\(\sqrt{\left(x-2\right)^2}=5\)
=> \(\left|x-2\right|=5\)
=> \(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=5+2=7\\x=-5+2=-3\end{matrix}\right.\)
Vậy \(x\in\left\{7;-3\right\}\)
\(\sqrt{\left(x-2\right)^2}=5\)
=> \(\left|x-2\right|=5\)
=> \(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=5+2=7\\x=-5+2=-3\end{matrix}\right.\)
Vậy \(x\in\left\{7;-3\right\}\)
Rút gọn biểu thức:
1) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+3\right)^2}\)
2) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
3) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)
4) \(4x+\sqrt{\left(x-12\right)^2}\left(x\ge2\right)\)
5) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
6) \(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)^2}\left(x\ge2y\right)\)
1) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
2) \(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)^2\left(x\ge2y\right)}\)
3) 4x + \(\sqrt{\left(x-12\right)^2}\left(x\ge2\right)\)
Tính DKXD của các căn bậc thức sau:
a)\(\sqrt{2x-4}\)
b)\(\sqrt{\dfrac{3}{-2x+1}}\)
c)\(\sqrt{\dfrac{-3x+5}{-4}}\)
d)\(\sqrt{-5\left(-2x+6\right)}\)
e)\(\sqrt{\left(x^2+2\right)\left(x-3\right)}\)
f)\(\sqrt{\dfrac{x^2+5}{-x+2}}\)
Bài 1: Rút gọn
\(3\sqrt{9a^6}-6a^3\) (với mọi a)
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(1-3x\right)^2}\) (Với \(\dfrac{1}{3}\) < x ≤ 1 )
\(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\)
\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
\(\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\) (với 1<x<2)
\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\) (với x ≥4)
\(\sqrt[3]{x^2-1}+\sqrt{3x^3-2}=3x-2\)
\(\sqrt{\left(2-x\right)\left(5-x\right)}=x+\sqrt{\left(2-x\right)\left(10-x\right)}\)
tìm giá trị nhỏ nhất:
\(a,\sqrt{\left(x-2000\right)^2}+\sqrt{\left(x-2001\right)^2}\)
\(b,\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+4\right)^2}+\sqrt{\left(x+5\right)^2}\)
\(c,\sqrt{\left(2x-1\right)^2}+\sqrt{\left(3x+4\right)^2}\)
1/ Cho biểu thức
\(P=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)với x>0, x\(\ne\)1
a) CMR: P=\(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) Tìm các giá trị của x để 2P=\(2\sqrt{5}+5\)
2/ Thu gọn biểu thức sau:
A= \(\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)
B= \(\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
giúp mình với ạ
1/ Cho biểu thức
\(P=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)với x>0, x\(\ne\)1
a) CMR: P=\(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) Tìm các giá trị của x để 2P=\(2\sqrt{5}+5\)
2/ Thu gọn biểu thức sau:
A= \(\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)
B= \(\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
giúp mình với ạ
Tính
A = \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
B = \(\sqrt{\left(4-\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
C = \(\sqrt{\left(1-\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
Giải phương trình:
a) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
b) \(\sqrt{x+1}+2\left(x+1\right)=x-1+\sqrt{1-x}+3\sqrt{1-x^2}\)
c) \(\left(\sqrt{1+x}-1\right)\left(\sqrt{1-x}+1\right)=2x\)
d) \(2\left(x^2+2x+3\right)=5\sqrt{\left(x+1\right)^3}\)
(Các bạn giúp mình sớm nhé! Thank!)