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\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+11\right)^2-25\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
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: \(\left(x-3\right)^2-\left(4x+5\right)^2+9\left(x+1\right)^2-6\left(x-3\right)\left(x+1\right)\)
\(=\left[\left(x-3\right)^2-6\left(x-3\right)\left(x+1\right)+9\left(x+1\right)^2\right]-\left(4x+5\right)^2\)
\(=\left(x-3-9x-9\right)^2-\left(4x+5\right)^2\)
\(=\left(8x+12\right)^2-\left(4x+5\right)^2\)
\(=\left(4x+7\right)\left(12x+17\right)\)
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: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)=24x^2\) (1)
\(\Leftrightarrow\left(x^2-5x-3x+15\right)\left(x-6\right)\left(x-10\right)=24x^2\)
\(\Leftrightarrow\left(x^2-8x+15\right)\left(x-6\right)\left(x-10\right)=24x^2\)
\(\Leftrightarrow\left(x^3-6x^2-8x^2+48x+15x-90\right)\left(x-10\right)=24x^2\)
\(\Leftrightarrow\left(x^3-14x^2+63x-90\right)\left(x-10\right)=24x^2\)
\(\Leftrightarrow x^4-10x^3-14x^3+140x^2+63x^2-630x-90x+900=24x^2\)
\(\Leftrightarrow x^4-2x^3-22x^3+44x^2+135x^2-270x-450x+900=0\)
\(\Leftrightarrow x^3\left(x-2\right)-22x^2\left(x-2\right)+135x\left(x-2\right)-450\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-22x^2+135x-450\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-15x^2-7x^2+105x+30x-450\right)=0\)
\(\Leftrightarrow\left(x-2\right)\cdot\left[x^2\cdot\left(x-15\right)-7x\left(x-15\right)+30\left(x-15\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-15\right)\left(x^2-7x+30\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-15=0\\x^2-7x+30=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=15\\x\notin R\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=15\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{2;15\right\}\)
PT\(\Leftrightarrow\)\(\left[\left(x-3\right)\left(x-10\right)\right]\left[\left(x-5\right)\left(x-6\right)\right]=24x^2\)
\(\Leftrightarrow\)\(\left(x^2-13x+30\right)\left(x^2-11x+30\right)=24x^2\)
Nhận thấy x=0 không là nghiệm của PT. Chia cả hai vế của phương trình cho \(x^2\) ta được:
PT\(\Leftrightarrow\)\(\left(x-13+\dfrac{30}{x}\right)\left(x-11+\dfrac{30}{x}\right)=24\)
Đặt \(x+\dfrac{30}{x}=t\) (1)
PT\(\Leftrightarrow\)\(\left(t-13\right)\left(t-11\right)=24\)
Tìm được \(\left[{}\begin{matrix}t=17\\t=7\end{matrix}\right.\)
Thay vào (1):\(\left[{}\begin{matrix}x^2-17x+30=0\\x^2-7x+30=0\end{matrix}\right.\)
Tìm được \(\left[{}\begin{matrix}x=15\\x=2\end{matrix}\right.\)
(x-3)(x-5)(x-6)(x-10)-24x2
=(x-3)(x-10)(x-5)(x-6)-24x2
=(x2-13x+30)(x2-11x+30)-24x2
Đặt x2-12x+30=k
Khi đó ta có:
(k-x)(k+x)-24x2=k2-x2-24x2=k2-25x2
=(k-5x)(k+5x)
=(x2-12x+30-5x)(x2-12x+30+5x)
=(x2-17x+30)(x2-7x+30)
=(x2-2x-15x+30)(x2-7x+30)
=(x-2)(x-15)(x2-7x+30)