Chứng minh rằng: nếu pt \(x^2+px+q=0\) có một nghiệm gấp \(k\) lần một nghiệm của pt \(x^2+mx+n=0\) thì các hệ số \(m,n,p,q\) thỏa mãn hệ thức sau:

\(\left(q-k^2n\right)^2+k\left(p-mk\right)\left(knp-qm\right)=0\)

\(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(P=\left(\dfrac{2\left(2\sqrt{x}-3\right)-\left(\sqrt{x}-1\right)}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(P=\left(\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right).\left(\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}\right)\)

\(P=\dfrac{\left(3\sqrt{x}-5\right)\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(2\sqrt{x}+1\right)}\)

\(P=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)

=))

\(P=\left(2+\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(P=\left(\dfrac{2\left(2\sqrt{x}-3\right)+\left(\sqrt{x}-1\right)}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(P=\left(\dfrac{4\sqrt{x}-6+\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right):\left(\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right):\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\dfrac{5\sqrt{x}-7}{2\sqrt{x}-3}\right).\left(\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}\right)\)

\(P=\dfrac{\left(5\sqrt{x}-7\right)\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(2\sqrt{x}+1\right)}\)

\(P=\dfrac{5\sqrt{x}-7}{2\sqrt{x}+1}\)

\(T=\dfrac{2\left(x-1\right)}{\sqrt{x}+1}+\dfrac{x-4}{\sqrt{x}-2}\)

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

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

\(T=2\sqrt{x}+2+\sqrt{x}+2\)

\(T=3\sqrt{x}+4\)

\(x=4\)

\(\Rightarrow T=3\sqrt{4}+4=3.2+4=10\)

\(2x+3\sqrt{x}+1=0\)

Đặt \(\sqrt{x}=a;a\ge0\)

\(\Leftrightarrow2a^2+3a+1=0\)

\(\Leftrightarrow2a^2+2a+a+1=0\)

\(\Leftrightarrow2a\left(a+1\right)+\left(a+1\right)=0\)

\(\Leftrightarrow\left(a+1\right)\left(2a+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-1\\a=-\dfrac{1}{2}\end{matrix}\right.\) ( ktm )

Vậy pt vô nghiệm

Dùng BĐT làm gì, dùng HĐT (anh học luôn mở rộng)

--> e dùng Schwars thử xem:v