\(\frac{20}{\sqrt{x}}+x\sqrt{x}+x=32\)
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E=\(\left(\frac{\sqrt{x^3}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x^3}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)+\left(\frac{x-1}{\sqrt{x}}\right)\cdot\left(\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
= \(\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\) +\(\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\right)\cdot\left(\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
=\(\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right)+\frac{1}{\sqrt{x}}\cdot\left(2x+2\right)\)
=\(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\frac{2x+2}{\sqrt{x}}\)
=\(\frac{2\sqrt{x}+2x+2}{\sqrt{x}}\)
ĐKXĐ:\(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có:
+)\(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}=\frac{2\sqrt{x}}{\sqrt{x}}=2\)
+)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{x-1}=\frac{2\left(x+1\right)}{x-1}\)
\(\Rightarrow\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\frac{x-1}{\sqrt{x}}.\frac{2\left(x+1\right)}{x-1}=\frac{2\left(x+1\right)\sqrt{x}}{x}\)
Thay vào E ta được: \(E=2+\frac{2\left(x+1\right)\sqrt{x}}{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tử số của phân số đầu phải là \(\sqrt{x}+2\) chứ không phải \(\sqrt{x+2}\), vì cái \(\sqrt{x}+2\) nó mới logic để rút gọn: )
\(Q=\left(\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}^3+8}-\dfrac{x-\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\left(\dfrac{x+4\sqrt{x}+4-x+\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\dfrac{\left(5\sqrt{x}+4\right).5.\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)\left(5\sqrt{x}+4\right)}\\ =\dfrac{5}{\sqrt{x}+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge0;x\ne4\)
\(B=\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2+\sqrt{x}-2-2\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{1}{x-4}\)
\(A=10-\left(2\sqrt{2}-3\sqrt{3}\right)\left(-2\sqrt{2}-3\sqrt{3}\right)\)
\(=10+\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)\)
\(=10+\left(8-27\right)=-9\)
\(AB=-1\Leftrightarrow\frac{-9}{x-4}=-1\Rightarrow x-4=9\Rightarrow x=13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x\ge0\)
Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)
\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)
Dòng thứ 2 qua dòng thứ 3 anh làm chậm lại được không ạ, tại tắt quá e không hiểu
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này chắc tác giả đánh sai tử thức của phân thức cuối cùng, biểu thức B phải là \(B=\frac{1}{C}\) trong đó \(C=\left(\frac{\sqrt{x}+3}{x+\sqrt{x}+1}-\frac{\sqrt{x}-3}{x\sqrt{x}-1}\right)\cdot\frac{x\sqrt{x}-\sqrt{x}+x^2-1}{\sqrt{x}}\)
Ta có \(C=\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\frac{\left(\sqrt{x}-3\right)}{x\sqrt{x}-1}\right)\cdot\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{x+\sqrt{x}}{x\sqrt{x}-1}\cdot\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\left(\sqrt{x}+1\right)^2\)
Thành thử ta được \(C=\left(\sqrt{x}+1\right)^2\)
Ta có \(x=98+20\sqrt{6}=\left(5\sqrt{2}+4\sqrt{3}\right)^2\to\sqrt{x}=5\sqrt{2}+4\sqrt{3}\to\)
hay \(C=\left(\sqrt{x}+1\right)^2=x+2\sqrt{x}+1=99+20\sqrt{6}+2\left(5\sqrt{2}+4\sqrt{3}\right)\)
\(=99+20\sqrt{6}+10\sqrt{2}+8\sqrt{3}\to B=\frac{1}{C}=\frac{1}{99+20\sqrt{6}+10\sqrt{2}+8\sqrt{3}}\)
Đề là tìm x hả bn?