phân tích đa thức thành nhân tử
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+8\)
Phân tích đa thức thành nhân tử: \(\left(x+5\right)^2+4\left(x+5\right)\left(x-5\right)+4\left(x^2-10x+25\right)=0\)
\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)
Phân tích đa thức thành nhân tử:
\(\left(x+3\right)^4+\left(x+5\right)^4-2\)
`(x+3)^4+(x+5)^4-2`
`={[(x+3)^2]^2-1^2}+{[(x+5)^2]^2 -1^2}`
`=[(x+3)^2-1^2][(x+3)^2+1]+[(x+5)^2-1^2][(x+5)^2+1]`
`=(x+3-1)(x+3+1)[(x+3)^2+1]+(x+5-1)(x+5+1)[(x+5)^2+1]`
`=(x+2)(x+4)[(x+3)^2+1]+(x+4)(x+6)[(x+5)^2+1]`
`=(x+4){(x+2)[(x+3)^2+1]+(x+6)[(x+5)^2+1]}`
`=(x+4)(2x^3+24x^2+108x+176)`
\(\left(x+3\right)^4+\left(x+5\right)^4-2\)
\(=\left[\left(x+3\right)^4-1\right]+\left[\left(x+5\right)^4-1\right]\)
\(=\left[\left(x^2+6x+9-1\right)\left(x^2+6x+9+1\right)\right]+\left[\left(x^2+10x+25-1\right)\left(x^2+10x+25+1\right)\right]\)
\(=\left(x^2+6x+8\right)\left(x^2+6x+10\right)+\left(x^2+10x+24\right)\left(x^2+10x+26\right)\)
\(=\left(x+2\right)\left(x+4\right)\left(x^2+6x+10\right)+\left(x+4\right)\left(x+6\right)\left(x^2+10x+26\right)\)
\(=\left(x+4\right)\left[\left(x+2\right)\left(x^2+6x+10\right)+\left(x+6\right)\left(x^2+10x+26\right)\right]\)
\(=\left(x+4\right)\left(x^3+6x^2+10x+2x^2+12x+20+x^3+10x^2+26x+6x^2+60x+156\right)\)
\(=\left(x+4\right)\left(2x^3+24x^2+108x+176\right)\)
\(=2\left(x+4\right)\left(x^3+12x^2+54x+88\right)\)
Phân tích đa thức thành nhân tử
\(5x\left(2x+3\right)+6x+9\)
\(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)
\(5x(2x+3)+6x+9\\=5x(2x+3)+3(2x+3)\\=(2x+3)(5x+3)\)
a: \(5x\left(2x+3\right)+6x+9\)
\(=5x\left(2x+3\right)+\left(6x+9\right)\)
\(=5x\left(2x+3\right)+3\left(2x+3\right)\)
\(=\left(2x+3\right)\left(5x+3\right)\)
b: \(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+4\right)\left(3x+48+5\right)\)
=(x+4)(3x+53)
Phân tích đa thức \(18x^3-\dfrac{8}{25}x\) thành nhân tử
a. \(\dfrac{2}{25}x\left(9x^2-4\right)=\dfrac{2}{25}x\left(3x-2\right)\left(3x+2\right)\)
b. \(2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)
Cách phân tích nào đúng, a hay b. Giải thích vì sao?
