọc tiếp=)

\(a,8\left(x-1\right)-4=6\left(x+2\right)-2\\ \Leftrightarrow8x-8-4=6x+12-2\\ \Leftrightarrow8x-8-4-6x-12+2=0\\ \Leftrightarrow2x-22=0\\ \Leftrightarrow x=11\)

Vậy `S={11}`

\(b,\dfrac{x}{2}-\dfrac{2x+1}{3}=\dfrac{2x-1}{5}\\ \Leftrightarrow b,15x-10\left(2x+1\right)=6\left(2x-1\right)\\ \Leftrightarrow15x-20x-10=12x-6\\ \Leftrightarrow-5x-10-12x+6-0\\ \Leftrightarrow-17x-4=0\\ \Leftrightarrow x=-\dfrac{4}{17}\)

Vậy `S={-4/17}`

\(c,ĐKXĐ:x\ne\pm2\\ \dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{x^2+2}{x^2-4}=0\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x^2+2}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2-2x}{\left(x+2\right)\left(x-2\right)}+\dfrac{x^2+2}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2+3x+2-x^2+2x+x^2+2}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2+5x+4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)

Vậy `S={-1;4}`

a) 8x - 8 -4 = 6x +12 -2
8x - 6x = 10 + 4
2x = 14
x = 7

b) \(\dfrac{15x}{30}\) - \(\dfrac{10\left(2x+1\right)}{30}\) = \(\dfrac{6\left(2x-1\right)}{30}\)
15x - 20x - 10 = 12x - 6
-5x -12x = 10 - 6
-17x =4
x = -4/17

`a)` Để `A` là phân số thì \(3n+3\ne0\)

                                         \(\Rightarrow n\ne-1\)

`b)`\(A=\dfrac{12n}{3n+3}\)

\(A=\dfrac{4n}{n+1}\)

\(A=4-\dfrac{4}{n+1}\)

Để `A` nguyên thì \(\dfrac{4}{n+1}\in Z\) hay \(n+1\in U\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

`@n+1=1=>n=0`

`@n+1=-1=>n=-2`

`@n+1=2=>n=1`

`@n+1=-2=>n=-3`

`@n+1=4=>n=3`

`@n+1=-4=>n=-5`

Vậy \(n\in\left\{0;-2;1;-3;3;-5\right\}\) thì `A` nguyên

`c)`\(A=4-\dfrac{4}{n+1}\)

\(n\in N\)

Để `A nhỏ nhất thì \(n+1\) nhỏ nhất

mà \(n\in N\Rightarrow n+1\ge1\)

\(\Rightarrow A\ge4-\dfrac{4}{1}=0\)

Dấu "=" xảy ra khi \(n=0\)

\(a,2\left(x-3\right)< x+7\\ \Leftrightarrow2x-6-x-7< 0\\ \Leftrightarrow x-13< 0\\ \Leftrightarrow x< 13\\ b,x-\dfrac{x+2}{6}\ge\dfrac{x-1}{3}+5+\dfrac{2x}{5}\\ \Leftrightarrow30x-5\left(x+2\right)\ge10\left(x-1\right)+30.5+6.2x\\ \Leftrightarrow30x-5x-10\ge10x-10+150+12x\\ \Leftrightarrow30x-5x-10-10x+10-150-12x\ge0\\ \Leftrightarrow3x-150\ge0\\ \Leftrightarrow3x\ge150\\ \Leftrightarrow x\ge50\)

`a)`\(2\left(x-3\right)< x+7\)

\(\Leftrightarrow2x-6< x+7\)

\(\Leftrightarrow x< 13\)

Vậy \(S=\left\{x|x< 13\right\}\)

`b)`\(x-\dfrac{x+2}{6}\ge\dfrac{x-1}{3}+5+\dfrac{2x}{5}\)

\(\Leftrightarrow\dfrac{30x-5x-10}{30}\ge\dfrac{10x-10+150+12x}{30}\)

\(\Leftrightarrow25x-10\ge22x+140\)

\(\Leftrightarrow3x\ge150\)

\(\Leftrightarrow x\ge50\)

Vậy \(S=\left\{x|x\ge50\right\}\)

\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)

\(=\dfrac{4.4^5}{3.3^5}.\dfrac{6.6^5}{2.2^5}=\dfrac{4^6}{3^6}.\dfrac{6^6}{2^6}=\left(\dfrac{6}{3}\right)^6.\left(\dfrac{4}{2}\right)^6=2^{12}\)

\(\Rightarrow2^x=2^{12}\Rightarrow x=12\)

a) \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x\)

\(\Leftrightarrow\cdot\dfrac{4^6.6^6}{3^6.2^6}=2^x\)

\(=>\left(\dfrac{4}{2}\right)^6.\left(\dfrac{6}{3}\right)^6=2^x\)

\(=>2^6.2^6=2^x\)

\(2^{12}=2^x\)

`=>x=12`

Vậy `x=12`

Thời gian bác Xuân đi trước:

`8` giờ `45` phút `-8` giờ `15` phút `=30` phút 

Mỗi phút bác Xuân hơn bác Thu:

`35` phút `-30` phút `=5` phút

Thời gian bác Thu đuỏi kịp:

`30 : 5 = 6 (giờ)`

Bác Thu đuổi kịp bác Xuân lúc:

`8` giờ `45` phút `+6` giờ `=14` giờ `45` phút

Từ \(b^2=ac\) và \(c^2=bd\) ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(1\right)\)

Mà \(\dfrac{a^3}{b^3}=\dfrac{a.a.a}{b.b.b}=\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)

Từ (1) và (2) ta có đpcm 

\(a,\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =\left(6x^2+14x+9x+21\right)-\left(6x^2-10x+33x-55\right)\\ =\left(6x^2+23x+21\right)-\left(6x^2+23x-55\right)\\ =6x^2+23x+21-6x^2-23x+55\\ =76\)

Vậy ...

\(b,\left(3x^2-2x+1\right)\left(x^2+2x+3\right)-4x\left(x^2-1\right)-3x^2\left(x^2+2\right)\\ =3x^4-2x^3+x^2+6x^3-4x^2+2x+9x^2-6x+3-4x^3+4x-3x^4-6x^2\\ =3\)

Vậy ...

`a)`\(\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)

\(=6x^2+9x+14x+21-6x^2-33x+10x+55\)

\(=76\) ( ko phụ thuộc vào biến )

`b)`\(\left(3x^2-2x+1\right)\left(x^2+2x+3\right)-4x\left(x^2-1\right)-3x^2\left(x^2+2\right)\)

\(=3x^4+6x^3+9x^2-2x^3-4x^2-6x+x^2+2x+3-4x^3+4x-3x^4-6x^2\)

\(=3\) ( ko phụ thuộc vào biến )

`a, 2/3 xx x = 2/7`

`x = 2/7 : 2/3`

`x= 2/7 xx 3/2`

`x=3/7`

`b, 2/5 : x= 3/7`

`x= 2/5 : 3/7`

`x= 2/5 xx 7/3`

`x=14/15`

2/3 x x= 2/7
x=2/7 : 2/3 
x= 3/7 
2/5 : x = 3/7
x= 3/7 x 2/5 
x= 6/35