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a) câu này dài quá à, mình ngại làm lắm
Áp dụng bđt này: \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
b)\(\left(1+x+x^2\right)\left(1-x\right)\left(1+x\right)\left(1-x+x^2\right)\)
\(=\left[\left(1+x^2\right)+x\right]\left(1-x^2\right)\left[\left(x^2+1\right)-x\right]\)
\(=\left[\left(1+x^2\right)^2-x^2\right]\left(1-x^2\right)\)
\(=\left(1+2x^2+x^4-x^2\right)\left(1-x^2\right)\)
\(=\left(x^4+x^2+1\right)\left(1-x^2\right)\)
a) =(a-b-c +a-b+c)( a-b-c -a+b-c)
= 2(a-b)(-2c)= -4c(a-b)
làm tặng câu a) thui
\(\left(a-b-c\right)^2-\left(a-b+c\right)^2\)
\(=\left(a-b-c-a+b-c\right)\left(a-b-c+a-b+c\right)\)
\(=\left(-2c\right)\left(-2b+2a\right)\)
\(=2\left(a-b\right)\left(-2c\right)\)
\(=-4c\left(a-b\right)\)
a) (x2 + 2xy + y2) : (x + y);
=(x+y)2:(x+y)
=x+y
b) (125x3 + 1) : (5x + 1);
=(5x+1)(25x2-5x+1):(5x+1)
=25x2-5x+1
c) (x2 – 2xy + y2) : (y – x).
=(x-y)2:(y-x)
=(y-x)2:(y-x)
=y-x
\(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=x^3-4x^2+16x+4x^2-16x+64\)
\(=x^3+64\)
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^2+3x^2y+9xy^2-3x^2y-9xy^2-27y^3\)
\(=\)\(x^2-27y^3\)
\(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3xy}+4y^2\right)\)
\(=\)\(\frac{x^3}{27}-\frac{2}{9xy}+\frac{4xy^2}{3}+\frac{2x^2y}{9}-\frac{4y}{3xy}+8y^3\)
làm nốt nha
a. A= x2-7x+20 = x2-2*\(\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{31}{4}\)=(x-\(\dfrac{7}{2}\))2+\(\dfrac{31}{4}\)>0 \(\forall x\)(đpcm)
b. B= 2x2+5x+14=2(x2+2*\(\dfrac{5}{4}x+\dfrac{25}{16}+\dfrac{87}{16}\))=2(x+\(\dfrac{5}{4}\))2+\(\dfrac{87}{8}\)>0(đpcm)
a) \(x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\)
\(\left(x^2+4x+4\right)\div\left(x+2\right)=x+2\)
b) \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(\left(x^3-1\right)\div\left(x-1\right)=x^2+x+1\)
c) \(x^3+6x^2+12x+8=x^3+3.x^2.2+3.x.2^2+2^3=\left(x+2\right)^3\)
\(\left(x^3+6x^2+12x+8\right)\div\left(x+2\right)=\left(x+2\right)^2\)
a) \(\left(x^2+2xy+y^2\right):\left(x+y\right)\)
\(=\left(x+y\right)^2:\left(x+y\right)\)
\(=x+y\)
b) \(\left(125x^3+1\right):\left(5x+1\right)\)
\(=\left(5x+1\right)\left(25x^2-5x+1\right):\left(5x+1\right)\)
\(=25x^2-5x+1\)
c) \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left(y-x\right)\)
\(=\left(y-x\right)^2:\left(y-x\right)\)
\(=y-x\)
a ) \(\left(x^2+2xy+y^2\right):\left(x+y\right)\)
\(=\left(x+y\right)^2:\left(x+y\right)\)
\(=\left(x+y\right)\)
b ) \(\left(125x^3+1\right)\left(5x+1\right)\)
\(=\left[\left(5x\right)^3+1\right]:\left(5x+1\right)\)
\(=\left(5x\right)^2-5x+1\)
\(=25x^2-5x+1\)
c ) \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left[-\left(x-y\right)\right]\)
\(=-\left(x-y\right)\)
\(=y-x\)
a) \(\left(x^2+2xy+y^2\right):\left(x+y\right)\\ =\left(x+y\right)^2:\left(x+y\right)\\ =\left(x+y\right)\)
b) \(\left(125x^3+1\right)\left(5x+1\right)\\=\left[\left(5x\right)^3+1\right]:\left(5x+1\right)\\ =\left(5x\right)^2-5x+1 \\ =25x^2-5x+1\)
c) \(\left(x^2-2xy+y^2\right):\left(y-x\right)\\ =\left(x-y\right)^2:\left[-\left(x-y\right)\right]\\ =-\left(x-y\right)\\ =y-x\)
\(\left(a^2-1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)=\left(a-1\right)\left(a^2+a+1\right)\left(a+1\right)\left(a^2-a+1\right)\)
\(=\left(a^3-1\right)\left(a^3+1\right)=a^6-1\)