\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{48}{49}.\dfrac{49}{50}=\dfrac{1}{50}\)

`a)25/(x+1)-1 1/6=-1/3-0,5`

`=>25/(x+1)=-1/3-1/2+1+1/6`

`=>25/(x+1)=1/3`

`=>75=x+1`

`=>x=74`

Vậy `x=74`

`b)(2x+25 3/5)^2-9/25=0`

`=>(2x+128/5)=9/25`

`**2x+128/5=3/5`

`=>2x=-125/5=-25`

`=>x=-25/2`

`**2x+128/5=-3/5`

`=>2x=-131/5`

`=>x=-131/10`

cám ơn bạn nhiều lắm ạ 

a: \(I_1 = \int \left( \tan(x) - \ln^{15}(\cos(x)) \right) dx\)

=>\(I_1 = \int \tan(x) \, dx - \int \ln^{15}(\cos(x)) \, dx\)

\(A=\int\tan(x)\,dx\)

\(=\int\frac{\sin(x)}{\cos(x)}\,dx\)

\(=-\int\frac{d(\cos(x))}{\cos(x)}=-\ln\vert{}\cos(x)\vert{}\)

\(B = \int \ln^{15}(\cos(x)) \, dx\)

Đặt \(u=\ln(\cos(x))\)

\(\implies du=\frac{-\sin(x)}{\cos(x)}dx=-\tan(x)dx\)

\(\int \tan(x) \ln^{15}(\cos(x)) \, dx = -\int u^{15} \, du = -\frac{u^{16}}{16} + C = -\frac{\ln^{16}(\cos(x))}{16} + C\)

Do đó: \(I_1 = -\ln\vert{}\cos(x)\vert{} - \int \ln^{15}(\cos(x)) \, dx + C\)

b: \(I_2 = \int \frac{x^4 + x^2 + 1}{2x^3 + 5x^2 - 7} \, dx\)

Ta có: \(x^4 + x^2 + 1 = \left( \frac{x}{2} - \frac{5}{4} \right)(2x^3 + 5x^2 - 7) + \left( \frac{25}{4}x^2 + \frac{7}{2}x - \frac{31}{4} \right)\)

=>\(\frac{x^4 + x^2 + 1}{2x^3 + 5x^2 - 7}=\frac{x}{2}-\frac{5}{4}+\frac{\frac{25}{4}x^2 + \frac{7}{2}x - \frac{31}{4}}{2x^3 + 5x^2 - 7}\)

\(=\frac{x}{2}-\frac{5}{4}+\frac{25x^2 + 14x - 31}{4(2x^3 + 5x^2 - 7)}\)

Đặt \(\frac{25x^2 + 14x - 31}{(x - 1)(2x^2 + 7x + 7)} = \frac{A}{x - 1} + \frac{Bx + C}{2x^2 + 7x + 7}\)

=>\(25x^2 + 14x - 31 = A(2x^2 + 7x + 7) + (Bx + C)(x - 1)\)

=>\(25x^2+14x-31=x^2\left(2A+B\right)+x\left(7A-B+C\right)+7A-C\)

=>\(\begin{cases}2A+B=25\\ 7A-B+C=14\\ 7A-C=-31\end{cases}\Rightarrow\begin{cases}2A+B=25\\ 7A-B+C-7A+C=14+31\\ 7A-C=-31\end{cases}\)

=>2A+B=25 và -B+2C=45 và 7A-C=-31

=>B=25-2A và -25+2A+2C=45 và 7A-C=-31

=>2A+2C=70 và 7A-C=-31 và B=25-2A

=>A+C=35 và 7A-C=-31 và B=25-2A

=>8A=4 và A+C=35 và B=25-2A

=>A=1/2; C=35-1/2=69/2; B=25-2*1/2=24

Do đó: \(\frac{25x^2 + 14x - 31}{4(2x^3 + 5x^2 - 7)} = \frac{1}{8(x - 1)} + \frac{24x + \frac{69}{2}}{4(2x^2 + 7x + 7)} = \frac{1}{8(x - 1)} + \frac{48x + 69}{8(2x^2 + 7x + 7)}\)

48x+69=12(4x+7)-15

=>\(\int \frac{48x + 69}{2x^2 + 7x + 7} dx = 12 \int \frac{4x + 7}{2x^2 + 7x + 7} dx - 15 \int \frac{dx}{2x^2 + 7x + 7}\)

\(= 12 \ln(2x^2 + 7x + 7) - \frac{15}{2} \int \frac{dx}{\left(x + \frac{7}{4}\right)^2 + \frac{7}{16}}\)

\(= 12 \ln(2x^2 + 7x + 7) - \frac{30}{\sqrt{7}} \arctan\left( \frac{4x + 7}{\sqrt{7}} \right)\)

=>\(I_2 = \int \left( \frac{x}{2} - \frac{5}{4} \right) dx + \frac{1}{8} \int \frac{dx}{x - 1} + \frac{1}{8} \int \frac{48x + 69}{2x^2 + 7x + 7} dx\)

\(=\frac{x^2}{4}-\frac{5x}{4}+\frac{1}{8}\ln\vert{}x-1\vert{}+\frac{3}{2}\ln(2x^2+7x+7)-\frac{15}{4\sqrt{7}}\arctan\left(\frac{4x + 7}{\sqrt{7}}\right)+C\)

a) x - 1/2 = 3/5

x = 3/5 + 1/2

x = 11/10

b) x - 1/2 = -2/3

x = -2/3 + 1/2

x = -1/6

c) 2/5 - x = 0,25

x = 2/5 - 0,25

x = 2/5 - 1/4

x = 3/20

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\)

Suy ra: \(x^2-4x+3-x^2=0\)

\(\Leftrightarrow-4x=-3\)

hay \(x=\dfrac{3}{4}\)(thỏa ĐK)

Vậy: \(S=\left\{\dfrac{3}{4}\right\}\)

\(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)

\(\Rightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A=1-\dfrac{1}{100}\)

\(\Rightarrow A=\dfrac{99}{100}\)

Đoạn suy ra đầu tiên cơ sở gì bạn suy ra được như vậy nhỉ?

=1/2+1/3+1/4+...+1/100

xét mẫu:có ssh là (100-2):1+1=99 số

tổng là (100+2)*99:2=5940

vậy ta có 1/5940

B = 1+1³+1⁴+...+1⁹⁸+1⁹⁹

B1 = 1³+1⁵+...+1⁹⁹+1¹⁰⁰

B1-B =(1³+1⁵+...+1⁹⁹+1¹⁰⁰) - ( 1+1³+1⁴+...+1⁹⁸+1⁹⁹ )

B0 = 1¹⁰⁰ - 1⁹⁹

B = 1