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a, A\(=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x-1}{\sqrt{x}}\) ĐK x>0 ;\(x\ne1;x\ne-1\)
\(A=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}}{x-1}\)
\(A=\frac{4x\sqrt{x}}{x-1}.\frac{\sqrt{x}}{x-1}\)=\(\frac{4x^2}{\left(x-1\right)^2}\)
b, Để A =2 \(\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Rightarrow4x^2=2\left(x-1\right)^2\)
<=> \(4x^2=2x^2-4x+2\)
<=> \(2x^2+4x-2=0\)
<=> \(x^2+2x-1=0\)
\(\Delta=1^2-1.\left(-1\right)\) = 2
=> \(\orbr{\begin{cases}x_1=-1-\sqrt{2}\left(loại\right)\\x_2=-1+\sqrt{2}\left(nhận\right)\end{cases}}\)
Vậy x=\(-1+\sqrt{2}\)thì A =2
c, Thay x =\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)=2
=>A = \(\frac{4.2^2}{\left(2-1\right)^2}=16\)
Vậy A=16 thì x=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(A=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)
ĐKXĐ : x > 1
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}+\frac{1}{\sqrt{x}-1}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left(\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\times\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\frac{x}{\sqrt{x}-1}\)
Để A = 9/2
=> \(\frac{x}{\sqrt{x}-1}=\frac{9}{2}\)( ĐK : x > 1 )
<=> 2x = 9( √x - 1 )
<=> 2x = 9√x - 9
<=> 2x + 9 = 9√x (1)
Bình phương hai vế
(1) <=> 4x2 + 36x + 81 = 81x
<=> 4x2 + 36x + 81 - 81x = 0
<=> 4x2 - 45x + 81 = 0
<=> 4x2 - 36x - 9x + 81 = 0
<=> 4x( x - 9 ) - 9( x - 9 ) = 0
<=> ( x - 9 )( 4x - 9 ) = 0
<=> \(\orbr{\begin{cases}x-9=0\\4x-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=\frac{9}{4}\end{cases}}\)( tm )
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 \(E=\frac{x}{\sqrt{x}-1}\)
\(Q=\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(1-\frac{1}{x-1}\right)\)
\(=\frac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}.\frac{x}{x-1}\)
\(=\frac{\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}}{\sqrt{\left(x-2\right)^2}}.\frac{x}{x-1}\)
\(=\frac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}\)
Nếu \(x\ge2\) thì
\(Q=\frac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}=\frac{2x\sqrt{x-1}}{\left(x-2\right)\left(x-1\right)}=\frac{2x}{\left(x-2\right)\left(\sqrt{x-1}\right)}\)
Nếu \(x< 2\) thì \(Q=\frac{1-\sqrt{x-1}+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}=\frac{2x}{\left(x-2\right)\left(x-1\right)}\)
Cảm ơn bạn nhiều nhưng mình thấy \(1-\frac{1}{x-1}=\frac{x-2}{x-1}\) mà bạn sao lại bằng \(\frac{x}{x-1}\)được
\(\left(\frac{x\sqrt{x}+x+2}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{x\sqrt{x}-x}\)
\(=\left(\frac{x\sqrt{x}+x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\frac{x\sqrt{x}-x}{1}\)
\(=\frac{x\sqrt{x}+x+2-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x\left(\sqrt{x}-1\right)}{1}\)
\(=\frac{x\sqrt{x}+x-\sqrt{x}+3}{\sqrt{x}+1}.x\)
\(=\frac{x^2\sqrt{x}+x^2-x\sqrt{x}+3x}{\sqrt{x}+1}\)
\(........?!\)