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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)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
a)\(=\left(x^2-7x-9x+63\right)+1\)
\(=x^2-7x-9x+63+1\)
=\(x^2-16x+64\)
\(=\left(x-8\right)^2\)
a: \(=x^2-16x+63+1\)
\(=x^2-16x+64=\left(x-8\right)^2\)
b: \(=\left(x^2+x\right)^2-2\left(x^2+x\right)+1+4\left(x^2+x\right)\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)
\(=\left(x^2+x+1\right)^2\)
c: \(=\left(x+2y-3-2\right)^2\)
\(=\left(x+2y-5\right)^2\)
d: \(=\left(x-y\right)^3-1-3\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y-1\right)^3\)
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: \(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
\(x^2+y^2=0\)
Mà \(x^2\ge0;y^2\ge0\)nên \(x^2+y^2\ge0\)
(Dấu "="\(\Leftrightarrow x=y=0\))
a) \(-2x\left(10x-3\right)+5x\left(4x+1\right)=25\)
\(-20x^2+6x+20x^2+5x=25\)
\(\Rightarrow6x+5x=25\)
\(\Rightarrow11x=25\)
\(\Rightarrow x=\dfrac{25}{11}\)
b) \(y\left(5-2y\right)+2y\left(y-1\right)=15\)
\(5y-2y^2+2y^2-2y=15\)
\(\Rightarrow5y-2y=15\)
\(\Rightarrow3y=15\)
\(\Rightarrow y=5\)
c)\(x\left(x+1\right)-\left(x+1\right)=35\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)=35\)
\(\Rightarrow x^2-1=35\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6;x=-6\)
d)\(x\left(x^2+x+1\right)-x^2\left(x+1\right)=0\)
\(x^3+x^2+x-x^3+x=0\)
\(\Rightarrow x^2+2x=0\)
\(\Rightarrow x\left(x+2\right)=0\)
\(\Rightarrow x=0;x=0-2=-2\)
Vậy \(x=0;x=-2\)