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a) Ta có: \(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{x^2+38x+4-3x^2+4x+2}{2x^2+17x+1}\)
\(=\dfrac{-2x^2+42x+6}{2x^2+17x+1}\)
c) Ta có: \(C=\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)
\(=\dfrac{-x+7x-4}{3x-2}\)
\(=\dfrac{6x-4}{3x-2}=2\)
a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
\(a.\left(3x-2\right)^2+\left(3x+2\right)^2-2\left(\left(3x-2\right)\left(3x+2\right)\right)\\=9x^2-12x+4+9x^2+12x+4-18x^2+8\\ =18x^2+8-18x^2+8\\ =16\)
1) \(3x^2-9=3\left(x^2-3\right)\)
2) \(\dfrac{1}{2}x^2-2y^2\)
\(=\dfrac{1}{2}\left(x^2-4y^2\right)\)
\(=\dfrac{1}{2}\left(x-2y\right)\left(x +2y\right)\)
3) \(3x^2-12y^2\)
\(=3\left(x^2-4y^2\right)\)
\(=3\left(x-2y\right)\left(x+2y\right)\)
4) \(\dfrac{1}{3}x^2y^2-3x^2\)
\(=\dfrac{1}{3}x^2\left(y^2-9\right)\)
\(=\dfrac{1}{3}x^2\left(y-3\right)\left(y+3\right)\)
#Toru
\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(3x+2\right)\left(3x-2\right)+x\\=\left[\left(3x+2\right)^2-2\left(3x+2\right)\left(3x-2\right)+\left(3x-2\right)^2\right]+x \\=\left(3x+2-3x+2\right)^2+x\\ \\=4^2+x\\ \\=x+16\)