Toán lp 9 

Thì mk khuyên bn sang h.vn để đc giải đáp tốt hơn

Nhìn thấy thì tiếc gì nhỉ =)

h.vn là gì bạn

\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\left(đkxđ\Leftrightarrow x\ge0\right).\)

\(=\frac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)\)

\(=x+\sqrt{x}-2\sqrt{x}-1=x-\sqrt{x}-1\)

\(P=x-\sqrt{x}-1=\sqrt{x}^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-1\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{5}{4}\)

\(\Rightarrow P_{min}=-\frac{5}{4}\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)

\(M=\frac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}}\)

a.Ta co:\(x^2-x=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=1\left(n\right)\end{cases}}\)

\(\Rightarrow M=\frac{1-2}{1}=-1\)

b.De \(M\in Z\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}}\in Z\Rightarrow\sqrt{x}-2⋮\sqrt{x}\Rightarrow x=4\)

Mình cảm ơn bạn nhiều nha ^^

\(hpt\Leftrightarrow\hept{\begin{cases}x+y+2\sqrt{xy}=16\\x+y+\sqrt{xy}=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{xy}=12\\x+y=-8\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}xy=144\\x+y=-8\end{cases}\Rightarrow}x+\frac{144}{x}=-8\)

pt vô nghiệm nên hệ pt vô nghiệm

\(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^2-2x+1\right)+25}\ge\sqrt{9}+\sqrt{25}=8\)

Do dấu "=" ko đồng thời xảy ra ở hai bđt nên pt vô nghiệm 

\(\sqrt{3\left(x+1\right)^2+9}-3+\sqrt{5\left(x^2-1\right)^2+25}-5=0\)

\(\Leftrightarrow\frac{3\left(x+1\right)^2}{\sqrt{3\left(x+2\right)^2+9}+3}+\frac{5\left(x+1\right)^2\left(x-1\right)^2}{\sqrt{5\left(x^2-1\right)^2+25}+5}=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(\frac{3}{\sqrt{3\left(x+2\right)^2+9}+3}+\frac{5\left(x-1\right)^2}{\sqrt{5\left(x^2-1\right)^2+25}+5}\right)=0\)

\(\left(\frac{3}{\sqrt{3\left(x+2\right)^2+9}+3}+\frac{5\left(x-1\right)^2}{\sqrt{5\left(x^2-1\right)^2+25}+5}\right)>0\left(\forall x\right)\)

\(\Rightarrow x=-1\)

Bạn kia làm sai rùi ạ chắc nhìn nhầm đề