\(\hept{\begin{cases}x-my=m^2+1\\mx+y=m^2+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-my-1=m^2\\mx+y-1=m^2\end{cases}}\)

\(\Rightarrow x-my-1=mx+y-1\)

\(\Leftrightarrow x-my=mx+y\)

\(\Leftrightarrow m\left(x+y\right)=x-y\)

\(\Leftrightarrow m=\frac{x-y}{x+y}\)

Ps Tham khảo nha

\(\sqrt{x^4}=10\left(\sqrt{50}:\sqrt{2}\right)\)

\(\Leftrightarrow\sqrt{x^4}=10\sqrt{25}\)

\(\Leftrightarrow\sqrt{x^4}=10.5\)

\(\Leftrightarrow\left|x^2\right|=50\)

\(\Leftrightarrow x^2=50\)

\(\Leftrightarrow x=\pm\sqrt{50}\)

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

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

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

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

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

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

b)Khi \(x=\frac{9}{4}\)

\(\Rightarrow\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=3\)

c)\(A=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)}< 1\)

\(\Leftrightarrow\sqrt{x}< \sqrt{x}-1\)(Voly)

=>ko có giá trị nào

\(\frac{a}{b^2+bc+c^2}+\frac{b}{c^2+ca+a^2}+\frac{c}{a^2+ab+b^2}=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{bc^2+abc+ba^2}+\frac{c^2}{ca^2+abc+cb^2}\)     (1)

Áp dụng BDT Cauchy-Schwarz: \(\left(1\right)\ge\frac{\left(a+b+c\right)^2}{ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+3abc}\)

Lại có: \(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+3abc=\left(ab+bc+ac\right)\left(a+b+c\right)\)

Thay vào -> dpcm