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a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
Mình đang lái xe đạp xuống dốc cao đang lên raaaat khổ mong mọi người giúp!PLS PLS PLS PLS PLS PLS PLS PLS
\(q^3\left(2x-1\right)+5\left(x-3\right)\)
\(=2xq^3-q^3+5\left(x-3\right)\)
\(=2xq^3-q^3+5x-15\)
Câu 1:
\(3x\left(12x+4\right)+9x\left(4x+3\right)\)
\(\Leftrightarrow3x\left(12x+4\right)+3x\left[3.\left(4x+3\right)\right]\)
\(\Leftrightarrow3x\left(12x+4\right)+3x\left(12x+6\right)\)
\(\Leftrightarrow3x\left[12x+4+12x+6\right]\)
\(\Leftrightarrow3x.\left(24x+10\right)\)
\(\Leftrightarrow72x^2+30x\)
Câu 2:
\(x\left(5+2x\right)+2x^2\left(x-1\right)\)
\(\Leftrightarrow5x+2x^2+2x^3-2x^2\)
\(\Leftrightarrow2x^3+5x\)
Đề 1
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Input:
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Inequality plot:
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Alternate forms:
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Expanded form:
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Solution:
- Approximate form
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Integer solution:
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\(\dfrac{8x^3y^2-6x^2y^3}{-2xy}=\dfrac{8x^3y^2}{-2xy}+\dfrac{6x^2y^3}{2xy}=-4x^2y+3xy^2\)
⇒ Chọn A.
\(\begin{array}{l} - 2{x^3}{y^4}:D = x{y^2}\\ \Rightarrow D = - 2{x^3}{y^4}:x{y^2} = - 2{x^2}{y^2}\end{array}\)
\(\begin{array}{l}\left( {10{x^5}{y^2} - 6{x^3}{y^4} + 8{x^2}{y^5}} \right):\left( { - 2{x^2}{y^2}} \right)\\ = \left( {10{x^5}{y^2}} \right):\left( { - 2{x^2}{y^2}} \right) - \left( {6{x^3}{y^4}} \right):\left( { - 2{x^2}{y^2}} \right) + \left( {8{x^2}{y^5}} \right):\left( { - 2{x^2}{y^2}} \right)\\ = - 5{x^3} + 3x{y^2} - 4{y^3}\end{array}\)
bạn sai ở hai chỗ: 5x2.y4: 10.x2.y= (1/2)y3
5.52.34.34 :10.52.3= 13,5 và (1/2).33=13,5
Cảm ơn nhiều nha!