`@` `\text {Ans}`

`\downarrow`

Thực hiện phép tính ;-;?

\((x+3) (x^2-3x+9) + (x-3) ( x^2+3x+9 )\)

`= x(x^2 - 3x + 9) + 3(x^2 - 3x + 9) + x(x^2 + 3x + 9) - 3(x^2 + 3x + 9)`

`= x^3 - 3x^2 + 9x + 3x^2 - 9x + 27 + x^3 + 3x^2 + 9x - 3x^2 - 9x - 27`

`= (x^3 + x^3) + (-3x^2 + 3x^2 + 3x^2 - 3x^2) + (9x - 9x + 9x - 9x) + (27 - 27)`

`= 2x^3`

=x³+3³+x³-3³

=2x³

\(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(B=x^3-9x^2+27x-27-\left(x^3-3x^2+9x+3x^2-9x+27\right)+\left(9x^2-1\right)\)

\(B=x^3-9x^2+27x-27-\left(x^3+27\right)+9x^2-1\)

\(B=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(B=27x-55\)

giải cho tui cả phần c nx đi

\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

Yêu cầu đề bài của bạn?

đề bài : Tính

C=(3x+2)³-18x.(3x+2)+(x-1)³-28x³+3x.(x-1)

a) A = x³+3x²+3x+9

\(=x^2\left(x+3\right)+3\left(x+3\right)=\left(x^2+3\right)\left(x+3\right)\)

b) B = x³-3x²+3x-9

\(=x^2\left(x-3\right)+3\left(x-3\right)=\left(x^2+3\right)\left(x-3\right)\)

c) C = x³-3x²+3x+7

\(=x\left(x^2-3x+3\right)+7\)

\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)

\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow3x+6+2x+2=5x+4\)

\(\Leftrightarrow3x+2x-5x=-6-2+4\)

\(\Leftrightarrow0x=-4\)

=> PT vô nghiệm 

\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)

\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow4x-2-15=9x-3\)

\(\Leftrightarrow4x-9x=2+15-3\)

\(\Leftrightarrow-5x=14\)

.....

mấy cái này mẫu nào dài cậu phân tích ra : 

VD : câu  3 : \(3x^2-4x+1\)

\(=3x^2-3x-x+1\)

\(=3x\left(x-1\right)-\left(x-1\right)\)

\(=\left(3x-1\right)\left(x-1\right)\)

r bắt đầu giải PHương trình :)) Mấy câu còn lại tương tự 

4; \(\frac{5}{x-2}+\frac{2}{x+4}=\frac{3x}{x^2+2x-8}.\)

\(\Leftrightarrow\frac{5\left(x+4\right)}{\left(x-2\right)\left(x+4\right)}+\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+4\right)}=\frac{3x}{\left(x-2\right)\left(x+4\right)}\)

\(\Leftrightarrow5x+20+2x-4=3x\)

\(\Leftrightarrow4x=-16\Leftrightarrow x=-2\left(TM\right)\)

KL ::

\(5;\frac{4}{x+6}+\frac{1}{x-3}=\frac{9}{x^2+3x-18}\)

\(\Leftrightarrow\frac{4\left(x-3\right)}{\left(x+6\right)\left(x-3\right)}+\frac{x+6}{\left(x-3\right)\left(x+6\right)}=\frac{9}{\left(x-3\right)\left(x+6\right)}\)

\(\Leftrightarrow4x+x=3+9-6\)

\(\Leftrightarrow5x=6\Leftrightarrow x=\frac{6}{5}\)

Ta có: (3x - 9)(2x - 7) = (3x - 9)(4x + 3)

<=> 6x² - 39x + 63 =  12x² - 27x - 27

<=> 6x² - 39x + 63 - 12x² + 27x + 27 = 0

<=> -6x² - 12x + 90 = 0

<=> -6(x² + 2x - 15) = 0

<=> x² + 5x  - 3x - 15 = 0

<=> x(x + 5) - 3(x + 5) = 0

<=> (x - 3)(x + 5) = 0

<=> \(\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Vậy pt có tập nghiệm là {3; -5}

(3x - 9)(2x -7) = (3x-9)(4x-3)

(3x-9)(2x-7) - (3x-9)(4x-3)=0

(3x-9)(2x-7 - 4x +3)=0

(3x-9)(-2x - 4)=0

3(x-3)(-2)(x+2)=0

(x-3)(x+2)=0

=> x-3=0 hoặc x+2=0

=> x=3 hoặc x=-2

c: Ta có: \(\left(2x-3\right)^2-\left(2x-3\right)\left(x-10\right)=7\)

\(\Leftrightarrow4x^2-12x+9-2x^2+20x+3x-30=7\)

\(\Leftrightarrow11x=28\)

hay \(x=\dfrac{28}{11}\)

d: Ta có: \(\left(3x-4\right)^2-9\left(x-3\right)\left(x+3\right)=8\)

\(\Leftrightarrow9x^2-24x+16-9x^2+81=8\)

\(\Leftrightarrow-24x=-89\)

hay \(x=\dfrac{89}{24}\)

f: Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x=-1\)

hay \(x=-\dfrac{1}{8}\)