ĐKXĐ: \(x\ge5\)
Ta có: \(\sqrt{x^2-25}-6=3\sqrt{x+5}-2\sqrt{x-5}\)
Đặt \(\sqrt{x-5}=a\left(a\ge0\right)\)
\(\sqrt{x+5}=b\left(b\ge0\right)\)
Pt trở thành: \(ab-6=3b-2a\)
\(\Leftrightarrow3b-2a-ab+6=0\)
\(\Leftrightarrow\left(b+2\right)\left(3-a\right)=0\)
Vì \(b\ge0\Rightarrow b+2>0\)
\(3-a=0\)
\(\Leftrightarrow3=\sqrt{x-5}\Leftrightarrow x-5=9\Leftrightarrow x=14\) (Tm ĐKXĐ)
Vậy...
1. Tính : \(\dfrac{12}{4-\sqrt{10}}\)-6\(\sqrt{\dfrac{5}{2}}\)+\(\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)
2,Rút gọn:A=(\(\dfrac{\sqrt{x}}{\sqrt{x}-5}\)-\(\dfrac{5}{\sqrt{x}+5}\)+\(\dfrac{10\sqrt{x}}{25-x}\)):\(\dfrac{3}{\sqrt{x}+5}\)
d) \(x-5\sqrt{x}+6=0\)
e) \(\sqrt{x-1}+\dfrac{3}{2}\sqrt{4x-4}-\dfrac{2}{5}\sqrt{25x-25}=4\)
f) \(\sqrt{x-5}+\sqrt{4x-20}-\dfrac{1}{3}\sqrt{9x-45}=6\)
\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
\(\sqrt{x^2-x}=\sqrt{3-x}\)
1,Tính \(\dfrac{12}{4-\sqrt{10}}-6\sqrt{\dfrac{5}{2}}+\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)
2,Rút gọn:A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{5}{\sqrt{x}+5}+\dfrac{10\sqrt{x}}{25-x}\right):\dfrac{3}{\sqrt{x}+5}\)
31 A=\(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\right)\)
a. rút gọn
b. tính A với x thỏa mãn \(x-5\sqrt{x}+6=0\)
thực hiện phép tính
a)\(\dfrac{3}{5}\)-\(\dfrac{1}{2}\)\(\sqrt{1\dfrac{11}{25}}\)
b)(5+2\(\sqrt{6}\))(5-2\(\sqrt{6}\))
c)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
d)\(\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)(với x,y>0)
giải phương trình
\(\sqrt{x^2-25}-6=2\sqrt{x-5}-3\sqrt{x+5}\)
1) \(\sqrt{x^2}=2x-5\)
2) \(\sqrt{25x^2-10x+1}=2x-6\)
3) \(\sqrt{25-10x+x^2}=2x-5\)
4) \(\sqrt{1-2x+x^2}=2x-1\)
5) \(\sqrt{4x^2+4x+1}=-x-3\)
25. 7\(\sqrt{x}\)- 6x- 2
26.\(x^2\)-\(\sqrt{x}\)+x -1
27. 2a - 5\(\sqrt{ab}\)+ 3b
28.\(\sqrt{ab}\)+2\(\sqrt{a}\)+ 3\(\sqrt{b}\)+6
