\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)

\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)

\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)

\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)

\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)

\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)

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

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

\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)

\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)

Đặt \(\hept{\begin{cases}\sqrt{2+x}=a\\\sqrt{2-x}=b\end{cases}\Rightarrow}a^2+b^2=4\)

\(\Rightarrow C=\frac{\sqrt{2ab}.\left(a^3-b^3\right)}{a^2+b^2+ab}=\frac{\sqrt{2ab}.\left(a-b\right)\left(a^2+b^2+ab\right)}{a^2+b^2+ab}\)

\(=\sqrt{2ab}.\left(a-b\right)=\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{2+x}-\sqrt{2-x}\right)\)

đây là toàn lp 3 hả bn

đây ko phải toán lớp 3

quên đây là toán lớp 1 

Câu 1:

a/ \(x\ge-11\)

Đặt \(\sqrt{x+11}=a\ge0\Rightarrow11=a^2-x\) pt trở thành:

\(x^2+a=a^2-x\Leftrightarrow x^2-a^2+a+x=0\)

\(\Leftrightarrow\left(x-a\right)\left(x+a\right)+x+a=0\Leftrightarrow\left(x-a+1\right)\left(x+a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-a+1=0\\x+a=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+1=a\\a=-x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{x+11}\left(1\right)\\-x=\sqrt{x+11}\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\\left(x+1\right)^2=x+11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-10=0\end{matrix}\right.\) \(\Rightarrow x=\frac{-1+\sqrt{41}}{2}\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}-x\ge0\\x^2=x+11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^2-x-11=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1-3\sqrt{5}}{2}\)

b/ \(x\ge-9\)

\(\sqrt{x+9}=x-9\Leftrightarrow\left\{{}\begin{matrix}x-9\ge0\\x+9=\left(x-9\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge9\\x^2-19x+72=0\end{matrix}\right.\) \(\Rightarrow x=\frac{19+\sqrt{73}}{2}\)

Câu 1:

a/ \(x\ge-11\)

Đặt \(\sqrt{x+11}=a\ge0\Rightarrow11=a^2-x\), pt đã cho trở thành:

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

TH1: \(x+a=0\Leftrightarrow x+\sqrt{x+11}=0\Leftrightarrow-x=\sqrt{x+11}\)

\(\Leftrightarrow\left[{}\begin{matrix}-x\ge0\\x^2=x+11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^2-x-11=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1-3\sqrt{5}}{2}\)

TH2: \(x-a+1=0\Leftrightarrow x+1=\sqrt{x+11}\) \(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\\left(x+1\right)^2=x+11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-10=0\end{matrix}\right.\) \(\Rightarrow x=\frac{-1+\sqrt{41}}{2}\)

b/ \(\sqrt{9+x}=x-9\Leftrightarrow\left\{{}\begin{matrix}x-9\ge0\\9+x=\left(x-9\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge9\\x^2-19x+72=0\end{matrix}\right.\) \(\Rightarrow x=\frac{19+\sqrt{73}}{2}\)

Câu 2:

a/

\(f\left(x\right)=\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x-1\right)\left(x-4\right)}=\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x-4\right)}\)

Lập bảng xét dấu ta được:

\(f\left(x\right)>0\) khi \(\left[{}\begin{matrix}x< -1\\x>4\\1< x< 3\end{matrix}\right.\)

\(f\left(x\right)< 0\) khi \(\left[{}\begin{matrix}-1< x< 1\\3< x< 4\end{matrix}\right.\)

\(f\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

\(f\left(x\right)\) ko xác định tại \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

b/ \(h\left(x\right)=\frac{-x^2+3x-1}{\left(x^2-2x+3\right)\left(x+2\right)}\)

Lập bảng xét dấu ta được:

\(f\left(x\right)>0\) khi \(\left[{}\begin{matrix}x< -2\\\frac{3-\sqrt{5}}{2}< x< \frac{3+\sqrt{5}}{2}\end{matrix}\right.\)

\(f\left(x\right)< 0\) khi \(\left[{}\begin{matrix}-2< x< \frac{3-\sqrt{5}}{2}\\x>\frac{3+\sqrt{5}}{2}\end{matrix}\right.\)

\(f\left(x\right)=0\) tại \(x=\frac{3\pm\sqrt{5}}{2}\)

\(f\left(x\right)\) ko xác định tại \(x=-2\)

\(B=\left(\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\right):\frac{x\sqrt{x}}{\left(4-x\right)^2}\)

\(=\frac{2x+18\sqrt{x}-x-9\sqrt{x}}{x-9}\cdot\frac{\left(4-x\right)^2}{x\sqrt{x}}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+9\right)\left(4-x\right)^2}{x\sqrt{x}\left(x-9\right)}\)

\(=\frac{\left(\sqrt{x}+9\right)\left(4-x\right)^2}{\sqrt{x}\left(x-9\right)}\)