\(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
a) ruts gon A
b) tim gtri x de \(A< \frac{1}{3}\)
a)\(\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+x\sqrt{x}}\right):\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
Rut gon
Tim x de B nguyen
cho P=\(\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\\ \)
a, rut gon P
b, tim x de P=\(\sqrt{x}-1\)
a/ \(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\sqrt{x^2}-1+\sqrt{x}-1}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{3\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{1}{\left(\sqrt{x}-1\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{\sqrt{x}-1}{1}\right)\)
=> \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}\)
b/ \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}=\sqrt{x}-1\)
<=> \(4\sqrt{x}=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
<=> \(4\sqrt{x}=x-1\). Bình phương 2 vế, ta được:
<=> 16x=(x-1)2
<=> 16x=x2-2x+1
<=> x2-18x+1=0
\(\Delta'=81-1=80=>\sqrt{\Delta'}=4\sqrt{5}\)
=> \(x_1=9-4\sqrt{5}\)
\(x_2=9+4\sqrt{5}\)
cho P=\(\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right).\left(\frac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\\ \)
a, rut gon
b, tim x de P<\(7-4\sqrt{3}\)
cho P=\(\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right).\left(\frac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\\ \)
a, rut gon
b, tim x de P<\(7-4\sqrt{3}\)
ĐKXĐ: \(x\ge0\)
a/ Đề \(=\left(\frac{1-\sqrt{x}^3}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+\sqrt{x}^3}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left[\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right]\left[\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right]\)
\(=\left(1+2\sqrt{x}+x\right)\left(1-2\sqrt{x}+x\right)\)
\(=\left(1+\sqrt{x}\right)^2\left(1-\sqrt{x}\right)^2\)
\(=\left[\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\right]^2=\left(1-x\right)^2\)
b/ \(P< 7-4\sqrt{3}\Leftrightarrow\left(1-x\right)^2< 7-4\sqrt{3}\)
\(\Rightarrow\left(1-x\right)^2< \left(2-\sqrt{3}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}1-x< 2-\sqrt{3}\Rightarrow x>-1+\sqrt{3}\\1-x< \sqrt{3}-2\Rightarrow x>3-\sqrt{3}\end{cases}}\)
Vậy \(x>3-\sqrt{3}\)
cho bieu thuc A = \(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
a. tim x de bieu thuc A co nghia ?rut gon A ?
b. tinh gia tri cua bieu thuc A tai x=7+4√3
a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
B=\(\left(\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}-1\right):\frac{\sqrt{x}-1}{\sqrt{x}-2}\)
a,Rut gon B
b,Tim cac gia tri cua x de 3B=\(2\sqrt{x}-7\)
cho bieu thuc A =\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
(x≥0;x≠1)
a. tim x de bieu thuc A co nghia ?rut gon A ?
b. tinh gia tri bieu thuc A tai x=7+4√3
a/ Ta có: A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+1\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right):\left(\sqrt{x}\right)=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b/ Ta có :\(x=7+4\sqrt{3}=3+4\sqrt{3}+4=\left(\sqrt{3}+2\right)^2
\)
\(\Rightarrow\sqrt{x}=|\sqrt{3}+2|=\sqrt{3}+2\)
Thay x vào A ta có:
A\(=\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}+2+1}{\sqrt{3}+2}=\frac{\sqrt{3}+3}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{3-\sqrt{3}}{1}=3-\sqrt{3}\)
Rút gọn
\(1.A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(2.B=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(3.C=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right).\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
\(A=\frac{\sqrt{x}+4}{\sqrt{x}+2};B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)
a. rut gon B
b. Tim x nguyen de P = B(A-1) nguyen
giup minh voi a