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1) \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
2) \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)
3) \(\left(x+\frac{1}{3}\right)^4=\left[\left(x+\frac{1}{3}\right)^2\right]^2=\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)^2=x^4+\frac{4}{9}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x+\frac{2}{9}x^2=x^4+\frac{2}{3}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x\)
4) \(\left(2x+y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
5) Sửa đề: \(\left(\frac{x}{2}-2y\right)^3=\frac{x^3}{8}-\frac{3x^2}{2}+6xy^2-8y^3\)
6) \(\left(\sqrt{2x-y}\right)^4=\left(2x-y\right)^2=4x^2-4xy+y^2\)
7) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
8) \(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)
\(C=\left(x^3+y^3\right)+3xy\left(x^2+y^2+2xy\left(x+y\right)\right)\)
\(C=\left(x^3+y^3+3x^2y+3xy^2-3x^2y-3xy^2\right)+3xy\left(x^2+y^2+2xy\right)\) (vì x+y=1)
\(C=\left(x+y\right)^3-3x^2y-3xy^2+3xy\left(x+y\right)^2\)
\(C=1^3-3xy\left(x+y\right)+3xy.1^2\) (vì x+y=1)
\(C=1-3xy+3xy\)(vì x+y=1)
\(C=1\)
\(D=2\left(\left(x+y\right)^3-3xy\left(x+y\right)\right)-3\left(\left(x+y\right)^2-2xy\right)\)
\(D=2\left(1^3-3xy\right)-3\left(1^2-2xy\right)\)(vì x+y=1)
\(D=2-6xy-3+6xy\)
\(D=-1\)
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)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
1, \(\left(x+2y\right)=x^2+4xy+4y^2\)
2, \(\left(4x-5y\right)^2=16x^2-40xy+25y^2\)
3, \(\left(2x-\frac{1}{2}\right)^2=4x^2-2x+\frac{1}{4}\)
4, \(\left(\frac{x}{2}-y\right)\left(\frac{x}{2}+y\right)=\frac{x^2}{4}+\frac{xy}{2}-\frac{xy}{2}-y^2=\frac{x^2}{4}-y^2\)
5, \(\left(x+\frac{1}{3}\right)^3=x^3+x^2+\frac{x}{3}+\frac{1}{27}\)
6, \(\left(x-2\right)\left(x^2+2x+4\right)=\left(x-2\right)\left(x+2\right)=x^2-4\)
a)
(x-2y)2 >= 0
Vx,y(y-2018)>=0
Vy=> P=(ghi lại đề) >= 0
vậy GTNN của p bằng 0
dấu "=" xảy ra (=) \(\hept{\begin{cases}x-2y=0\\y-2018=0\end{cases}}\left(=\right)\hept{\begin{cases}x=2y\\y=2018\end{cases}}\left(=\right)\hept{\begin{cases}y=2018\\x=4036\end{cases}}\)
b) (x+y-3)4 >= 0
Vx,y(x-2y)2 >=
Vx,y=> Q=(ghi lại đề) >= 2018
vậy GTNN của Q bằng 2018
dấu "=" xảy ra (=) \(\hept{\begin{cases}x+y-3=0\\x-2y=0\end{cases}}\left(=\right)\hept{\begin{cases}x=2y\\3y=3\end{cases}}\left(=\right)\hept{\begin{cases}y=1\\x=2\end{cases}}\)
c)
(2x + 1/6)4>= 0
Vx=> N=(ghi lại đề) >= -2
vậy GTNN của N bằng -2
dấu "=" xảy ra (=) 2x+1/6=0
(=) 2x=-16
(=) x=-1/12
#Học-tốt
Menhera-Kun chữ V rồi gạch đó là j