(x+8)(2x+15)(2x^2+35x+120

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+11\right)^2-25\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(y=x^2+7x+10\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(x^2+7x+10+1-5)(x^2+7x+10+1+5)\\=(x^2+7x+6)(x^2+7x+16)\\=(x^2+x+6x+6)(x^2+7x+16)\\=[x(x+1)+6(x+1)](x^2+7x+16)\\=(x+1)(x+6)(x^2+7x+16)\\Toru\)

Rút gọn thôi chứ phân tích sao được ._.

( x - 3 )² - ( 4x + 5 )² - 9( x + 1 )² - 6( x - 3 )( x + 1 )

= x² - 6x + 9 - ( 16x² + 40x + 25 ) - 9( x² + 2x + 1 ) - 6( x² - 2x - 3 )

= x² - 6x + 9 - 16x² - 40x - 25 - 9x² - 18x - 9 - 6x² + 12x + 18

= -30x² - 52x - 7

Sửa đề lại 1 chút là phân tích được mà bn Quỳnh:))

Ta có: \(\left(x-3\right)^2-\left(4x+5\right)^2+9\left(x+1\right)^2-6\left(x-3\right)\left(x+1\right)\)

\(=\left[\left(x-3\right)^2-6\left(x-3\right)\left(x+1\right)+9\left(x+1\right)^2\right]-\left(4x+5\right)^2\)

\(=\left(x-3-9x-9\right)^2-\left(4x+5\right)^2\)

\(=\left(8x+12\right)^2-\left(4x+5\right)^2\)

\(=\left(4x+7\right)\left(12x+17\right)\)

đặt ẩn đi

seo mak mt của tui mấy cái hàm cừ nhàu lên hết á

==

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)=24x^2\)

\(\Leftrightarrow x^4-24x^3+203x^2-720x+900=24x^4\)

\(\Leftrightarrow x^4-24x^3+203x^2-720x+900-24x^2=0\)

\(\Leftrightarrow x^4-24x^3+179x^3-720x+900=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-15\right)\left(x^2-7x+30\right)=0\)

có: \(x^2-7x+30\ne0\), nên:

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=15\end{cases}}\)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

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

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

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

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

\(=3xy\left(x^2y-2\right)\)

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

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

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

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

\(=\left(2x-1\right)^2-\left(3y\right)^2\)

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

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-2\right)\)

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)=24x^2\)

\(\Leftrightarrow\left[\left(x-5\right)\left(x-6\right)\right]\cdot\left[\left(x-3\right)\left(x-10\right)\right]=24x^2\)

\(\Leftrightarrow\left(x^2-11x+30\right)\left(x^2-13x+30\right)-24x^2=0\)

Đặt: \(x^2-13x+30=t\)

Lúc này PT trở thành:

\(t\left(t+2x\right)-24x^2=0\)

\(\Leftrightarrow t^2+2tx-24x^2=0\)

\(\Leftrightarrow t^2+6tx-4tx-24x^2=0\)

\(\Leftrightarrow t\left(t+6x\right)-4x\left(t+6x\right)=0\)

\(\Leftrightarrow\left(t+6x\right)\left(t-4x\right)=0\)

\(\Leftrightarrow\left(x^2-7x+30\right)\left(x^2-17x+30\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-7x+30=0\\x^2-17x+30=0\end{matrix}\right.\)

Ta có: \(x^2-7x+30=\left(x-\dfrac{7}{2}\right)^2+\dfrac{71}{4}>0\)(vô nghiệm)

=> \(x^2-17x+30=0\)

\(\Leftrightarrow\) \(\left(x-15\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-15=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=2\end{matrix}\right.\)

Vậy x = 2 hoặc x = 15