giúp mk bài này để mk ôn thi hsg vs nha
Cho \(x=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{20}}\)
Tính \(P=\left(1+x^5-x^7\right)^{2016}\)
Bài 1 Tính
1)\(\sqrt{\left(-0,3\right)^2}\)
2)\(-\frac{1}{2}\sqrt{\left(0,3\right)^2}\)
3)\(\sqrt{a^{10}}\)(a>0)
4)\(\sqrt{\left(2-x^2\right)}\)(x<2)
5)\(\sqrt{x^2+2x+1}\)
6)\(\sqrt{\left(1-\sqrt{2}\right)^2}\)
7)\(\sqrt{11+6\sqrt{2}}\)
8)\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}\)
9)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
Làm hộ e, mk vs ạ e, mk xin cảm ơn bn, a, cj ạ
\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)
\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)
\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)
\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)
\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))
\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)
\(=-2\)
\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
\(\sqrt{\left(-0,3\right)^2}\)
\(=\left|0,3\right|\)
\(=0,3\)
Giúp mình làm bài này với
Bài 1: Tính
A=\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
B=\(\frac{\sqrt{2+\sqrt{3}}}{2}\div\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
C=\(\frac{1}{\sqrt{2}+2}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+...+\frac{1}{99\sqrt{100}+100\sqrt{99}}\)
Bài 2: Giải phương trình:
a. \(\sqrt{4x-20}+\sqrt{x-5}-\frac{1}{3}\sqrt{9x-45}=4\)
b.\(\frac{2\sqrt{x}-7}{3}=\sqrt{x}-1\)
c.\(5\sqrt{x-1}-\sqrt{36x-36}-\sqrt{9x-9}=\sqrt{8x+12}\)
Bài 3: Rút gọn
\(M=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\times\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a-1}}\right)\)
a. Tìm a để M>0
b. Tìm a để M<0
Bài 1:
Cho \(x=\frac{\sqrt{\left(4+2\sqrt{3}\right)}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}-2}\)
Tính \(P=\left(x^2+x+1\right)^{2013}+\left(x^2+x-1\right)^{2013}\)
Giải giúp mk vs ạ thanks m.n nhìu nà !!!!! :)))
Ta có
\(x=\frac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}-2}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.4.\sqrt{5}-8}-2}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\frac{1}{5-4-2}=-1\)
Thế vào ta được
\(P=\left(x^2+x+1\right)^{2013}+\left(x^2+x-1\right)^{2013}\)
\(=\left(1-1+1\right)^{2013}+\left(1-1-1\right)^{2013}=1-1=0\)
Tìm điều kiện để căn thức có nghĩa:
1,\(\sqrt{x-2\sqrt{x-1}}\)
2,\(\sqrt{x-1-2\sqrt{x-2}}\)
3,\(\sqrt{\left|2m+1\right|-1}\)
4,\(\sqrt{2\sqrt{2-\left|3n-\sqrt{2}\right|}}\)
5,\(\frac{\sqrt{5x^2-4x-1}}{\left|x^2-3x\right|+\left|x^2-9\right|}\)
6,\(\frac{\sqrt{x^2-x-6}}{\left|x^2-3x\right|+\left|x^2-9\right|}\)
Ai giải giúp mk vs đi ạ!!! Mk camon m.n nhìu :)))
cho P=\(\frac{x\sqrt{x}+5\sqrt{x}-12}{x-\sqrt{5}-6}-\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+2}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\)
a, Nêu ĐKXĐ và RGBT
b, Tìm MinP
mn lm giúp mk vs
a, \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}=4+\sqrt{11}-3\sqrt{7}\)
b, \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
Các bạn giúp mk vs, Mk cần gấp.
a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
Tính A=\(\left(\frac{2}{\sqrt{5}-3}-\frac{2}{\sqrt{5}+3}\right)×\frac{\sqrt{3}-3}{1-\sqrt{3}}+3\sqrt{27}\)
B=\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)×\left(11+\sqrt{6}\right)\)
Tìm x để E=\(\sqrt{x-5}+\sqrt{7}\)nhỏ nhất
Tìm x để F=\(\frac{4-\sqrt{x}}{\sqrt{x}+2}\)lớn nhất
Giải giúp mik vs đap cần gấp . Cảm ơn mn. Giải cho mik bài 1 cx đc
1/ Rút gọn
A=\(\sqrt{7}-4\sqrt{3}+\sqrt{4}-2\sqrt{3}\)
B=\(\left(2+\frac{5-\sqrt{5}}{\sqrt{5}-1}\right)\) \(\left(2-\frac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
C=\(\left(\sqrt{3}+1\right)\) \(\left(\frac{\sqrt{14}-6\sqrt{3}}{5+\sqrt{3}}\right)\)
2/Cho P=\(\left(\sqrt{x-\frac{1}{\sqrt{x}}}\right)\):\(\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
a/ cmr: P>0, V x >0, x\(\ne\)1
b/Tính GT P khi x\(\frac{2}{2+\sqrt{3}}\)
\(A=\left(\sqrt{x}+2\right):\left(\frac{x+8}{x\sqrt{x}+8}+\frac{\sqrt{x}}{x-2\sqrt{x}+4}-\frac{1}{2+\sqrt{x}}\right)ĐK:x\ge0.\)
Giúp mk bài này vs làm ko ra
\(A=\left(\sqrt{x}+2\right):\left(\frac{x+8}{x\sqrt{x}+8}+\frac{\sqrt{x}}{x-2\sqrt{x}+4}-\frac{1}{2+\sqrt{x}}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+\sqrt{x}\left(\sqrt{x}+2\right)-\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+8+x+2\sqrt{x}-x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left(\frac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right)\)
\(=\left(\sqrt{x}+2\right):\left[\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\right]\)
\(=\left(\sqrt{x}+2\right):\frac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}+2}\)
\(=x-2\sqrt{x}+4\)
=.= hok tốt!!