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\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
a,(x-y)^2-2(x+y)+1 b, x^2-y^2+4x+4 c, 4x^2-y^2+8(y-2)
=(x-y-1)^2 =(x^2+4x+4)-y^2 =4x^2-y^2+8y-16
=(x+2)^2-y^2 =4x^2-(y^2-8y+16)
=(x+2-y)(x+2+y) =4x^2-(y-4)^2
a) (x+y)2-2(x+y)+1=(x+y-1)2
b) x2-y2+4x+4 = (x2+4x+4)-y2=(x+2)2-y2=(x+y+2)(x-y+2)
c)4x2-y2+8(y-2) = 4x2-(y2-8y+16) = (2x)2-(y-4)2=(2x+y-4)(2x-y+4)
d)x3-2x2+2x-4 = x2(x-2)+2(x-2) = (x-2)(x2+2)
e)xy-4+2x-2y=x(y+2) - 2(y+2) = (x-2)(y+2)
\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)
\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)
\(\Leftrightarrow4x+4x>-1\)
\(\Leftrightarrow8x>-1\)
\(\Leftrightarrow x>-\frac{1}{8}\)
\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)
\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-6x^2< 1+3\)
\(\Leftrightarrow-2x^2< 4\)
\(\Leftrightarrow x^2>2\)
\(\Leftrightarrow x>\pm\sqrt{2}\)
a) \(4x^2-12x=-9\)
\(\Leftrightarrow4x^2-12x+9=0\)
\(\Leftrightarrow\left(2x-3\right)^2=0\)
\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)
b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(25-4x^2\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7+5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)
c)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=0\\x=2\end{array}\right.\)
d) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-\frac{23}{17}\end{array}\right.\)
a) \(\left(3-2x\right)\left(x+1\right)+x\left(2x-1\right)=3x+3-2x^2-2x+2x^2-x=3\)
b) \(\frac{x^2+9}{x^2+3x}+\frac{6}{x+3}=\frac{x^2+9}{x\left(x+3\right)}+\frac{6x}{x\left(x+3\right)}=\frac{x^2+6x+9}{x\left(x+3\right)}=\frac{\left(x+3\right)^2}{x\left(x+3\right)}=\frac{x+3}{x}\)
c)\(\frac{2+x}{2-x}+\frac{4x^2}{4-x^2}+\frac{x-2}{2+x}=\frac{\left(x+2\right)^2}{\left(2-x\right)\left(2+x\right)}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}+\frac{-\left(x-2\right)^2}{\left(2+x\right)\left(2-x\right)}\)
\(=\frac{x^2+4x+4+4x^2-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}=\frac{4x^2+8x}{\left(x+2\right)\left(2-x\right)}=\frac{4x\left(x+2\right)}{\left(x+2\right)\left(2-x\right)}=\frac{4x}{2-x}\)
d) \(\left(x^3+4x^2+6x+4\right):\left(x+2\right)\)
\(=\left(x^3+2x^2+2x^2+4x+2x+4\right):\left(x+2\right)\)
\(=\left[x^2\left(x+2\right)+2x\left(x+2\right)+2\left(x+2\right)\right]:\left(x+2\right)\)
\(=\left(x^2+2x+2\right)\left(x+2\right):\left(x+2\right)=x^2+2x+2\)
a) \(A=\dfrac{\left(-2\right)^5}{\left(-2\right)^3}=\left(-2\right)^{5-3}=\left(-2\right)^2=4\)
b) \(y\ne0:B=\dfrac{\left(-y\right)^7}{\left(-y\right)^3}=\left(-y\right)^{7-3}=\left(-y\right)^4=y^4\)
c) \(x\ne0:C=\dfrac{\left(x\right)^{12}}{\left(-x\right)^{10}}=\left(x\right)^{12-10}=\left(x\right)^2=x^4\)
d) \(x\ne0:D=\dfrac{2x^6}{\left(2x\right)^3}=\dfrac{2x^6}{8x^3}=\dfrac{1}{4}\left(x\right)^{6-3}=\dfrac{1}{4}\left(x\right)^3\)
e) \(x\ne0:E=\dfrac{\left(-3x\right)^5}{\left(-3x\right)^2}=\left(-3x\right)^{5-2}=\left(-3x\right)^3=-27x^3\)
f) \(x,y\ne0:F=\dfrac{\left(xy^2\right)^4}{\left(xy^2\right)^2}=\left(xy^2\right)^{4-2}=\left(xy^2\right)^2=x^2y^4\)
i) \(x\ne-2:I=\dfrac{\left(x+2\right)^9}{\left(x+2\right)^6}=\left(x+2\right)^{9-6}=\left(x+2\right)^3\)
1/ (x2 - 2)(x2 + 2x + 2)
2/ x2 - (x2 + 2)2 = (x - x2 - 2)(x + x2 + 2)