\(7=4+3=4+\sqrt{9}< 4+\sqrt{13}\)

Giả sử A(\(x_0,y_0\)) là điểm cố định mà hàm số đi qua

\(\Rightarrow y_0=mx_0-x_0-3\Leftrightarrow mx_0-x_0-y_0-3=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_0=0\\-x_0-y_0-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0=0\\y_0=-3\end{matrix}\right.\)

Vậy A(\(0,-3\))

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

b) Khi \(x=7+4\sqrt{3}\Rightarrow\frac{7+4\sqrt{3}}{\sqrt{\left(2+\sqrt{3}\right)^2}+1}=\frac{7+4\sqrt{3}}{3+\sqrt{3}}\)

c)\(M=\frac{1}{2}\Leftrightarrow\frac{x}{\sqrt{x}+1}=\frac{1}{2}\Leftrightarrow\sqrt{x}=2x-1\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\x^2=4x^2-4x+1\Leftrightarrow3x^2-4x+1=0\Leftrightarrow\left(3x-1\right)\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\\left[{}\begin{matrix}x=\frac{1}{3}\left(l\right)\\x=1\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

1 x 10 + 9 x 10

= 10 + 9 x 10

= 10 + 90

= 100

Ủng hộ tíck mik nha !!!

a. \(\sqrt{50}-\sqrt{3}.\sqrt{6}+\frac{\sqrt{22}}{\sqrt{11}}=5\sqrt{2}-3\sqrt{2}+\sqrt{2}=3\sqrt{2}\)

b. \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\sqrt{7+4\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}=\sqrt{2}\)

\(164+165+166+...+1000\)

\(=\frac{\left(\left\{1000+164\right\}.\left[1000-164\right]:1+1\right)}{2}\)

\(=\frac{974268}{2}=487134\)

\(\Rightarrow x^4+y^4+z^4-3xyz=0\)

\(\Rightarrow x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=0\)

\(\Rightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)

\(\Rightarrow\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3xy\left(x+y\right)-3xyz=0\)

\(\Rightarrow\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2-3xy\right)=0\)

\(\Rightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-y=0\\x-z=0\\y-z-0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\x=z\\y=z\end{cases}\Rightarrow}x=y=z=1}\)

a) \(m+2>0\Leftrightarrow m>-2\)

b) \(m+2< 0\Leftrightarrow m< -2\)

a) \(m+5\ne0\Leftrightarrow m\ne-5\)

b)\(m+5>0\Leftrightarrow m>-5\)

c) Thay \(x=2,y=3\)ta được

\(3=2\left(m+5\right)+2m-10\Leftrightarrow4m=3\Leftrightarrow m=\frac{3}{4}\)

d) Đồ thị cắt tung độ bằng 9

\(\Rightarrow9=\left(m+5\right).0+2m-10\Leftrightarrow m=\frac{19}{2}\)

e)Đồ thị đi qua hoành độ bằng 10

\(\Rightarrow0=10\left(m+5\right)+2m-10\Leftrightarrow12m=-40\Leftrightarrow m=\frac{-10}{3}\)