Bài 3. (2 điểm) Phân tích đa thức sau thành nhân tử:
a) $5x^2+10xy-4x-8y$;
b) $4x^2+4x-y^2+1$.
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`x^3 - y^6 `
`= x^3 - (y^2)^3`
`= (x - y^2) (x^2 + xy^2 + y^4) `
\(\left(4x-1\right)\left(4x+1\right)=\left(4x-1\right)^2\)
\(\Leftrightarrow\left(4x-1\right)\left(4x+1\right)-\left(4x-1\right)^2=0\)
\(\Leftrightarrow\left(4x-1\right)\left[\left(4x+1\right)-\left(4x-1\right)\right]=0\)
\(\Leftrightarrow2\left(4x-1\right)=0\)
\(\Leftrightarrow4x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
(4\(x\) - 1).(4\(x\) + 1) = (4\(x\) - 1)2
(4\(x-1\)).(4\(x\) + 1) - (4\(x\) - 1)2 = 0
(4\(x\) - 1).(4\(x\) + 1 - 4\(x\) + 1) = 0
(4\(x\) - 1).[(4\(x\) - 4\(x\)) + (1 +1)] = 0
(4\(x\) - 1).[0 + 2] = 0
(4\(x\) - 1).2 = 0
4\(x\) - 1 = 0
4\(x\) = 1
\(x=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4}\)
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a; A = 5\(x^2\) + 10\(xy\) - 4\(x\) - 8y
A = (5\(x^2\) + 10\(xy\)) - (4\(x\) - 8y)
A = 5\(x\).(\(x\) + 2\(y\)) - 4.(\(x+2y\))
A = (\(x+2y\)).(5\(x\) - 4)
B = 4\(x^2\) + 4\(x\) - y2 + 1
B = (4\(x^2\) + 4\(x\) + 1) - y2
B = [(2\(x\))2 + 2.2\(x\).1 + 12] - y2
B = [2\(x\) + 1]2 - y2
B = (2\(x+1\) - y)(2\(x+1\) - y)