ĐKXĐ :\(2\le x\le6\)
\(\Rightarrow6-x+x-2+2\sqrt{\left(6-x\right)\left(x-2\right)}=4\)
\(\Rightarrow2\sqrt{\left(6-x\right)\left(x-2\right)}=0\)
\(\Rightarrow\sqrt{\left(6-x\right)\left(x-2\right)}=0\)
\(\Rightarrow6-x=0\Rightarrow x=6\)
hoặc \(x-2=0\Rightarrow x=2\)
Vậy x = 2 ; x = 6
1,\(\sqrt{x-5}+\sqrt{x+4}=3\)
2,\(\sqrt{x+4-4\sqrt{x}}+\sqrt{x+6-6\sqrt{x}}=1\)
3,\(\sqrt{x+3}-\sqrt{x-4}=1\)
4,\(\sqrt{15-x}+\sqrt{3-x}=6\)
5,\(\sqrt{10-x}+\sqrt{x+3}=5\)
6,\(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\)
7,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)
8,\(\sqrt{x^2-5x+6}+\sqrt{x-2-3\sqrt{x-3}}=3\)
9,\(2x^2-x+4=2\sqrt{2x+3}\)
Tìm `ĐKXĐ`:
\(\sqrt{\dfrac{-5}{6+x}}\)
\(\sqrt{\dfrac{-2}{6-x}}\)
\(\sqrt{\dfrac{-x+3}{-6}}\)
\(\sqrt{\dfrac{7x-1}{-9}}\)
\(\sqrt{\dfrac{x+2}{x^2+2x+1}}\)
\(\sqrt{\dfrac{x-2}{x^2-2x+4}}\)
Phân tích thành nhân tử
\(x+\sqrt{x}\)
\(x-\sqrt{x}\)
\(a+3\sqrt{a}-10\)
\(x\sqrt{x}+\sqrt{x}-x-1\)
\(x+\sqrt{x}-2\)
\(x-5\sqrt{x}+6\)
\(x\sqrt{x}-1\)
\(x\sqrt{x}-x+\sqrt{x}-1\)
\(x+2\sqrt{x}-15\)
\(x-2\sqrt{x}-3\)
\(a+\sqrt{a}-6\)
\(x-16\)
\(x+2\sqrt{x}+1\)
\(x-1\)
\(x-2\sqrt{x}+1\)
\(a\sqrt{a}+1\)
\(a+\sqrt{a}-2\)
\(2x-5\sqrt{x}+3\)
\(x-9\)
\(x+\sqrt{x}-6\)
\(a,x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(b,\sqrt{x^2+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)
\(c,\sqrt{x+3+4\sqrt{x-1}}+\sqrt{8-6\sqrt{x-1}+x}=1\)
\(d,\sqrt{3-x+x^2}-\sqrt{2+x-x^2}=1\)
\(e,\sqrt{4x-9}+2\sqrt{3x^2-5x+2}=\sqrt{3x-2}+\sqrt{x-1}\)
\(\sqrt{x^2-5x-6}=x-2\)
\(\sqrt{x^2-8x+16}=4-x\)
\(\sqrt{x^2-2x}=2-x\)
\(\sqrt{2x+27}-6=x\)
\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
\(\sqrt{x^2-x}=\sqrt{3-x}\)
Tìm điều kiện xác định và rút gọn biểu thức:
D=\(\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)
E=\(\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{2-\sqrt{x}}{3-\sqrt{x}}-\dfrac{\sqrt{x}-8}{x-\sqrt{x}-6}\right):\left(1-\dfrac{\sqrt{x}+6}{2\sqrt{x}+4}\right)\)
((x)/(\sqrt(x)-2)-(x-1)/(\sqrt(x)+2)-(\sqrt(x)+6)/(x-4))-:((\sqrt(x)+2)/(\sqrt(x)-2)-1)
Cho A =\(\frac{x^3+2x^2+3x+x^2\sqrt{4-x^2}+6}{\sqrt{x+3}+3}:\frac{x^2\left(\sqrt{x+2}+\sqrt{2-x}\right)+3\sqrt{x+2}}{2\sqrt{x+3}-3\sqrt{x+2}-\sqrt{x^2+5x+6}}\)
Rút gọn a
1. Thu gọn A = \(\left(\dfrac{x}{x\sqrt{x}-4\sqrt{x}}-\dfrac{6}{3\sqrt{x}-6}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
2. Tìm x để A<2
