Phương trình hoành độ giao điểm:

\(-x^2=2mx-3m+4\Leftrightarrow x^2+2mx-3m+4=0\) (1)

(P) tiếp xúc (d) khi và chỉ khi (1) có nghiệm kép

\(\Leftrightarrow\Delta'=m^2-\left(-3m+4\right)=0\Leftrightarrow m^2+3m-4=0\)

\(\Rightarrow\left[{}\begin{matrix}m=1\\m=-4\end{matrix}\right.\)

Với \(m=1\Rightarrow x_{1,2}=-\dfrac{2m}{2}=-m=-1\)

Với \(m=-4\Rightarrow x_{1,2}=-m=4\)

Xem lại đề đi bạn

bài giả nè các anh chị

\(P=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}=\dfrac{x\sqrt{2}}{\sqrt{2x}\left(\sqrt{2}+\sqrt{x}\right)}+\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)}\)

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

\(\sqrt{4x^2+xy+4y^2}=\sqrt{\dfrac{7}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x^2+2xy+y^2\right)}\ge\sqrt{\dfrac{7}{4}\left(x+y\right)^2+\dfrac{1}{2}\left(x+y\right)^2}=\dfrac{3}{2}\left(x+y\right)\)

Tương tự: 

\(\sqrt{4y^2+yz+4z^2}\ge\dfrac{3}{2}\left(y+z\right)\); \(\sqrt{4z^2+zx+4x^2}\ge\dfrac{3}{2}\left(z+x\right)\)

Cộng vế:

\(B\ge\dfrac{3}{2}\left(x+y+y+z+z+x\right)=3\left(x+y+z\right)\)

\(B\ge3.\dfrac{1}{3}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=1\)

\(B_{min}=1\) khi \(x=y=z=\dfrac{1}{9}\)

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

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

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

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

\(a,P=\frac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-2\)

\(P=\frac{3x+3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1-2x-2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

\(b,P=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)

\(\sqrt{x}-1\ge-1\)

\(\Leftrightarrow\frac{2}{\sqrt{x}-1}\le-2\)

\(\Leftrightarrow P\le1-2=-1\)

\(c,P=4\)\(\Leftrightarrow\sqrt{x}+1=4\sqrt{x}-4\Leftrightarrow3\sqrt{x}=5\Leftrightarrow x=\frac{25}{9}\)

d, Do câu b, \(P\le-1\)\(\Rightarrow P< -3\)

Tối qua bạn có hỏi câu a và mk đã giải đc là:\(P=\frac{3}{\sqrt{x}-1}\)

b)\(P=1\Rightarrow\frac{3}{\sqrt{x}-1}=1\)

\(\Leftrightarrow\sqrt{x}-1=3\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=4^2\)

\(\Leftrightarrow x=16\)(TM)

Vậy khi x=16 thì P=1