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) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)

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

\(=\left|x-1\right|+\left|x-3\right|\ge\left|\left(x-1\right)+\left(3-x\right)\right|=2\)

Vậy\(A_{min}=2\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\)

\(TH1:\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\Leftrightarrow1\le x\le3\)

\(TH1:\hept{\begin{cases}x-1\le0\\3-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge3\end{cases}}\left(L\right)\)

Vậy \(A_{min}=2\Leftrightarrow1\le x\le3\)

mik giải ko ghi đề nha

x+2+2x+1=4x

2+1=4x-x-2x

3=1x

3:1=x

3=x

vậy x=3

\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow-2x=15-8=7\)

\(\Leftrightarrow x=\frac{-7}{2}\)

Vậy \(x=\frac{-7}{2}\)

Mình làm cho bạn 2 câu khó hơn còn mấy câu còn lại dungf phương pháp quy đồng rồi chuyển vế là tính được mà

c, <=> [(x-1)/2009 ]-1 +[ (x-2)/2008] -1 = [(x-3)/2007]-1 +[(x-4)/2006]-1

<=> (x-2010)/2009 + (x-2010)/2008 = (x-2010)/2007 + (x-2010)/2006

<=> (x-2010)*(1/2009+1/2008-1/2007-1/2006)=0

=> x-2010=0 => x=2010

d, TH1 : cả hai cùng âm

=>> 2X-4 <O => X< 2 

Và 9-3x<0 =>> x> 3 

=>> loại 

Th2 cả hai cùng dương

2x-4>O => x>2 

Và 9-3x>O => x<3 

=>> 2<x<3 (tm)

b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18

4x 2 -4x+1-4x 2+25=18

26-4x=18

4x=8

x=2

a,27x-18=2x-3x^2

<=> 3x^2-2x+27-18x=0

<=> 3x^2-20x+27=0

\(\Delta\)= 20^2-4-12.27

tính \(\Delta\)rồi tìm x1 ,x2

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

a) \(f\left(x\right)=-x^4+3x^3-\frac{1}{3}x^2+2x+5\)

\(g\left(x\right)=x^4+3x^3-\frac{2}{3}x^2-2x-10\)

b) \(f\left(x\right)+g\left(x\right)=-x^4+3x^3-\frac{1}{3}x^2+2x+5+x^4+3x^3-\frac{2}{3}x^2-2x-10\)

                                \(=6x^3-x^2-5\)

c) +) Thay x=1 vào đa thức f(x) + g(x) ta được :

       \(6.1^3-1^2-5=0\)

Vậy x=1 là nghiệm của đa thức f(x) + g(x)

+) Thay x=-1 vào đa thức f(x) + g(x) ta được :

    \(6.\left(-1\right)^3-\left(-1\right)^2-5=-10\)

Vậy x=-1 ko là nghiệm của đa thức f(x) + g(x)