a/ \(\sqrt{x^2-x-6}=\sqrt{x-3}\)\(\Rightarrow x^2-x-6=x-3\Rightarrow x^2-2x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\Rightarrow x=3\)hoặc \(x=-1\)
Vậy x = 3, x = -1
b/ \(\sqrt{x^2-x}=x-5\Rightarrow x^2-x=x^2-10x+25\Rightarrow9x=25\Rightarrow x=\frac{25}{9}\)
Vậy x = 25/9
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,\(2x^2-x+4=2\sqrt{2x+3}\)
Phân tích thành nhân tử
\(x+\sqrt{x}\)
\(x-\sqrt{x}\)
\(a+3\sqrt{a}-10\)
\(x\sqrt{x}+\sqrt{x}-x-1\)
\(x+\sqrt{x}-2\)
\(x-5\sqrt{x}+6\)
\(x\sqrt{x}-1\)
\(x\sqrt{x}-x+\sqrt{x}-1\)
\(x+2\sqrt{x}-15\)
\(x-2\sqrt{x}-3\)
\(a+\sqrt{a}-6\)
\(x-16\)
\(x+2\sqrt{x}+1\)
\(x-1\)
\(x-2\sqrt{x}+1\)
\(a\sqrt{a}+1\)
\(a+\sqrt{a}-2\)
\(2x-5\sqrt{x}+3\)
\(x-9\)
\(x+\sqrt{x}-6\)
1. \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+3-4\sqrt{x-1}}\left(2< x< 5\right)\)
2. \(\frac{6}{1-\sqrt{3}}-\frac{3\sqrt{3}-1}{\sqrt{3}+1}+\sqrt{3}\)
3. \(\sqrt{29-12\sqrt{5}+\sqrt{24-8\sqrt{3}}}\)
4. \(\sqrt{\frac{4}{9-4\sqrt{5}}}-\sqrt{\frac{4}{9+4\sqrt{5}}}\)
5. \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{x}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\)
6. \(\frac{6-\sqrt{6}}{\sqrt{6}-1}-9\sqrt{\frac{2}{3}}-\frac{4}{2-\sqrt{6}}\)
7. \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(\sqrt{x}-1\right)^2}{2}\left(x\ge0,x\ne1\right)\)
Bài: Rút gọn biểu thức sau
1)(1-\(\dfrac{\sqrt{x}}{1-\sqrt{x}}\)):(\(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)-\(\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\)+\(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\))
2)(\(\dfrac{1}{\sqrt{x}+1}\)-\(\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\)):(\(\dfrac{1}{\sqrt{x}-1}\)-\(\dfrac{2}{x-1}\))
\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
\(\sqrt{x^2-x}=\sqrt{3-x}\)
Bài 1:Tìm ĐKXĐ:
a.\(\sqrt{3x}\)
b.\(\sqrt{\dfrac{x-1}{x+3}}\)
Bài 2:Thực hiện phép tính:
C=\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
Bài 3:
A=(1-\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)):(\(\dfrac{1}{\sqrt{x}-2}-\dfrac{2}{x-4}\)) với x>0;x≠4
a.Rút gọn A
b.Tính giá trị của A khi x =\(\dfrac{1}{4}\)
c. Chứng minh A<2
d.Tìm giá trị nguyên của x để A nguyên.
Trả lời giúp mình với ạ!Mình cảm ơn nhiều!
rút gọn
\((1-\dfrac{\sqrt{x}}{\sqrt{x}+1})\div(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\)
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=\(\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{2-\sqrt{x}}{3-\sqrt{x}}-\dfrac{\sqrt{x}-8}{x-\sqrt{x}-6}\right):\left(1-\dfrac{\sqrt{x}+6}{2\sqrt{x}+4}\right)\)
1\(\sqrt{5+2\sqrt{8}}-\sqrt{5-2\sqrt{8}}\) 2)\(\dfrac{\sqrt{x^2+2\sqrt{3x}+3}}{x^2-3}\) 3) \(\dfrac{\sqrt{x^2-5x+6}}{\sqrt{x-2}}\) 4)\(\dfrac{\sqrt{\left(x-4\right)^2}}{x^2-5x+4}\) 5) \(\dfrac{3x+1}{\sqrt{9x^2+6x+1}}\)
Cho bt: P=\(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}.\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
a, Rút gọn P
b, P khi x = 6-2\(\sqrt{5}\)
giải hộ e với e đang cần gấp để đối chiếu kết quả!
