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b) (x+2).(x+3).(x+4).(x+5) - 24
= [(x+2).(x+5)].[(x+3).(x+4)] - 24
= [ x2 + 7x + 10].[x2 + 7x + 12] - 24
= [x2 + 7x + 11 -1].[x2 + 7x + 11+1] - 24
= (x2 +7x+11)2 - 12 - 24
= (x2 +7x+11)2 - 25
= (x2 +7x + 11 - 25).(x2 +7x + 11 + 25)
= (x2 + 7x - 14).(x2 + 7x + 36)
a) \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(m=x^2+x+1\)ta có :
\(A=m\left(m+1\right)-12\)
\(A=m^2+m-12\)
\(A=m^2+4m-3m-12\)
\(A=m\left(m+4\right)-3\left(m+4\right)\)
\(A=\left(m+4\right)\left(m-3\right)\)
Lại thay m vào ta có :
\(A=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
b) \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(A=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(A=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(p=x^2+7x+11\)ta có :
\(A=\left(p-1\right)\left(p+1\right)-24\)
\(A=p^2-1^2-24\)
\(A=p^2-25\)
\(A=\left(p-5\right)\left(p+5\right)\)
Lại thay p vào A ta có :
\(A=\left(x^2+7x+5\right)\left(x^2+7x+15\right)\)
a) \(\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
Đặt \(x^2+x=y\) ta được:
\(y^2-14y+24\)
\(=x\left(y-12\right)-2\left(y-12\right)\)
\(=\left(y-2\right)\left(y-12\right)\)
Thay ngược trở lại:
\(\left(x^2+x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x-3\right)\left(x+4\right)\)
d) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+10\right)+1\)
Đặt \(x^2+5x+4=a\) được:
\(a\left(a+6\right)+1\)
\(=a^2+6a+1\)
\(=a^2+2.a.3+3^2-8\)
\(=\left(a+3\right)^2-\left(\sqrt{8}\right)^2\)
\(=\left(a+3-\sqrt{8}\right)\left(a+3+\sqrt{8}\right)\)
Mấy câu kia tương tự.
a)
\((x^2+x)^2+4x^2+4x=(x^2+x)^2+4(x^2+x)\)
\(=(x^2+x)(x^2+x+4)\)
\(=x(x+1)(x^2+x+4)\)
b) \(x(x+1)(x+2)(x+3)+1\)
\(=[x(x+3)][(x+1)(x+2)]+1\)
\(=(x^2+3x)(x^2+3x+2)+1\)
\(=(x^2+3x)^2+2(x^2+3x)+1\)
\(=(x^2+3x+1)^2\)
c)
\((x+2)(x+3)(x+4)(x+5)-24\)
\(=[(x+2)(x+5)][(x+3)(x+4)]-24\)
\(=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=a\)
Khi đó biểu thức bằng:
\(a(a+2)-24=a^2+2a-24=a^2-4a+6a-24\)
\(=a(a-4)+6(a-4)=(a-4)(a+6)\)
\(=(x^2+7x+10-4)(x^2+7x+10+6)\)
\(=(x^2+7x+6)(x^2+7x+16)\)
\(=(x^2+x+6x+6)(x^2+7x+16)\)
\(=(x+1)(x+6)(x^2+7x+16)\)
Nhân 2 vế với 2 rồi chuyển vế và rút gọn
Bạn Tên Là Long
a/ \(\Leftrightarrow2x^3+9x^2-27=0\)
\(\Leftrightarrow2x^3+12x^2+18x-3x^2-18x-27=0\)
\(\Leftrightarrow2x\left(x^2+6x+9\right)-3\left(x^2+6x+9\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x+3\right)^2=0\)
\(\Leftrightarrow...\)
b/ \(\Leftrightarrow x^3-3x^2+3x-1+x^3+x^3+3x^2+3x+1=x^3+6x^2+12x+8\)
\(\Leftrightarrow x^3-3x^2-3x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+x+1\right)=0\)
c/ \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt \(x^2+x=t\)
\(t\left(t-2\right)-24=0\Leftrightarrow t^2-2t-24=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x=6\\x^2+x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+4=0\end{matrix}\right.\)
