⇔\(x^2-x+1+1-2\sqrt{x}=0\)
⇔\(\left(x^2-2\sqrt{x}+1\right)\)\(-x+1=0\)
⇔\(\left(x-1\right)^2\)\(-\left(x-1\right)\)=0
⇔\(\left(x-1\right)\)\(\left(x-2\right)=0\)
⇔x-1 = 0 hoặc x-2 = 0
+) x-1 = 0 ⇔ x=1
+) x-2 = 0 ⇔ x=2
⇔\(x^2-x+1+1-2\sqrt{x}=0\)
⇔\(\left(x^2-2\sqrt{x}+1\right)\)\(-x+1=0\)
⇔\(\left(x-1\right)^2\)\(-\left(x-1\right)\)=0
⇔\(\left(x-1\right)\)\(\left(x-2\right)=0\)
⇔x-1 = 0 hoặc x-2 = 0
+) x-1 = 0 ⇔ x=1
+) x-2 = 0 ⇔ x=2
Giải pt:
1) x - 2\(\sqrt{x - 1}\) = 16
2) \(\sqrt{1\:-\:x}\) - \(\sqrt{x - 3}\) = 0
3) \(\sqrt{x - 4}\) + 2 = 7
4) x - \(\sqrt{x - 2\sqrt{x\:-\:1}}\) = 0
5) \(\sqrt{x - 2}\) - \(\sqrt{x^2 - 2x}\) = 0
6) \(\sqrt{3\:-\:2\sqrt{2}}\) - \(\sqrt{x^2 + 2x\sqrt{2}+2}\) = 0
\(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-3}{x-\sqrt{x}-2}\right):\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{2}{\sqrt{x}-2}\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x+\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-\sqrt{x}+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+3+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+2\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+2}\)
\(A=\frac{x-2\sqrt{x}}{x^3+1}+\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\frac{1+2x-2\sqrt{x}}{x^2\sqrt{x}_{ }^2}\)
\(B=\frac{\frac{1}{\sqrt{x+2}}-\sqrt{x-2}}{\frac{1}{\sqrt{x-2}}-\frac{1}{\sqrt{x+2}}}:\frac{\sqrt{x-2}\sqrt{x^2-4}}{\left(x+2\right)\sqrt{x-2}-\left(x-2\right)\sqrt{x+2}}+x^2+1\\ x>2\)
giải hộ em ạ, em cảm ơn :>
\(\left(\dfrac{2}{2-\sqrt{x}}+\dfrac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right)\)
\(\left(\dfrac{\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}\right).\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\) RÚT GỌN
Chứng minh đẳng thức sau:
1) \(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}\)
2) \(\left(\sqrt{x}-\dfrac{x}{x+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{x\sqrt{x}-\sqrt{x}}\right)=x\sqrt{x}\left(x>0;x\ne1\right)\)
a) \(8\sqrt{x+2}\) + \(\sqrt{11-x}\) - \(2\sqrt{22+9x-x^2}\)+ 4 =0
b) \(\sqrt{1+4x}\)+ \(2\sqrt{2-x}\)+\(2\sqrt{\left(1+4x\right)\left(2-x\right)}\)=3
c) \(\sqrt{8+\sqrt{x}}\)+\(\sqrt{5-\sqrt{x}}\)=5
d) \(\sqrt{x^4-1}\)-2 =\(\sqrt{x-1}\)- \(2\sqrt{x^3+x^2+x+1}\)
Rút gọn biểu thức
1) x + 3 + \(\sqrt{x^2-6x+9}\) (x \(\le\) 3)
2) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\) (-2 \(\le\) x \(\le\) 0)
3) \(\sqrt{x^{2^{ }}+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
4) \(\dfrac{\sqrt{x^2-2x+1}}{x-1}\) (x > 1)
5) |x - 2| + \(\dfrac{\sqrt{x^2-4x+4}}{x-2}\) (x < 2)
6) 2x - 1 - \(\dfrac{\sqrt{x^2-10x+25}}{x-5}\)
Tìm x
a. \(\left(\sqrt{2x+17}-\sqrt{2x+1}\right)^2=\frac{16}{x}\)
b. \(7+2\sqrt{x}-x=\left(2+\sqrt{x}\right)\sqrt{7-x}\)
c. \(\frac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\frac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
d. \(x+5+3\sqrt{x-1}=\sqrt{x^2+x-2}+4\sqrt{x+2}\)
e. \(21x-25+2\sqrt{x-2}=19\sqrt{x^2-x-2}+\sqrt{x+1}\)
P=(\(\dfrac{x-2}{x+2\sqrt{x}}\)+\(\dfrac{1}{\sqrt{x+2}}\)).\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với x >0 và x≠1
A=(\(\dfrac{2}{\sqrt{x}-2}\)+\(\dfrac{3}{2\sqrt{x+1}}\)-\(\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\)):\(\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\) với x > 0 và x≠4
A=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x-1}}\)+\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) với x≥0 và x≠1
V=(\(\dfrac{1}{\sqrt{x}+2}\)+\(\dfrac{1}{\sqrt{x}-2}\))\(\dfrac{\sqrt{x}+2}{\sqrt{x}}\) với x>0 và x≠4
A=(\(\dfrac{1}{x-1}\)+\(\dfrac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\))(\(\dfrac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}\)-1)
P=\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\)+\(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}\)+\(\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
MỌI NGƯỜI GIÚP ĐỠ MÌNH RÚT GỌN MẤY BIỂU THỨC NÀY VỚI Ạ . EM XIN CẢM ƠN
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)(\(\dfrac{1}{\sqrt{x}+1}\)-\(\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\)):(\(\dfrac{1}{\sqrt{x}-1}\)-\(\dfrac{2}{x-1}\))