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A/\(\left(2x^3+y^2-7xy\right)4xy^2.\)
\(=8x^4y^2+4xy^4-28x^2y^3\)
B/\(\left(2x^3-x-1\right)\left(5x-2\right)\)
\(=10x^4-5x^2-5x-4x^3+2x+2\)
\(=10x^4-5x^3-3x-4x^3+2\)
C/\(\left(2x^2-3\right)\left(4x^4+6x^2+9\right)\)
\(=\left(2x^2-3\right)\left(2x+3\right)^2\)
D/\(\left(3x^2-2y\right)^3-\left(2x^2-y\right)^3\)
( Bài này áp dụng hằng đẳng thức là làm được ạ )
\(\frac{x^2-3x-x+3}{x-3}=\frac{x\left(x-3\right)-\left(x-3\right)}{x-3}=\frac{\left(x-3\right)\left(x-1\right)}{x-3}=x-1\)( ĐK: \(x\ne3\))
\(\frac{2x^3-5x^2-4x+3}{2x-1}=\frac{\left(2x^3-x^2\right)-\left(4x^2-2x\right)-\left(6x-3\right)}{2x-1}=\frac{x^2\left(2x-1\right)-2x\left(2x-1\right)-3\left(2x-1\right)}{2x-1}=\frac{\left(2x-1\right)\left(x^2-2x-3\right)}{2x-1}=x^2-2x-3\)( ĐK: \(x\ne\frac{1}{2}\))
Tham khảo nhé~
\(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\frac{x^2\left(3x^2-2x+1\right)-2x\left(3x^2-2x+1\right)-5\left(3x^2-2x+1\right)}{3x^2-2x+1}\)
\(=\frac{\left(3x^2-2x+1\right)\cdot\left(x^2-2x-5\right)}{3x^2-2x+1}\)
\(=x^2-2x-5\)
\(\frac{2x^3-9x^2+19x-15}{x^2-3x+5}\)
\(=\frac{2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)}{x^2-3x+5}\)
\(=\frac{\left(x^2-3x+5\right)\left(2x-3\right)}{x^2-3x+5}\)
\(=2x-3\)
P/s : Phá ngoặc ra là ok :
a )
\(\left[4x-2\left(x-3\right)\right].\left(-3x\right)\)
\(=\left[4x-2x+6\right]\left(-3x\right)\)
\(=-12x^2+6x^2-18x\)
b )
\(3\left[x-3\left(4-2x\right)+8\right]\)
\(=3\left[x-12+6x+8\right]\)
\(=3\left[7x-4\right]\)
\(=21x-12\)
c )
\(5\left(3x^2-4y^3\right)+9\left(2x^2-y^3\right)\)
\(=15x^2-20y^3+18x^2-9y^3\)
\(=33x^2-29y^3\)
d )
\(3x^2\left(2y-1\right)-2x^2\left(5y-3\right)\)
\(=6x^2y-3x^2-10x^2y+6x^2\)
\(=-4x^2y+3x^2\)
\(3x^4-4x^3+2x\left(x^3-2x^2+7x\right)\)
\(=3x^4-4x^3+2x^4-4x^3+14x^2\)
\(=5x^4-8x^3+14x^2\)
3x4 - 4x3 + 2x(x3 - 2x2 + 7x )
= 3x4 - 4x3 + 2x4 _ 4x3 + 14x2
= 5x4 - 8x3 + 14x2
a: \(=2x^2-x+5\)
b: \(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c: \(=-x^3+\dfrac{3}{2}-2x\)
d: \(=-2x^2+4xy-6y^2\)
e: \(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
a)\(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=2x^2\left(5x^2-2x+1\right)-3x\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
a. \(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
b. \(\left(2x^4-x^3+3x^2\right):\left(\frac{1}{3}x^2\right)\)
\(=\left(2x^4-x^3+3x^2\right).\frac{3}{x^2}\)
\(=0,6x^2-3x+0,9\)
a) \(\left(2x-3\right)\left(x^2-2x+1\right)+2\left(2-x\right)^3\)
\(=2x\left(x^2-2x+1\right)-3\left(x^2-2x+1\right)+2\left(2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\right)\)
\(=2x^3-4x^2+2x-3x^2+6x-3+2\left(8-12x+6x^2-x^3\right)\)
\(=2x^3-4x^2+2x-3x^2+6x-3+16-24x+12x^2-2x^3\)
\(=\left(2x^3-2x^3\right)+\left(-4x^2-3x^2+12x^2\right)+\left(2x+6x-24x\right)+\left(-3+16\right)\)
\(=5x^2-16x+13\)
b)
2x^3 - 7x^2 + 2x + 3 x^2 - 4x + 3 2x^3 - 8x^2 + 6x x^2 - 4x + 3 2x + 1 - x^2 + 4x + 3 0
Vậy \(\left(2x^3-7x^2+2x+3\right):\left(x^2-4x+3\right)=2x+1\)
Câu b thêm dấu " - " ở chỗ 2x3 - 7x2 + 2x +3 và 2x3 - 8x2 + 6x nhé :)))