Phân tích đa thức thành nhân tử
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
Ta có:
(x + 2)(x + 3)(x + 4)(x + 5) - 24
= [(x + 2)(x + 5)][(x + 3)(x + 4)] - 24
= (x2 + 5x + 2x + 10)(x2 + 4x + 3x + 12) - 24
= (x2 + 7x + 10)(x2 + 7x + 12) - 24
Đặt x2 + 7x + 10 = k
=> k(k + 2) - 24 = k2 + 2k - 24 = k2 + 6x - 4x - 24
= k(k + 6) - 4(k + 6)
= (k - 4)(k + 6)
=> (x + 2)(x + 3)(x + 4)(x + 5) - 24
= (x2 + 7x + 10 - 4)(x2 + 7x + 10 + 6)
= (x2 + 7x + 6)(x2 + 7x + 16)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)
Đặt \(x^2+7x+11=t\)thay vào (1) ta được:
\(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)Thay \(t=x^2+7x+11\)ta được:
\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)
\(=\left[x\left(x+1\right)+6\left(x+1\right)\right]\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(\left(x+2\right).\left(x+5\right).\left(x+3\right).\left(x+4\right)-24\)
\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Dat \(x^2+7x+10=y\)
ta co :\(y\left(y+2\right)-24=y^2+2y-24\)
\(y^2+6y-4y-24=\left(y^2+6y\right)-\left(4y+24\right)\)
\(y\left(y+6\right)-4\left(y+6\right)=\left(y-4\right)\left(y+6\right)\)
tra lai bien ta co
\(\left(x^2+7x+10-4\right)\left(x^2+7x+12+6\right)\)
\(\left(x^2+7x+6\right)\left(x^2+7x+18\right)=\left(x^2+6x+x+6\right)\left(x^2+7x+18\right)\)
\(\left(x+1\right)\left(x+6\right)\left(x^2+7x+18\right)\)
Phân tích đa thức thành nhân tử:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10+16+8\right)+16\)
\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)
\(=\left(x^2+10x+16+4\right)^2\)
\(=\left(x^2+10+20\right)^2\)
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right)
\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\)
\(\left(1\right)\Rightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\)
\(=\left(x^2+10x+20\right)^2\)
phân tích đa thức thành nhân tử:
\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\)
\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\)
\(=\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+16\)
\(=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\)(1)
Đặt \(x^2-10x+20=t\)thay vào (1) ta được :
\(\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-16+16\)
\(=t^2\)Thay \(t=x^2-10x+20\)ta được :
\(\left(x^2-10x+20\right)^2\)
\(=\left(x^2-2.5.x+25-25+20\right)^2\)
\(=\left[\left(x-5\right)^2-5\right]^2\)
\(=\left(x-5-\sqrt{5}\right)^2\left(x-5+\sqrt{5}\right)^2\)
phân tích đa thức thành nhân tử :
a, \( \left(x-5\right)^2-4\left(x-3\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2\)
(x - 5)2 - 4(x - 3)2 + 2(2x - 1)(x - 5) + (2x - 1)2
= [(x - 5)2 + 2(2x - 1)(x - 5) + (2x - 1)2) - [2(x - 3)]2
= (x - 5 + 2x - 1)2 - (2x - 6)2
= (3x - 6)2 - (2x - 6)2
= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)
( x - 5 )2 - 4( x - 3 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2
= [ ( x - 5 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2 ] - 22( x - 3 )2
= ( x - 5 + 2x - 1 )2 - ( 2x - 6 )2
= ( 3x - 6 )2 - ( 2x - 6 )2
= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )
= x( 5x - 12 )
\(\left(x-5\right)^2-4\left(x-3\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2\)
\(=\left(x-5\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2-4\left(x-3\right)^2\)
\(=\left(x-5+2x-1\right)^2-\left(2x-6\right)^2\)
\(=\left(3x-6\right)^2-\left(2x-6\right)^2\)
\(=\left[\left(3x-6\right)-\left(2x-6\right)\right].\left[\left(3x-6\right)+\left(2x-6\right)\right]\)
\(=\left(3x-6-2x+6\right)\left(3x-6+2x-6\right)\)
\(=\left(5x-12\right)x\)
Phân tích đa thức thành nhân tử
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ \)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Let \(t=x^2+7x+10\) we have:
\(=t\left(t+2\right)-24=t^2+2t-24\)
\(=\left(t-4\right)\left(t+6\right)=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Phân tích đa thức sau thành nhân tử:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+18\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)-16\)
\(=\left(x^2+10x+20\right)^2-16+16=\left(x^2+10x+20\right)^2\)
Chúc bạn học tốt.
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(\Rightarrow\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+6\right)\left(x+8\right)\right]+16\)
\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(\Rightarrow\left(x^2+10x+16\right)\left[\left(x^2+10x+16\right)+8\right]+16\)
\(\Rightarrow\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+4^2\)
\(\Rightarrow\left(x^2+10x+20\right)^2\)