a)  Áp dụng AM-GM ta có:

\(2x+\frac{6}{x}\ge2\sqrt{2x.\frac{6}{x}}=2\sqrt{12}=4\sqrt{3}\)

Dấu "=" xảy ra  <=> \(x=\sqrt{3}\)

b)   \(\frac{4x^2-2x+25}{x}\ge18\)

<=>  \(4x^2-2x+25\ge18x\)

<=>  \(4x^2-20x+25\ge0\)

<=>  \(\left(2x-5\right)^2\ge0\)  luôn đúng

Dấu "=" xảy ra  <=>  \(x=2,5\)

a) Vì x > 0

Nên áp dụng BĐT Cô-si ta có: \(2x+\frac{6}{x}\ge2\sqrt{2x.\frac{6}{x}}=2\sqrt{12}=4\sqrt{3}\)

Vậy => ĐPCM

b) Ta có: \(\frac{4x^2-2x+25}{x}=\frac{\left(2x\right)^2-2.2x.\frac{1}{2}+\frac{1}{4}+\frac{99}{4}}{x}=\frac{\left(2x-\frac{1}{2}\right)^2+\frac{99}{4}}{x}\)

P/s: phân tích tới đây thôi, mình chưa nghĩ ra

ai đó giúp mình với !

\(x-y=3\)  =>   \(x=3+y\)

\(P=xy=\left(3+y\right)y=y^2+3y=\left(y+1,5\right)^2-2,25\ge-2,25\)

Dấu "=" xảy ra  <=>  \(y=-1,5\)=>  \(x=1,5\)

Vậy MIN  \(P=-2,25\)khi   \(x=1,5;\)\(y=-1,5\)

ĐK:  \(x\ge0\)

\(x-6\sqrt{x}+9\)

\(=\left(\sqrt{x}\right)^2-2.\sqrt{x}.3+3^2\)

\(=\left(\sqrt{x}-3\right)^2\)

p/s: do không rõ đề với lại bạn có nhắc tới HĐT nên mk nghĩ đề là như z, sai thì bỏ qua nhé

Kb nha

rảnh vl

\(x^3+27y^3=1-9xy\left(x+3y\right)\)

<=>  \(x^3+27y^3+9xy\left(x+3y\right)=1\)

<=>  \(\left(x+3y\right)^3=1\)

<=>  \(x+3y=1\)

Vậy  \(M=1\)

\(x^3+27x^3=1-9xy\left(x+3y\right)\))

\(=\left(x+3y\right)\left(x^2-3xy+9y^2\right)=1-9xy\left(x+3y\right)\)

=\(\left(x+3y\right)\left(x^2-3xy+9y^2\right)-1+9xy\left(x+3y\right)=0\)

=\(\left(x+3y\right)\left(x^2-3xy+9y^2+9xy\right)-1=0\)

=\(\left(x+3y\right)\left(x^2+6xy+9y^2\right)-1=0\)

=\(\left(x+3y\right)\left(x+3y\right)^2-1=0\)

=\(\left(x+3y\right)\left(x+3y\right)^2=1\)

\(\Rightarrow x+3y=\left(x+3y\right)^2=1\)

\(\Rightarrow x+3y=1\)

a) \(x^2-3xy+x-3y=x\left(x-3y\right)+\left(x-3y\right)=\left(x-3y\right)\left(x+1\right)\)

b) \(x^2-6x-y^2+9=x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

c) \(7x^3y-14x^2y+7xy=7xy\left(x^2-2x+1\right)=7xy\left(x-1\right)^2\)

\(x^2-3xy+x-3y=\left(x^2+x\right)-\left(3xy+3y\right)=x\left(x+1\right)-3y\left(x+1\right)=\left(x+1\right)\left(x-3y\right)\)

\(x^2-6x-y^2+9=\left(x^2-2.x.3+3^2\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

\(7x^3y-14x^2y+7xy=\left(7x^3y-7x^2y\right)-\left(7x^2y-7xy\right)=7x^2y.\left(x-1\right)-7xy.\left(x-1\right)\)

\(=\left(x-1\right).\left(7x^2y-7xy\right)=7xy.\left(x-1\right).\left(x-1\right)=7xy.\left(x-1\right)^2\)

x³-3x²-x+2

=x²(x+3)-(x+3)

=(x+3)(x²-1)

=(x+3)(x-1)(x+1)

học tốt nha !!!

\(x^3-3x^2-x+2=x^2\left(x-3\right)+2\)\(2\)\(=\left(x^2+2\right)\left(x-3\right)\)

Mik k chắc cho lắm . chúc bạn học tốt:D

<=>(x²+y²+z²+2xy+2yz+2xz)+(x²+2x+1)+(y²+4y+4)=0

<=>(x+y+z)²+(x+1)²+(y+2)²=0

Mà \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\Rightarrow\left(x+y+z\right)^2+\left(x+1\right)^2+\left(y+2\right)^2\ge0}\)

=>\(\hept{\begin{cases}x+y+z=0\\x+1=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}z=3\\x=-1\\y=-2\end{cases}}}\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow a.d=b.c\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{5a}{5c}=\frac{5b}{5d}=\frac{5a+5b}{5c+5d}=\frac{5a-5b}{5c-5d}\)

\(\Rightarrow\frac{5a+5b}{5c+5d}=\frac{5a-5b}{5c-5d}\)