\(\sqrt{3x^2-17x+4}=3x-2\)

\(\Leftrightarrow\hept{\begin{cases}3x-2\ge0\\3x^2-17x+4=9x^2-12x+4\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{2}{3}\\6x^2+5x=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge\frac{2}{3}\\x\left(6x+5\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{2}{3}\\x=0\left(tm\right);x=-\frac{5}{6}\left(ktm\right)\end{cases}}\)

Vậy pt có nghiệm x = 0

à xin lỗi dạo này lú :( x = 0 ktm nhé ._. 

Vậy pt vô nghiệm 

\(\sqrt{x}=2\)   

\(\hept{\begin{cases}2\ge0\left(llđ\right)\\x=2^2\end{cases}}\)   luôn luôn đúng 

x = 4 

\(P^2=\left(\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\right)^2\)

\(\le\left(1^2+1^2+1^2\right)\left(4a+3+4b+3+3c+3\right)\)

\(=63\)

\(\Rightarrow P\le\sqrt{63}=3\sqrt{7}\).

Dấu \(=\)khi \(\hept{\begin{cases}4a+3=4b+3=4c+3\\a+b+c=3\end{cases}}\Leftrightarrow a=b=c=1\).

vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\)\(\frac{2}{\sqrt{x}+3}\le\frac{2}{3}\)

vậy A đạt GTLN ki và chỉ khi \(\sqrt{x}=0\)suy ra x=0

a) Có \(BC^2=15^2=225\)

\(AB^2+AC^2=9^2+12^2=81+144=225\)

do đó \(BC^2=AB^2+AC^2\)

Theo định lí Pythaogre đảo suy ra tam giác \(ABC\)vuông tại \(A\).

b) \(AH=\frac{AB.AC}{BC}=\frac{9.12}{15}=7,2\left(cm\right)\)

\(HB=\frac{AB^2}{BC}=\frac{9^2}{15}=5,4\left(cm\right)\)

\(HC=BC-HB=15-5,4=9,6\left(cm\right)\)

Ta có: E = \(\frac{x}{\sqrt{x}-1}=\frac{x-1+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}-1}\)

E = \(\sqrt{x}-1+\frac{1}{\sqrt{x}-1}+2\ge2\sqrt{\left(\sqrt{x}-1\right)\cdot\frac{1}{\sqrt{x}-1}}+2=2+2=4\)(bđt cosi)

Dấu "=" xảy ra <=> \(\sqrt{x}-1=\frac{1}{\sqrt{x}-1}\) <=> \(\left(\sqrt{x}-1\right)^2=1\)

<=> \(\orbr{\begin{cases}\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{cases}}\) <=> \(\orbr{\begin{cases}x=4\left(tm\right)\\x=0\left(ktm\right)\end{cases}}\)

Vậy MinE = 4 <=>. x = 4