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P/s : sửa đề
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(P=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}-3x}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(P=\frac{-3\sqrt{x}\left(1+\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{-3\sqrt{x}}{\sqrt{x}+3}\)
b) \(P< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6\sqrt{x}+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\frac{-5\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)
Mà \(2\left(\sqrt{x}+3\right)>0\)
\(\Rightarrow-5\sqrt{x}+3< 0\)
\(\Leftrightarrow-5\sqrt{x}< -3\)
\(\Leftrightarrow\sqrt{x}>\frac{3}{5}\)
\(\Leftrightarrow x>\frac{9}{25}\)
Vấy .................
c) \(P.\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+2\sqrt{x}-2+x=2\)
\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}-2-2+x=0\)
\(\Leftrightarrow-\sqrt{x}-4+x=0\)
\(\Leftrightarrow-\sqrt{x}\left(1-\sqrt{x}\right)=4\)
Còn lại lập bảng tự tìm giá trị của x là ra .( Chú ý : đối chiếu ĐKXĐ )
d)
\(P.\left(\sqrt{x}+3\right)+x\left(\sqrt{x}-m\right)=x-\sqrt{x}\left(3+m\right)\)
\(\Leftrightarrow\frac{-3\sqrt{x}}{\sqrt{x}+3}\left(\sqrt{x}+3\right)+x\sqrt{x}-xm=x-3\sqrt{x}-m\sqrt{x}\)
\(\Leftrightarrow-3\sqrt{x}+x\sqrt{x}-xm-x+3\sqrt{x}+m\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(x+m\right)-x\left(m+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left[x+m-m\sqrt{x}-\sqrt{x}\right]=0\)
\(\Leftrightarrow\sqrt{x}\left[m\left(1-\sqrt{x}\right)-\sqrt{x}\left(1-\sqrt{x}\right)\right]=0\)
\(\Leftrightarrow\sqrt{x}=0;m-\sqrt{x}=0;1-\sqrt{x}=0\)
+) \(\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
+) \(1-\sqrt{x}=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
+) \(m-\sqrt{x}=0\)
\(\Leftrightarrow\orbr{\begin{cases}m-\sqrt{0}=0\\m-\sqrt{1}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}m=0\\m=1\end{cases}}}\)
Vậy ..................
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne9\end{cases}}\)
\(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow C=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{9-x}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3\sqrt{x}+9}{9-x}:\frac{2\sqrt{x}+4}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3}{3-\sqrt{x}}\cdot\frac{x-3\sqrt{x}}{2\sqrt{x}+4}\)
\(\Leftrightarrow C=\frac{-3}{2\sqrt{x}+4}\)
b) Để \(-\frac{3}{2\sqrt{x}+4}< -1\)
\(\Leftrightarrow\frac{1+2\sqrt{x}}{2\sqrt{x}+4}< 0\)
Vì \(\hept{\begin{cases}1+2\sqrt{x}>0\\2\sqrt{x}+4>0\end{cases}\Leftrightarrow C>0}\)
Vậy để C <-1 <=> \(x\in\varnothing\)
c) \(A=\frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow A^2=3+2+2\sqrt{5}=5+2\sqrt{5}\)
\(B=\sqrt{5}+1\)
\(\Leftrightarrow B^2=5+1+2\sqrt{5}=6+2\sqrt{5}\)
Vì \(5+2\sqrt{5}< 6+2\sqrt{5}\)
\(\Leftrightarrow A^2< B^2\)
\(\Leftrightarrow A< B\)
Vậy \(\frac{1}{\sqrt{3}-\sqrt{2}}< \sqrt{5}+1\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
Câu 4: a) ĐK: \(x^2\ge9\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
b) ĐK: \(x^2-3x+2\ge0\Leftrightarrow\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)
c) Đk: \(-3\le x< 5\)
d) x + 3 và 5 - x đồng dấu. Xét hai trường hợp:
\(\left\{{}\begin{matrix}x+3\ge0\\5-x>0\left(\text{do mẫu phải khác 0}\right)\end{matrix}\right.\Leftrightarrow-3\le x< 5\)
\(\left\{{}\begin{matrix}x+3< 0\\5-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -3\\x>5\end{matrix}\right.\) do x ko thể đồng thời thỏa mãn cả hai nên loại.
Câu 1:
a) Đặt \(A=x+\sqrt{\left(x+2\right)^2}\cdot\left(x-2\right)\)
\(A=x+\left|x+2\right|\cdot\left(x-2\right)\)
+) Với \(x\ge-2\):
\(A=x+\left(x+2\right)\left(x-2\right)=x+x^2-4\)
+) Với \(x< -2\):
\(A=x-\left(x+2\right)\left(x-2\right)=x-x^2+4\)
b) \(B=\sqrt{m^2-6m+9-2m}\)
\(B=\sqrt{m^2-8m+9}\)
Bạn xem lại đề nhé :)
c) \(C=1+\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(C=1+\sqrt{x-1}\)
d) \(D=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(D=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(D=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(D=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
+) Xét \(x\ge8\):
\(D=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
+) Xét \(4< x< 8\):
\(D=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
Vậy....