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\(A=0.5\cdot4\sqrt{3-x}-\sqrt{3-x}-2\sqrt{3}+1=\sqrt{3-x}-2\sqrt{3}+1\) (xác định khi x=<3)
a)thay \(x=2\sqrt{2}\)vào a ra có
\(\sqrt{3-2\sqrt{2}}-2\sqrt{3}+1=\sqrt{\left(\sqrt{2}-1\right)^2}-2\sqrt{3}+1\)
\(=\sqrt{2}-1+2\sqrt{3}+1=\sqrt{2}+2\sqrt{3}\)
Để A=1<=> \(\sqrt{3-x}-2\sqrt{3}+1=1\\ \Leftrightarrow\sqrt{3-x}-2\sqrt{3}+1-1=0\\ \Leftrightarrow\sqrt{3-x}-2\sqrt{3}=0\\ \Leftrightarrow3-x=12\Leftrightarrow x=-9\)
Lời giải:
Đặt \(\frac{1}{x-1}=a; \frac{1}{y-1}=b\) thì HPT trở thành:
\(\left\{\begin{matrix} a-3b=-1\\ 2a+4b=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=\frac{1}{2}\\ b=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{x-1}=\frac{1}{2}\\ \frac{1}{y-1}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=y=3\)
Vậy HPT có nghiệm $(x,y)=(3,3)$
\(\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}\)
\(=xyz.\left [ \frac{1}{yz(1+x^2)}+\frac{2}{xz(1+y^2)}+\frac{3}{xy(1+z^2)} \right ]\)
\(=xyz.\left [ \frac{1}{yz+x(x+y+z)}+\frac{2}{xz+y(x+y+z)}+\frac{3}{xy+z(x+y+z)} \right ]\)
\(=xyz.\left [ \frac{1}{(x+y)(x+z)}+\frac{2}{(x+y)(y+z)}+\frac{3}{(x+z)(y+z)} \right ]\)
\(=xyz.\frac{y+z+2(z+x)+3(x+y)}{(x+y)(y+z)(z+x)}=\frac{xyz(5x+4y+3z)}{(x+y)(y+z)(z+x)}\)
Nhập biểu thức ở dấu Σ cho dễ nhìn em ạ
dấu đấy là dấu j ạ
Gõ ở đó e nhé
\(P=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-2x}{x-9}:\dfrac{\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-x-3\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)\cdot\sqrt{x}\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(-\sqrt{x}+5\right)}=\dfrac{-x}{-\sqrt{x}+5}\)
\(A=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}+\dfrac{x+9}{25-x}\right):\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)
\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(x-25\right)}\cdot\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}=\dfrac{1-\sqrt{x}}{\sqrt{x}-5}\)
\(B=\left(\dfrac{1}{x-4}-\dfrac{1}{x-4\sqrt{x}+4}\right):\dfrac{\sqrt[2]{x}}{2\sqrt{x}-x}\)
\(=\dfrac{\sqrt{x}-2-\sqrt{x}-2}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}\cdot\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(=\dfrac{-4}{x-4}\)