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mình biết nội quy rồi nên đưng đăng nội quy
ai chơi bang bang 2 kết bạn với mình
mình có nick có 54k vàng đang góp mua pika
ai kết bạn mình cho
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6, (x^2 +1) -4x^2 = x^2 + 1 - 4x^2 = 1 - (4x^2 - x^2) = 1 - 3x^2 = (1-\(\sqrt{3}\)x)(1+\(\sqrt{3}\)x)
7, x^2 - 4x -5 = x^2 - 2.x.2 + 4 - 9 = (x^2 - 2.x.2 +4) - 3^2 = (x-2)^2 - 3^2 = (x-2-3)(x-2+3) = (x-5)(x+1)
8, x^5 - 3x^4 + 3x^3 - x^2 = x^2(x^3 -3x^2 + 3x -1) = x^2(x-1)^3
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\(x^5-3x^4-x^3-x^2+3x+1\)
\(=\left(x^5-x^2\right)-\left(3x^4-3x\right)-\left(x^3-1\right)\)
\(=x^2\left(x^3-1\right)-3x\left(x^3-1\right)-\left(x^3-1\right)\)
\(=\left(x^3-1\right)\left(x^2-3x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left[\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\right]\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-\frac{3}{2}-\frac{\sqrt{13}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{13}}{2}\right)\)
\(x^5-3x^4-x^3-x^2+3x+1\)\(1\)\(=\left(x^5-x^4\right)-\left(2x^4-2x^3\right)-\left(3x^3-3x^2\right)-\left(4x^2-4x\right)-\left(x-1\right)\)
\(=x^4\left(x-1\right)-2x^3\left(x-1\right)-3x^2\left(x-1\right)-4x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4-2x^3-3x^2-4x-1\right)\)
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câu b) )Đa thức <=>
3.x^3 - x^2 - 6x^2 + 2x +15x - 5
= x^2.(3x - 1) - 2x(3x-1) + 5(3x-1)
= (3x-1).(x^2 - 2x +5)
Tick nhé
\(=x^5-2x^4+x^3-x^4+2x^3-x^2\)
\(=x^3\left(x^2-2x+1\right)-x^2\left(x^2-2x+1\right)\)
\(=\left(x^2-2x+1\right)\left(x^3-x^2\right)\)
\(=\left(x-1\right)^2x^2\left(x-1\right)=\left(x-1\right)^3x^2\)
\(=x^2\left(x^3-1\right)-3x^3\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1-3x\right)\)
\(=x^2\left(x-1\right)\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)\left(x-1\right)^2\)
\(=x^2\left(x-1\right)^3\)