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Bài 2:
a)A= \(6x^2\)\(-11x+3\)
<=>A=\(6x^2\)\(-2x-9x+3\)
<=>A=(\(6x^2\)\(-2x\))-\(\left(9x-3\right)\)
=>A=\(2x\left(3x-1\right)\)\(-3\left(3x+1\right)\)
<=>A=\(2x\left(3x-1\right)+3\left(3x-1\right)\)
=>A=(3x-1)(2x+3)
b)(x2+x+1)(x2+x+2)-12
Đặt t=x2+x+1
t(t+1)-12=t2+t-12
=(t-3)(t+4)=(x2+x+1-3)(x2+x+1+4)
=(x2+x-2)(x2+x+5)
=(x-1)(x+2)(x2+x+5)
c)(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=t2+8t+15
=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+15)
=(x2+8x+10)(x2+8x+22)
d)(x+2)(x+3)(x+4)(x+5)-24
=(x2+7x+10)(x2+7x+12)-24
Đặt t=x2+7x+10
t(t+2)-24=(t-4)(t+6)
=(x2+7x+10-4)(x2+7x+10+6)
=(x2+7x+6)(x2+7x+16)
=(x+1)(x+6)(x2+7x+16)
a/ Đặt x2 + 4x + 8 = a
Thì đa thức ban đầu thành
a2 + 3ax + 2x2 = (a2 + 2ax + x2) + (ax + x2)
= (a + x)2 + x(a + x) = (a + x)(a + 2x)
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 \(12x^2+11x-1=a\)
\(\Rightarrow B=a.\left(a+3\right)-4\)
\(B=a^2+3a-4\)
\(B=\left(a^2-a\right)+\left(4a-4\right)\)
\(B=a.\left(a-1\right)+4.\left(a-1\right)\)
\(B=\left(a-1\right)\left(a+4\right)\)
\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-5\\x=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)
Vậy .........
\(b,\left(x^2-4\right)+\left(x-2\right)\left(3-2x=0\right)\)
\(\Leftrightarrow x^2-4-2x^2+7x-6=0\)
\(\Leftrightarrow-x^2+7x-10=0\)
\(\Leftrightarrow-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
Vậy ..................
\(c,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x=1\)
\(d,x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow2x^2-7x-4x+14=0\)
\(\Leftrightarrow2x^2-11x+14=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
Vậy ............
\(e,\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow4x^2-20x+25-x^2-4x-4=0\)
\(\Leftrightarrow3x^2-24x+21=0\)
\(\Leftrightarrow3\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=1\end{cases}}\)
Vậy .....................
\(f,x^2-x-\left(3x-3\right)=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy ..............
\(A=\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)-18\)
\(=\frac{1}{4}\left[\left(2x+1\right)\left(x+1\right)^2.4\left(2x+3\right)\right]-72\)
\(=\frac{1}{4}\left[\left(2x+1\right)\left(2x+3\right)\left(2x+2\right)^2\right]-72\)
\(=\frac{1}{4}\left[\left(4x^2+8x+3\right)\left(4x^2+8x+4\right)-72\right]\)
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.
Đặt A=(x+2)(x+3)(x+4)(x+5)-24
= (x+2)(x+5)(x+3)(x+4)-24
= (x^2+7x+10)(x^2+7x+12)-24
Đặt x^2+7x+11 = a thay vào A ta được :
A=(a-1)(a+1)=a^2-25 = a^2 - 5^2 = (a-5)(a+5) ( 2)
Thế a vào (2) ta được :
A=(x^2+7x+11-5)(x^2+7x+11+5)
= (x^2+7x+6)(x^2+7x+16)