`@` `\text {Ans}`

`\downarrow`

`(8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)-33`

`\Leftrightarrow 8x(3x+2) -3(3x+2) - 4x(x+4) + 7(x+4) = 2x(5x-1) + 5x-1 - 33`

`\Leftrightarrow 24x^2 + 16x - 9x - 6 - 4x^2 - 16x - 7x - 28 = 10x^2 - 2x + 5x - 1 - 33`

`\Leftrightarrow 20x^2 -16x - 34 = 10x^2 + 3x - 34`

`\Leftrightarrow 20x^2 - 16x - 34 - 10x^2 - 3x + 34 = 0`

`\Leftrightarrow 10x^2 - 19x = 0`

`\Leftrightarrow x(10x - 19)=0`

`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\)

`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x=19\end{matrix}\right.\)

`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)

Vậy, `x={0; 19/10}.`

Với bài này bn áp dụng bài phần tử của tập hợp nhé! 

(8-4):6=129

Gọi 129 là x

X-7=59

Gọi  59 làc

Vậy phần bài này là phần tử

Đs 78/9

`a)2x^2+3(x-1)(x+1)=5x(x+1)`

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

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`b)(x-3)^3+3-x=0` nhỉ?

`<=>(x-3)^3-(x-3)=0`

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

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`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

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`d)3(2x-3)+2(2-x)=-3`

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

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`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`

3, 2x(x^2-8x+16)-(x+5)(x^2-4)+2(x^2+10x+25)-x+1

=2x^3-16x^2+32x-(x^3-4x+5x^2-20)+2x^2+20x+50-x+1

=2x^3-16x^2+32x-x^3+4x-5x^2+20+2x^2+20x+50-x+1

=x^3-19x^2+55x+71

Không biết nãy bị lỗi ở đâu, mình gửi lại:<

a. 6x² - (2x + 5)(3x - 2) = 7

<=> 6x² - 6x² + 4x - 15x + 10 = 7

<=> -11x = -3

<=> \(x=\dfrac{3}{11}\)

b. (5 - x)(25 + 5x + x²) + x(x² - 7) = 25

<=> 125 - x³ + x³ - 7x = 25

<=> -7x = 25 - 125

<=> -7x = -100

<=> \(x=\dfrac{100}{7}\)

c. (7 - 2x)² + (3 + 2x)(3 - 2x) = 30

<=> 49 - 28x + 4x² + 9 - 4x² = 30

<=> 4x² - 4x² - 28x = 30 - 49 - 9

<=> -28x = -28

<=> x = 1

Bài 3:

a:

ĐKXĐ: x>=-5

\(x^2-7x=6\sqrt{x+5}-30\)

=>\(x^2-4x-3x+12=6\sqrt{x+5}-18\)

=>\(\left(x-4\right)\left(x-3\right)=6\left(\sqrt{x+5}-3\right)\)

=>\(\left(x-4\right)\left(x-3\right)=6\cdot\frac{x+5-9}{\sqrt{x+5}+3}\)

=>\(\left(x-4\right)\left(x-3-\frac{6}{\sqrt{x+5}+3}\right)=0\)

=>x-4=0

=>x=4(nhận)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{x+4\sqrt{x}+4}=5x+2\)

=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

=>\(5x+2=\sqrt{x}+2\)

=>\(5x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)

b: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)

=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\) =4

=>|x-1|+|x+2|=4(1)

TH1: x<-2

(1) sẽ trở thành: -x-2+1-x=4

=>-2x-1=4

=>-2x=5

=>x=-5/2(nhận)

TH2: -2<=x<1

(1) sẽ trở thành: x+2+1-x=4

=>3=4(vô lý)

TH3: x>=1

(1) sẽ trở thành: x+2+x-1=4

=>2x+1=4

=>2x=3

=>x=3/2(nhận)

c: ĐKXĐ: x>=1

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)

=>\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)

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

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\)

=>\(\left|\sqrt{x-1}-1\right|=2-\sqrt{x-1}-1=1-\sqrt{x-1}\)

=>\(\sqrt{x-1}-1\le0\)

=>\(\sqrt{x-1}\le1\)

=>0<=x-1<=1

=>1<=x<=2

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

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

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

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

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

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

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