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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: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(A=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+10=y\)
\(\Rightarrow\)\(A=y.\left(y+2\right)-24\)
\(A=y^2+2y+1-25\)
\(A=\left(y+1\right)^2-5^2\)
\(A=\left(y+1-5\right)\left(y+1+5\right)\)
\(A=\left(y-4\right)\left(y+6\right)\)
\(\Rightarrow A=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)
\(A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)
\(A=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)
Đặt \(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(B=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=a\)
\(\Rightarrow B=a.\left(a+3\right)-4\)
\(B=a^2+3a-4\)
\(B=\left(a^2-a\right)+\left(4a-4\right)\)
\(B=a.\left(a-1\right)+4.\left(a-1\right)\)
\(B=\left(a-1\right)\left(a+4\right)\)
\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
a) \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(= \left(x^2+8x+7\right)\left(x^2+5x+3x+15\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+7=t\), ta đc:
⇒ \(t\left(t+8\right)+15\) = \(t^2+8t+15=\left(t+5\right)\left(t+3\right)\)
b)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x+2=t\)
⇒\(t\left(t-3\right)-4\)=\(\left(t-4\right)\left(t+1\right)\)
c) tương tự nha
\(P\left(x\right)=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left[\left(4x+1\right)\left(3x+2\right)\right].\left[\left(12x-1\right)\left(x+1\right)\right]-4\)
\(=\left(12x^2+8x+3x+2\right).\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right).\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x=t\), ta có:
\(\left(t+2\right)\left(t-1\right)-4\)
\(=t^2-t+2t-2-4=t^2+t-6\)
\(=t^2-2t+3t-6\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
Thay \(t=12x^2+11x\), ta được:
\(P\left(x\right)=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
Đs...
\(A=\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)-18\)
\(=\frac{1}{4}\left[\left(2x+1\right)\left(x+1\right)^2.4\left(2x+3\right)\right]-72\)
\(=\frac{1}{4}\left[\left(2x+1\right)\left(2x+3\right)\left(2x+2\right)^2\right]-72\)
\(=\frac{1}{4}\left[\left(4x^2+8x+3\right)\left(4x^2+8x+4\right)-72\right]\)
Đặt: \(4x^2+8x+3=t\)
Ta có: \(A=\frac{1}{4}\left[t^2+t-72\right]\)
\(=\frac{1}{4}\left[\left(t+9\right)\left(t-8\right)\right]\)
\(=\frac{1}{4}\left[\left(4x^2+8x+12\right)\left(4x^2+8x-5\right)\right]\)
\(=\left(x^2+2x+3\right)\left[4x^2+8x-5\right]\)
\(=\left(x^2+2x+3\right)\left(2x-1\right)\left(2x+5\right)\)
\(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left[\left(4x+1\right)\left(3x+2\right)\right]\left[\left(12x-1\right)\left(x+1\right)\right]-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x+2=a\)
Khi đó: \(B=a\left(a-3\right)-4\)
\(=a^2-3a-4=\left(a+1\right)\left(a-4\right)\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
\(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
\(=x^4+x^2+4-2x^3-4x+4x^2+x^2-4x+4\)
\(=x^4-2x^3+6x^2-8x+8\)
\(=x^4-2x^3+2x^2+4x^2-8x+8\)
\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)=\left(x^2-2x+2\right)\left(x^2+4\right)\)
\(3x^4-5x^3-18x^2-3x+5\)
\(=3x^4+x^3-x^2-6x^3-2x^2+2x-15x^2-5x+5\)
\(=x^2\left(3x^2+x-1\right)-2x\left(3x^2+x-1\right)-5\left(3x^2+x-1\right)\)
\(=\left(3x^2+x-1\right)\left(x^2-2x-5\right)\)
Bài này thật sự khó và hay đấy.
=(4x+1)(3x+2)(12x-1)(x+1)-4
=(12x2+11x+2)(12x2+11x-1)-4
đặt a=12x2+11x+2
khi đó đa thức trở thành:
a(a-3)-4
=a2-3a-4
=a2+a-4a-4
=a(a+1)-4(a+1)
=(a+1)(a-4)
thay x vào là ok
