Ta có \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Leftrightarrow\left(x-2\sqrt{xy}+y\right)+\left(y-2\sqrt{yz}+z\right)+\left(z-2\sqrt{zx}+x\right)\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\) (luôn đúng)

Vậy \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)

Theo BĐT Cauchy-Schwarz dạng Engel

\(A=\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{1}{2}\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+y}=\dfrac{y}{y+z}=\dfrac{z}{z+x}\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\end{matrix}\right.\)

\(\Leftrightarrow a=b=c=\dfrac{1}{3}\)

`A(x)=7x^3-5x^2-7x+3-7x^3+5x^2+17x+27`

`A(x)=(7x^3-7x^3)-(5x^2-5x^2)+(-7x+17x)+(3+27)`

`A(x)=10x+30`

Cho `A(x)=0`

`=>10x+30=0`

`=>10x=-30`

`=>x=-3`

Vậy nghiệm của đa thức `A(x)` là `x=-3`

Đặt \(A\left(x\right)=0\)

\(\rightarrow7x^3-5x^2-7x+3-7x^3+5x^2+17x+27=0\)

\(\Leftrightarrow10x+30=0\)

\(\Leftrightarrow10x=-30\)

\(\Leftrightarrow x=-3\)

Vậy \(x=-3\) là nghiệm của đa thức \(A\left(x\right)\)

A(x) = 7x³ - 5x² - 7x + 3 - 7x³ + 5x² +17x +27

= 10x + 30

đặt A(x) = 0

<=> 10x + 30 = 0

<=> 10x = -30

<=> x = -3

vậy x = -3 là nghiệm của A(x)

1,\(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{x}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{5}-1\right|+\sqrt{5}+1\right)\)\(=\dfrac{1}{\sqrt{2}}\left|\sqrt{5}-1+\sqrt{5}+1\right|=\dfrac{1}{\sqrt{2}}.2\sqrt{5}\)\(=\sqrt{10}\)

2, \(\sqrt{x-3}-2\sqrt{x^2-3x}=0\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\left(ktm\right)\end{matrix}\right.\)

Vậy pt có nghiệm x=3

3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\left(đk:x>-\dfrac{5}{7}\right)\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow x=6\left(tm\right)\)

4, \(x-5\sqrt{x}+4=0\)(đk: \(x\ge0\))

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)

Vậy...

1) Bạn tự làm

2) ĐK: \(x\ge3\)

PT \(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\2\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)

  Vậy ...

3) ĐK: \(x>-\dfrac{5}{7}\)

PT \(\Rightarrow9x-7=7x+5\) \(\Leftrightarrow x=6\)

  Vậy ...

4) ĐK: \(x\ge0\)

PT \(\Leftrightarrow x-4\sqrt{x}-\sqrt{x}+4=0\)

      \(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)

  Vậy ...

a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)

\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)

\(\Leftrightarrow-10x^2-10x=0\)

\(\Leftrightarrow-10x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c: Ta có: \(x^3+3x^2+3x+28=0\)

\(\Leftrightarrow\left(x+1\right)^3=-27\)

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

hay x=-4

\(\Leftrightarrow\left(x+2\right)^3=343\)

=>x+2=7

hay x=5

\(x^3-3x^2+3x-1=-8\)

\(\Leftrightarrow x-1=-2\)

hay x=-1