Tính: \(A=\dfrac{\sqrt{6+2\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}-\sqrt{6-2\left(\sqrt{6}-\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}\)
Tính
a)\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}-\dfrac{6}{\sqrt{ }6}\)
b) \(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)
a) \(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}-\dfrac{6}{\sqrt{6}}=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}-\dfrac{6}{\sqrt{6}}\)
\(=\dfrac{1}{6\sqrt{6}}-\dfrac{6}{\sqrt{6}}=-\dfrac{35}{6\sqrt{6}}\)
b)\(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)
\(=6+2\sqrt{30}+5+6-2\sqrt{30}+5=22\)
\(a:\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)
b : \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)
c : \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right).\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)
d : \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)
\(=\left|\sqrt{3}-2\right|+\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{3}\)
b) \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{2}\)
\(=\sqrt{2}-\sqrt{2}\)
\(=0\)
c) \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\cdot\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)
\(=\left[2-\dfrac{\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\right]\cdot\left[2-\dfrac{\sqrt{5}\left(3-\sqrt{5}\right)}{3-\sqrt{5}}\right]\)
\(=\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)\)
\(=4-4\sqrt{5}+5\)
\(=9-4\sqrt{5}\)
d) \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=6-121\)
\(=-115\)
Tính :
a) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\)
b) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21+12\sqrt{3}}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
d) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
e) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
f) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
g) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)-\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
a : \(\sqrt{5+2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
b : \(\dfrac{\sqrt{7}-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
a: \(\sqrt{5+2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\left|\sqrt{2}-\sqrt{3}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
b: Sửa đề: \(\dfrac{7-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
\(=\dfrac{\sqrt{7}\left(\sqrt{7}-2\right)}{-\left(\sqrt{7}-2\right)}+\dfrac{6\left(\sqrt{7}-1\right)}{6}+18-12\)
\(=-\sqrt{7}+\sqrt{7}-1+6=5\)
Tính a=\(\dfrac{\sqrt[3]{10+6\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-5}\)
b, a= \(\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\) CMR \(\dfrac{64}{\left(a^2-3\right)^3}-3a\) ∈ Z
a: Sửa đề: căn 6+2căn 5-căn 5
\(a=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}=\dfrac{2}{1}=2\)
b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)
=>a^3-3a-4=0
=>a^3-3a=4
\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)
\(=\left(\dfrac{a^3-3a}{a^2-3}\right)^3-3a=a^3-3a\)
=4
RÚT GỌN BIỂU THỨC
A= \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\)\(\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
B= \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\)\(\left(\sqrt{6}+11\right)\)
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
1)\(\dfrac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)
2)\(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)
3)\(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)
2) Ta có: \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)
\(=3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)-\sqrt{6}\)
\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)
\(=-11\)
3) Ta có: \(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)
\(=\left(\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}\right)\left(4-2\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{5}\right)\left(4-2\sqrt{3}\right)\)
\(=4\sqrt{6}-6\sqrt{2}+4\sqrt{5}-2\sqrt{15}\)
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)
\(3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3\)
Xét \(a\ge0\Rightarrow\) biểu thức \(=9a^3-6a^3=3a^3\)
Xét \(a< 0\Rightarrow\) biểu thức \(=-9a^3-6a^3=-15a^3\)
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(1-3x\right)^2}=\left|x-1\right|+\left|1-3x\right|\)
\(=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x\)
\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2\)
\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16\)
\(\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5\)
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|\)
Xét \(x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow\)biểu thức \(=\sqrt{x-4}+2+\sqrt{x-4}-2\)
\(=2\sqrt{x-4}\)
Xét \(x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow\) biểu thức \(=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
thực hiện phép tính ( rút gọn biểu thức )
a) \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{3\sqrt{6}}{\sqrt{2}}+\dfrac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
b) \(\left(\dfrac{2-2\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{6}-3}{\sqrt{3}-\sqrt{2}}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
a: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-3\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)
\(=\sqrt{3}-3\sqrt{3}+\sqrt{3}=-\sqrt{3}\)
b: \(=\left(\left(2-2\sqrt{5}\right)\left(\sqrt{5}+2\right)+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(2\sqrt{5}+4-10-4\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(-2\sqrt{5}+\sqrt{3}-6\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=-20+2\sqrt{15}+\sqrt{15}-3-6\sqrt{5}+6\sqrt{3}\)
\(=-23+3\sqrt{15}-6\sqrt{5}+6\sqrt{3}\)
* Tính
a. A=\(\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
b. B=\(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)
=1
b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+1\)
=1