Bài 1: Rút gọn:
a) \(A=4\sqrt{3+2\sqrt{2}}-\sqrt{57+40\sqrt{2}}\)
b) \(B=\sqrt{1100}-7\sqrt{44}+2\sqrt{176}-\sqrt{1331}\)
c) \(C=\sqrt{\left(1-\sqrt{2002}\right)^2}.\sqrt{2003+2\sqrt{2002}}\)
d) \(D=\sqrt{72}-\sqrt{5\frac{1}{3}}+4,5\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
Bài 2: Cho biểu thức:
\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)
a) Tìm ĐKXĐ
b) Tính giá trị của A với x=3
c) Tìm giá trị của x để |A|=\(\frac{1}{2}\)
Bài 1:
a, \(4\sqrt{3+2\sqrt{2}}-\sqrt{57+40\sqrt{2}}\)
\(=4\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(4\sqrt{2}+5\right)^2}\)
\(=4\left(\sqrt{2}+1\right)-4\sqrt{2}-5\)
\(=4\sqrt{2}+4-4\sqrt{2}-5=-1\)
b, \(B=\sqrt{1100}-7\sqrt{44}+2\sqrt{176}-\sqrt{1331}\)
\(=10\sqrt{11}-14\sqrt{11}+8\sqrt{11}-11\sqrt{11}=-7\sqrt{11}\)
c, \(C=\sqrt{\left(1-\sqrt{2002}\right)^2}.\sqrt{2003+2\sqrt{2002}}\)
\(=\left(1-\sqrt{2002}\right).\sqrt{\left(\sqrt{2002}+1\right)^2}\)
\(=\left(1-\sqrt{2002}\right).\left(\sqrt{2002}+1\right)=-2001\)
Câu d bạn kiểm tra lại đề bài nhé.
Bài 2:
\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}+\frac{\sqrt{x}}{1-x}\)
a, ĐK: \(x\ge0,x\ne1\)
b, ĐK: \(x\ge0,x\ne1\)
\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}+\frac{\sqrt{x}}{1-x}\)
\(=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{2\sqrt{x}+2-2\sqrt{x}+2}{4\left(x-1\right)}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{4-4\sqrt{x}}{4\left(x-1\right)}=\frac{4\left(1-\sqrt{x}\right)}{4\left(1-x\right)}=\frac{1-\sqrt{x}}{1-x}\)
Thay \(x=3\left(TM\right)\)vào A ta có: \(A=\frac{1-\sqrt{3}}{3-1}=\frac{1-\sqrt{3}}{2}\)
Vậy với \(x=3\)thì \(A=\frac{1-\sqrt{3}}{2}\)
c, \(\left|A\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}A=\frac{1}{2}\\A=-\frac{1}{2}\end{cases}}\)
TH1: \(A=\frac{1}{2}\)\(\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=\frac{1}{2}\Leftrightarrow2-2\sqrt{x}=x-1\)\(\Leftrightarrow x-1-2+2\sqrt{x}=0\)\(\Leftrightarrow x+2\sqrt{x}-3=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(TM\right)\\\sqrt{x}=-3\left(L\right)\end{cases}}}\)
TH2: \(A=-\frac{1}{2}\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=-\frac{1}{2}\)\(\Leftrightarrow2-2\sqrt{x}=1-x\Leftrightarrow-x+1-2+2\sqrt{x}=0\)\(\Leftrightarrow-x-1+2\sqrt{x}=0\Leftrightarrow x-2\sqrt{x}+1=0\)\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=0\Leftrightarrow\sqrt{x}=-1\left(L\right)\)
Vậy với \(x=1\)thì \(\left|A\right|=\frac{1}{2}\)
