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Lê Thị Tuyết
A=\(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x-2}}\)
Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)
C=\(\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{1+x}{\sqrt{x}}\)
Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0
a) \(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
Lời giải:
a)
\(A=\frac{\sqrt{3}-1+\sqrt{3}+1}{(\sqrt{3}+1)(\sqrt{3}-1)}+2-\sqrt{3}=\frac{2\sqrt{3}}{3-1}+2-\sqrt{3}=\sqrt{3}+2-\sqrt{3}=2\)
b)
\(B=\left(\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right):\frac{\sqrt{x}}{(\sqrt{x}-1)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}.(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)^2}{\sqrt{x}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{x}=\frac{x-1}{x}\)
a, Với \(x\ge0;x\ne1\)
\(Q=\left(\frac{x-1}{\sqrt{x}-1}-\frac{x\sqrt{x}-1}{x-1}\right):\left(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1-\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
a/ Bạn tự giải
b/ \(B=\frac{1}{\sqrt{x}-1}-\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x+\sqrt{x}+1-x-2-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{-\sqrt{x}}{x+\sqrt{x}+1}\)
c/ \(C=-AB=\frac{\left(x+\sqrt{x}+1\right)\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Do \(\sqrt{x}\ge0\Rightarrow C\ge0\)
\(C=\frac{\sqrt{x}+1-1}{\sqrt{x}+1}=1-\frac{1}{\sqrt{x}+1}< 1\)
\(\Rightarrow0\le C< 1\)
Mà C nguyên \(\Rightarrow C=0\Rightarrow x=0\)
1: Ta có: \(Q=\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}-\frac{x\sqrt{x}-1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\frac{x-2\sqrt{x}+x\sqrt{x}-x\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\frac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\frac{x-2\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}:\frac{x-1}{x+\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}\cdot\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
2: Ta có: \(\frac{1}{Q}=4\sqrt{x}-4\)
\(\Leftrightarrow Q=\frac{1}{4\sqrt{x}-4}\)
\(\Leftrightarrow\frac{x+\sqrt{x}+1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{4\sqrt{x}-4}\)
\(\Leftrightarrow\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=\left(x+\sqrt{x}+1\right)\left(4\sqrt{x}-4\right)\)
\(\Leftrightarrow x+x\sqrt{x}-\sqrt{x}-1=4x\sqrt{x}-4\)
\(\Leftrightarrow x+x\sqrt{x}-\sqrt{x}-1-4x\sqrt{x}+4=0\)
\(\Leftrightarrow x-3x\sqrt{x}-\sqrt{x}+3=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-\left(3x\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(x\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left[\sqrt{x}-3\left(x+\sqrt{x}+1\right)\right]=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-3x-3\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(-3x-2\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)=0\)(vì \(-3x-2\sqrt{x}-3\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\sqrt{x}=1\)
hay x=1(không thỏa mãn ĐKXĐ)
Vậy: Không có giá trị nào của x thỏa mãn \(\frac{1}{Q}=4\sqrt{x}-4\)
Chào em, em có thể kam khảo tại link:
Câu hỏi của Lê Thu Hà - Toán lớp 9 - Học toán với OnlineMath
Nếu link bị chặn em copy và dán tại:
https://olm.vn/hoi-dap/question/1261852.html
Câu hỏi của Lê Thu Hà - Toán lớp 9 - Học toán với OnlineMath
a) Rút gọn E
\(E=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-\sqrt{x}}{x-\sqrt{x}}\right)\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}+\frac{2-x}{\sqrt{x}-\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left[\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(E=\frac{x}{\sqrt{x}-1}\)
Vậy \(E=\frac{x}{\sqrt{x}-1}\)
\(P=\dfrac{x\sqrt{x}-x-\sqrt{x}-2}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)^2}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)}{\left(x+\sqrt{x}+1\right)}.\dfrac{\left(1-x^2\right)\left(x-1\right)}{2}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(x-1\right)\left(1-x^2\right)}{2\left(x+\sqrt{x}+1\right)}\)
