ĐK : x > 0
\(B=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\)
\(B=\frac{x-1}{x+\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\)
\(B=\frac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\)
\(B=\frac{x-\sqrt{x}}{x+\sqrt{x}}\) \(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Rút gọn:
\(A=\left(\frac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right):\left(\frac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\) với \(x\ge0;x\ne1\)
\(B=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right):\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\) với \(x>0;x\ne1\)
B=\(\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(\frac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{x+1}\right)\)
a) Rút gọn B
b) Chứng tỏ 0>B>2
tính giá trị biểu thức
1) A = \(\frac{15\sqrt{x}-11}{x-2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\) tại \(x=3-2\sqrt{2}\)
2) \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\) tại \(x=7-2\sqrt{6}\)
3) \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\) tại \(x=7-4\sqrt{3}\)
Bài 1 : Cho A=\(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
B=\(\frac{1}{x+\sqrt{x}}-\frac{1}{x-\sqrt{x}}+\frac{2\sqrt{x}}{x-1}\)
a) Rút gọn A và B
b) tìm x để A > B
Rút gọn biểu thức:
1) \(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\cdot\left(x-1\right)}{\sqrt{x}-1}\)
2) \(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
3) \(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
4) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
A = (\(\sqrt{x}-\frac{x+2}{x-1}\)) : ( \(\frac{\sqrt{x}}{\sqrt{x}+1}\)- \(\frac{\sqrt{x}-4}{1-x}\))
B= (\(\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}-1}\)) :\(\sqrt{x}+\frac{\sqrt{x}}{\sqrt{x}-1}\))
C=(\(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\)) .(\(\frac{1-x}{\sqrt{2}}\))2
Mọi ng giúp em nhanh với ạ, em cảm ơn ạ.
1/Rút gọn
A=\(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{xy}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\left(\sqrt{x^3+x}\right)}\)(x>0; y>0; x#y)
B= \(\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)( x>0)
C=\(\left(\frac{x+1}{\sqrt{x}}+2\right).\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x\sqrt{x}+1\right)}\)(x>0)
D=\(\left(\frac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right):\left(x-1\right)-\frac{2}{\sqrt{x}-1}\)(x>=0; x#1)
giúp em với ạ em đang cần gấp ạ
Rút gọn biểu thức sau :( chú ý đặt ĐKXĐ trước khi trước khi thực hiện rút gọn)
a,P= \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}\)
b, D=\(\frac{\sqrt{x}+4}{1-7\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{x+1}}+\frac{24\sqrt{x}}{7x+6\sqrt{x}-1}\)
(\(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}-x+\sqrt{x}-1}\))\(:\left(\frac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\right)\)
Rút gọn và tìm GTNN của biểu thức:
\(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right)+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
