Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)
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=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\) \(ĐKXĐ:x\ge0;x\ne1;x\ne4\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{x-1}\right]\)
\(A=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left[\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
vậy \(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b)theo bài ra: \(A=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right).\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-\left(\sqrt{3}\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1-\sqrt{3}=0\\\sqrt{x}-1+\sqrt{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(1-\sqrt{3}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\x=3-2\sqrt{3}+1\end{cases}}\)
vậy......
\(K=\left(\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\left(\sqrt{a}-1\right)\)
\(=\frac{a-1}{\sqrt{a}}\Rightarrow\left\{{}\begin{matrix}m=1\\n=-1\end{matrix}\right.\Rightarrow m^2+n^2=2\)
\(A=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\Rightarrow\left\{{}\begin{matrix}m=0\\n=-2\end{matrix}\right.\Rightarrow m-n=2\)
Cảm ơn bạn nha ;)