a) thay m=-1 vào x²(2m-1)x-m=0 ta có:

x²+(-3)x+1=0\(\Delta\)=5

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{5}}{2}\\x=\frac{3-\sqrt{5}}{2}\end{cases}}\)

b) A=\(x_1^2+x_2^2-x_1x_2=\left(x_1+x_2\right)^2-2x_1x_2-x_1x_2=\left(x_1+x_2\right)^2-3x_1x_2\)

Vi-et \(\hept{\begin{cases}x_1+x_2=1-2m\\x_1x_2=-m\end{cases}}\)

=> \(A=\left(1-2m\right)^2-3\left(-m\right)=4m^2-4m+1+3m=4m^2-m+1\)

\(\sqrt[3]{a^2+\frac{1}{a^2}}+\sqrt[3]{b^2+\frac{1}{b^2}}\ge2\sqrt[6]{\frac{a^2}{b^2}+\frac{b^2}{a^2}+a^2b^2+\frac{1}{a^2b^2}}\ge2\sqrt[6]{2+a^2b^2+\frac{1}{a^2b^2}}\)

đến đây thì ta thấy từ giả thuyết có \(a+b=\frac{2}{3}\Rightarrow a^2b^2\le\frac{1}{81}\)

Xét:

\(a^2b^2+\frac{1}{a^2b^2}=\left(a^2b^2+\frac{1}{6561a^2b^2}\right)+\frac{6560}{6561a^2b^2}\ge\frac{6562}{81}\)

\(\Rightarrow\sqrt[3]{a^2+\frac{1}{a^2}}+\sqrt[3]{b^2+\frac{1}{b^2}}\ge2\sqrt[3]{\frac{82}{9}}\)

Đẳng thức xảy ra khi \(a=b=\frac{1}{3}\)

em nghĩ bài này tìm giá trị lớn nhất ạ

\(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2=\left(1\cdot\sqrt{a+b}+1\cdot\sqrt{b+c}+1\cdot\sqrt{c+a}\right)^2\)

áp dụng bđt Cauchy-Schwartz, ta có:

\(P^2\le\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left[1^2+1^2+1^2\right]\)

\(P^2\le2\cdot3=6\)

Vậy \(P\le\sqrt{6}\)

dấu "="xảy ra <=> \(a=b=c=\frac{1}{3}\)

Ta có : 

\(P=x-4\sqrt{x}+\frac{x+16}{\sqrt{x}+3}+10\)

\(\Rightarrow P-4=\left(x-4\sqrt{x}+4\right)+\frac{x+16}{\sqrt{x}+3}-4+10\)

\(\Rightarrow P-4=\left(\sqrt{x}-2\right)^2+\frac{x+16-4\left(\sqrt{x}+3\right)}{\sqrt{x}+3}+10\)

\(\Rightarrow P-4=\left(\sqrt{x}-2\right)^2+\frac{x-4\sqrt{x}+4}{\sqrt{x}+3}+10\)

\(\Rightarrow P-4=\left(\sqrt{x}-2\right)^2+\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+3}+10\)

\(\Rightarrow P-4\ge10\)

\(\Rightarrow P\ge4\)

\(\Rightarrow GTNN\left(P\right)=4\Rightarrow\left(\sqrt{x}-2\right)^2=10\Rightarrow x=4\)