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g: \(=\left|\sqrt{6}-1\right|=\sqrt{6}-1\)
h: \(=\left|2\sqrt{3}-1\right|=2\sqrt{3}-1\)
l: \(=\left|2-\sqrt{3}\right|-2=2-\sqrt{3}-2=-\sqrt{3}\)
j: \(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|\)
\(=\sqrt{2}-1+2-\sqrt{2}\)
\(=1\)
b) \(\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\left|2\sqrt{6}-3\right|-\left|5-2\sqrt{6}\right|\)
\(=2\sqrt{6}-3-5+2\sqrt{6}\)
\(=4\sqrt{6}-8\)
c) \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|3-\sqrt{6}\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
\(a,\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+2-\sqrt{2}\)
\(=\left|\sqrt{2}-1\right|+2-\sqrt{2}\)
\(=\sqrt{2}-1+2-\sqrt{2}\)
\(=1\)
\(---\)
\(b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}-5+2\sqrt{6}\)
\(=\left|2\sqrt{6}-3\right|-5+2\sqrt{6}\)
\(=2\sqrt{6}-3-5+2\sqrt{6}\)
\(=4\sqrt{6}-8\)
\(---\)
\(c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|\sqrt{6}-3\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
#\(Toru\)
Câu 1:
a: \(\dfrac{2}{5}\sqrt{75}-0,5\cdot\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\cdot\sqrt{12}\)
\(=\dfrac{2}{5}\cdot5\sqrt{3}-0,5\cdot4\sqrt{3}+10\sqrt{3}-\dfrac{2}{3}\cdot2\sqrt{3}\)
\(=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}\)
\(=10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\cdot3\sqrt{3}-2\sqrt{3}}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{9-6}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+3-\sqrt{6}\)
\(=\dfrac{\sqrt{3}}{\sqrt{2}}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
c: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
=\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
Bài 2:
a: 
b: Phương trình hoành độ giao điểm là:
\(3x+2=-x-4\)
=>4x=-6
=>x=-3/2
Thay x=-3/2 vào y=-x-4, ta được:
\(y=-\left(-\dfrac{3}{2}\right)-4=\dfrac{3}{2}-4=-\dfrac{5}{2}\)
Vậy: \(A\left(-\dfrac{3}{2};-\dfrac{5}{2}\right)\)
c: Vì (d2)//(d) nên \(\left\{{}\begin{matrix}a=-1\\b\ne-4\end{matrix}\right.\)
Vậy: (d2): y=-x+b
Thay x=-2 và y=5 vào (d2), ta được:
\(b-\left(-2\right)=5\)
=>b+2=5
=>b=5-2=3
Vậy: (d2): y=-x+3
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(=\left|3+\sqrt{5}\right|+\left|3-\sqrt{5}\right|\)
\(=3+\sqrt{5}+3-\sqrt{5}=6\)
c: \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)
\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)
\(=2\sqrt{3}-3+2\sqrt{3}+3=4\sqrt{3}\)
d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)
\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)
\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)+3}\)
\(=\sqrt{16-3+3}=\sqrt{16}=4\)
e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{3}\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
32, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
=\(\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{33-2.3.2\sqrt{6}}\)
=\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{24-2.3.2\sqrt{6}+9}\)
=\(\left|3-\sqrt{6}\right|+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
=\(3-\sqrt{6}+\left|2\sqrt{6}-3\right|\)=\(3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
33, \(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=\left|\sqrt{5}-1\right|+\sqrt{5}+1=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
34, \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
=\(\sqrt{8-2.\sqrt{3}.\sqrt{5}}-\sqrt{23-2.2.\sqrt{5}.\sqrt{3}}\)
=\(\sqrt{5-2\sqrt{3}.\sqrt{5}+3}-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.\sqrt{3}+3}\)
=\(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}\)
=\(\left|\sqrt{5}-\sqrt{3}\right|-\left|2\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}-2\sqrt{5}+\sqrt{3}=-\sqrt{5}\)
35,\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
=\(\sqrt{16-2.4.\sqrt{15}+15}+\sqrt{15-2.3.\sqrt{15}+9}\)
=\(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
=\(\left|4-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
=\(4-\sqrt{15}+\sqrt{15}-3\)
=1
36, \(\sqrt{49-5\sqrt{96}}-\sqrt{49+5\sqrt{96}}\)
=\(\sqrt{49-2.5.\sqrt{24}}-\sqrt{49+2.5\sqrt{24}}=\sqrt{25-2.5.\sqrt{24}+24}-\sqrt{25+2.5.\sqrt{24}+24}=\sqrt{\left(5-\sqrt{24}\right)^2}-\sqrt{\left(5+\sqrt{24}\right)^2}\)
=\(\left|5-\sqrt{24}\right|-\left|5+\sqrt{24}\right|=5-\sqrt{24}-5-\sqrt{24}=-2\sqrt{24}\)
37, \(\sqrt{3+2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
=\(\left|\sqrt{2}+1\right|+\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{2}+1+\sqrt{3}-\sqrt{2}=\sqrt{3}+1\)
a: Ta có: \(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1-2-\sqrt{3}\)
=-3
b: Ta có: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
=1
c: Ta có: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
