1,Tìm GTNN của bt P =(x+7).\(\frac{1}{\sqrt{x}+3}\)
1,Cho biểu thức:
A=\((\frac{1}{\sqrt{x}+3}+\frac{3}{x-9}).\frac{\sqrt{x}-3}{\sqrt{x}}\)
a,Rút gọn
b,Tìm x để A=\(\frac{1}{5}\)
c,Tìm GTNN của bt P=(x+7).A
a) ĐKXĐ: \(x>0;x\ne9\)
\(A=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}+3}\)
b) \(A=\frac{1}{5}\) \(\Rightarrow\)\(\frac{1}{\sqrt{x}+3}=\frac{1}{5}\)
\(\Rightarrow\)\(\sqrt{x}+3=5\)
\(\Leftrightarrow\)\(\sqrt{x}=2\)
\(\Leftrightarrow\)\(x=4\)(t/m ĐKXĐ)
Vậy...
1,Cho biểu thức:
A=\((\frac{1}{\sqrt{x}+3}+\frac{3}{x-9}).\frac{\sqrt{x}-3}{\sqrt{x}}\)
a,Rút gọn
b,Tìm x để A=\(\frac{1}{5}\)
c,Tìm GTNN của bt P=(x+7).A
CHO BT: P=\(\left(\frac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
a) rg p
b) tính gt p biết x = \(\frac{53}{9-2\sqrt{7}}\)
c) tìm gtnn của \(\frac{1}{p}\)
CHO BT: P=\(\left(\frac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
a) rg p
b) tính gt p biết x = \(\frac{53}{9-2\sqrt{7}}\)
c) tìm gtnn của \(\frac{1}{p}\)
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{2\sqrt{x}+x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b: Thay \(x=9+2\sqrt{7}\) vào P, ta được:
\(P=\dfrac{\sqrt{9+2\sqrt{7}}+1}{9+2\sqrt{7}+\sqrt{9+2\sqrt{7}+1}}\simeq0,25\)
CHO BT: P=\(\left(\frac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
a) rg p
b) tính gt p biết x = \(\frac{53}{9-2\sqrt{7}}\)
c) tìm gtnn của \(\frac{1}{p}\)
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
CHO BT: P=\(\left(\frac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
a) rg p
b) tính gt p biết x = \(\frac{53}{9-2\sqrt{7}}\)
c) tìm gtnn của \(\frac{1}{p}\)
CHO BT: P=\(\left(\frac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
a) rg p
b) tính gt p biết x = \(\frac{53}{9-2\sqrt{7}}\)
c) tìm gtnn của \(\frac{1}{p}\)
Tìm GTNN của bt P=\(\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\left(x>0,x\ne1\right)\)
CHO BT: P=\(\left(\frac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
a) rg p
b) tính gt p biết x = \(\frac{53}{9-2\sqrt{7}}\)
c) tìm gtnn của \(\frac{1}{p}\)
c) \(\frac{1}{P}=1+\frac{x}{\sqrt{x}+1}\)\(=1+\frac{x-1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)
\(=1+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)
\(=1+\sqrt{x}-1+\frac{1}{\sqrt{x}+1}\)
\(=-1+\sqrt{x}+1+\frac{1}{\sqrt{x}+1}\)\(\ge-1+2\sqrt{\left(\sqrt{x}+1\right)\left(\frac{1}{\sqrt{x}+1}\right)}=1\)
Dau "=" xay ra khi x = 0