d/ \(\Leftrightarrow\left(x-7\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)-72=0\)
\(\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\)
Đặt \(x^2-9x+14=0\)
\(t\left(t+6\right)-72=0\Leftrightarrow t^2+6t-72=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-9x+14=6\\x^2-9x+14=-12\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-9x+8=0\\x^2-9x+26=0\end{matrix}\right.\)
a) \(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left(x+2\right)^2=3^2\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=3-2=1\)
a) ( x + 2 )2 = 9
=> ( x + 2 ) 2 = 9
=> ( x + 2 )2 = 32
=> x + 2 = + 3
=> \(\orbr{\begin{cases}x+2=-3\\x+2=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
Vậy x = -1; 5
b) ( x + 2 )2 - x2 + 4 = 0
=> ( x + 2 )2 - ( x2 - 4 ) = 0
=> ( x + 2 )2 - ( x + 2 ) ( x - 2 ) = 0
=> ( x + 2 ) ( x + 2 - x + 2 ) = 0
=> ( x + 2 ) . 4 = 0
=> x + 2 = 0
=> x = - 2
Vậy x = - 2
c) 5 ( 2x - 3 )2 - 5 ( x + 1 )2 - 15( x + 4 ) ( x - 4 ) = - 10
=> 5 ( 4x2 - 12x + 9 ) - 5 ( x2 + 2x + 1 ) - 15 ( x2 - 42 ) = - 10
=> 20x2 - 60x + 45 - 5x2 - 10x - 5 - 15x2 + 240 = -10
=> - 70x + 280 = - 10
=> - 70x = - 290
=> x = \(\frac{29}{7}\)
Vậy x = \(\frac{29}{7}\)
d) x ( x + 5 ) ( x - 5 ) - ( x + 2 ) ( x2 - 2x + 4 ) = 3
=> x ( x2 - 25 ) - ( x3 - 8 ) = 3
=> x3 - 25x - x3 + 8 = 3
=> - 25x + 8 = 3
=> - 25x = -5
=> x = \(\frac{1}{5}\)
Vậy x = \(\frac{1}{5}\)
a)x4+2x3+5x2+4x-12
=(x4+2x3+x2)+(4x2+4x)-12
=(x2+x)2+4(x2+x)-12
Đặt t=x2+x
=t2+4t-12=(t-2)(t+6)
=(x2+x-2)(x2+x+6)
=(x-1)(x+2)(x2+x+6)
b)(x+1)(x+2)(x+3)(x+4)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt x2+5x+4=t
t(t+2)+1=t2+2t+1
=(t+1)2=(x2+5x+4+1)2
=(x2+5x+5)2
c)(x+1)(x+3)(x+5)(x+7)+15
=(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+5)
=(x2+8x+10)(x+2)(x+6)
d)(x+1)(x+2)(x+3)(x+4)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=x2+5x+4
t(t+2)-24=(t-4)(t+6)
=(x2+5x+4-4)(x2+5x+4+6)
=x(x+5)(x2+5x+10)
a ) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+\dfrac{3}{2}-\dfrac{1}{2}\right)\left(x^2+x+\dfrac{3}{2}+\dfrac{1}{2}\right)-12\)
\(=\left(x^2+x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}-12\)
\(=\left(x^2+x+\dfrac{3}{2}\right)^2-\dfrac{49}{4}\)
\(=\left(x^2+x+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x^2+x+\dfrac{3}{2}+\dfrac{7}{2}\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(=\left[x\left(x+2\right)-\left(x+2\right)\right]\left(x^2+x+5\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)
b ) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-24\)
\(=\left(x^2+7x+11\right)^2-1-24\)
\(=\left(x^2+7x+11\right)^2-25\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left[x\left(x+1\right)+6\left(x+1\right)\right]\left(x^2+7x+16\right)\)
\(=\left(x+6\right)\left(x+1\right)\left(x^2+7x+16\right)\)
a ) Đặt \(t=x^2+x+1\)
b ) \(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)\)
Đặt \(t=x^2+7x+10\)