Áp dụng tính chất dãy tỉ số bằng nhau ta có: \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}=k\)

\(\Rightarrow\hept{\begin{cases}a=kx;b=ky;c=kz\Rightarrow a^2=k^2x^2;b^2=k^2y^2;c^2=k^2z^2\\a+b+c=k\left(x+y+z\right)\end{cases}}\)

Có: \(\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{x^2+y^2+z^2}{\left(kx^2+ky^2+kz^2\right)^2}=\frac{x^2+y^2+z^2}{k^2\left(x^2+y^2+z^2\right)^2}=\frac{1}{k^2\left(x^2+y^2+z^2\right)}\)

\(=\frac{1}{k^2x^2+k^2y^2+k^2z^2}=\frac{1}{a^2+b^2+c^2}\)(đpcm)

3x=2y=0

=>x=y=0

=>9x^2-4y^2=0-0=0

GTNN:=5

\(x^2+5x+7=x^2+5x+\frac{25}{4}-\frac{5}{4}=\left(x+\frac{5}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)-5/4. GTNN =-5/4 .Hình như thế

bn cho thêm đk đi ko đủ nha

= 16x³ -16x² + 4x² - 4x + 7x - 7

= 16x²(x-1)+4x(x-1)+7(x-1)

=(x-1)(16x²+4x+7)

a) 9x² - 6x + 2 = (3x)² - 2.3x.1 + 1² + 1 = (3x - 1)² + 1 mà\(\left(3x+1\right)^2\ge0\Rightarrow\left(3x+1\right)^2+1\ge1>0\)

b) x² + x + 1 = x² + 2.x.\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)mà\(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

c) 2x² + 2x + 1 =\(\left(\sqrt{2}x\right)^2+2\sqrt{2}x.\frac{1}{\sqrt{2}}+\left(\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}\ge\frac{1}{2}>0\)

a)    \(9x^2-6x+2=\left(\left(3x\right)^2-2.3x.1+1\right)+1=\left(3x-1\right)^2+1>0\)

b)   .\(x^2+x+1=\left(\left(x^2\right)+2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

c)    \(2x^2+2x+1=x^2+\left(x^2+2x+1\right)=x^2+\left(x+1\right)^2>0\)