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a: \(=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)
b: \(=\dfrac{\left(5x+1\right)\left(25x^2-5x+1\right)}{5x+1}=25x^2-5x+1\)
c: \(=\dfrac{2x^3-x^2+x+6x^2-3x+3}{2x^2-x+1}=x+3\)
a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
Bài giải:
a) (x2 + 2xy + y2) : (x + y) = (x + y)2 : (x + y) = x + y.
b) (125x3 + 1) : (5x + 1) = [(5x)3 + 1] : (5x + 1)
= (5x)2 – 5x + 1 = 25x2 – 5x + 1.
c) (x2 – 2xy + y2) : (y – x) = (x – y)2 : [-(x – y)] = - (x – y) = y – x
Hoặc (x2 – 2xy + y2) : (y – x) = (y2 – 2xy + x2) : (y – x)
= (y – x)2 : (y – x) = y - x.
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)
\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)
=1/5-1=-4/5
c \(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)
d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)
\(=20x^3-30x^2+15x+4\)
\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
\(\left(a-b\right)^2-\left(b-a\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+1\right)\)
\(5\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)\)
\(=\left(a+b\right)\left[5\left(a+b\right)-\left(a-b\right)\right]\)
\(=\left(a+b\right)\left[5a+5b-a+b\right]\)
\(=\left(a+b\right)\left[4a+6b\right]\)
\(\frac{x^2+2xy+y^2}{x+y}=\frac{\left(x+y\right)^2}{x+y}=x+y\)
\(\frac{125x^3+1}{5x+1}=\frac{\left(5x\right)^3+1}{5x+1}=\frac{\left(5x+1\right)\left(25x^2-5x+1\right)}{5x+1}=25x^2-5x+1\)
\(\frac{2x^3+5x^2-2x+3}{2x^2-x+1}=\frac{\left(2x^3-x^2+x\right)+\left(6x^2-3x+3\right)}{2x^2-x+1}\)
\(=\frac{x\left(2x^2-x+1\right)+3.\left(2x^2-x+1\right)}{2x^2-x+1}=\frac{\left(2x^2-x+1\right)\left(x+3\right)}{2x^2-x+1}=x+3\)
Tham khảo nhé~