phan tich da thuc thanh nhan tu
a)x4+x2+x+1
b)11x4+4x2-5
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\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
\(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
Ta có : x5 + x + 1
= x5 + x4 - x4 - x3 + x3 + x2 - x2 - x + x + 1
= (x5 + x4) - (x4 + x3) + (x3 + x2) - (x2 + x) + (x + 1)
= x5(x + 1) - x4.(x + 1) + x3(x + 1) - x2(x + 1) + (x + 1)
= (x + 1)(x5 - x4 + x3 - x2 + 1)
\(x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1 = x3(x2-x+1)+x2(x2-x+1)-(x2-x+1) = (x2-x+1)(x3+x2-1) \)
hc tốt nha !!!!!!!!!
\(x^5+x+1=x^5-x^2+x^2+x+1=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
a) \(x^7+x^2+1\)
\(=x^7-x+x+x^2+1\)
\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^4+x\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^2-x\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^2-x+1\right)\left(x^2+x+1\right)\)
b) \(x^7+x^5+1\)
\(=x^7+x^6+x^5-x^6+1\)
\(=\left(x^7+x^6+x^5\right)-\left(x^6-1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x^3+1\right)\left(x^3-1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x^4-x^3+x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^3-x^2+1\right)\left(x^2+x+1\right)\)
\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left[\left(x-1\right)\left(x-7\right)\right].\left[\left(x-3\right)\left(x-5\right)\right]-20\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)
Đặt \(x^2-8x+11=t\) \(\Rightarrow\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20=\left(t-4\right)\left(t+4\right)-20=t^2-16-20=t^2-36=\left(t-6\right)\left(t+6\right)\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left(x^2-8x+11-6\right)\left(x^2-8x+11+6\right)=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)
b ( x^2 + 3x + 2)( x^2 + 7x + 12) - 24
= [ x^2 +x + 2x + 2) ( x^2 +3x + 4x + 12) - 24
= [x(x+1) + 2 (x + 1) [x(x+3) + 4(x+3) ] - 24
= ( x + 1)(x+2) (x+3)(x+4) - 24
= ( x + 1).(x+4) (x+2)(x+3) - 24
=(x^2 + 5x + 4)(x^2+5x+6) - 24
Đặt x^2 + 5x +4 =y ta có:
= y(y+2) - 24
= y^2 + 2y - 24
= y^2 + 2y + 1 - 25
= ( y + 1)^2 - (5)^2
= ( y + 1 - 5 )( y + 1 + 5)
= ( y- 4)(y +6)
Thay y trở lại là đc
đúng nha