Rút gọn
\(\dfrac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13+3}}\right):\sqrt{\sqrt{13}+3}}\)
Cho \(x=\frac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13+3}}\right):\sqrt{\sqrt{13}+2}}\)
Tính \(A=x^2+2017x-2018\)
Ta có
\(\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)^2\)
\(=27+10\sqrt{2}+27-10\sqrt{2}-2\sqrt{\left(27+10\sqrt{2}\right)\left(27-10\sqrt{2}\right)}\)
\(=54-2\sqrt{529}=8\)
\(\Rightarrow\) \(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}=\sqrt{8}=2\sqrt{2}\)
Xét tử số
\(\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}\)
\(=\left(\sqrt{27+10\sqrt{2}}.\sqrt{27-10\sqrt{2}}\right)\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)
\(=23\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)
\(=23.2\sqrt{2}=46\sqrt{2}\)
Lại có \(\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2\)
\(=\sqrt{13}-3+\sqrt{13}+3+2\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}+3\right)}\)
\(=2\sqrt{13}+2\sqrt{4}=2\sqrt{13}+4\)
ta bình phương mẫu số
\(\left(\frac{\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}}{\sqrt{\sqrt{13}+2}}\right)^2=\frac{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2}{\sqrt{13}+2}\)
\(=\frac{2\sqrt{13}+4}{\sqrt{13}+2}=2\)
Vậy mẫu \(=\sqrt{2}\)
Vậy \(x=\frac{46\sqrt{2}}{\sqrt{2}}=46\) thay vào ta đc A = 92880
1) Rút gọn:
a) A = \(\sqrt{5-2\sqrt{3-\sqrt{3}}}-\sqrt{3+\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b) B = \(\sqrt{13+\sqrt{2}+5\sqrt{1+2\sqrt{2}}}+\sqrt{13+\sqrt{2}+5\sqrt{1+2\sqrt{2}}}\)
c) C = \(\dfrac{\sqrt{21+3\sqrt{5}}+\sqrt{21-3\sqrt{5}}}{\sqrt{21}+6\sqrt{11}}+\sqrt{11-6\sqrt{2}}\)
d) D = \(\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right).\sqrt{\dfrac{2+2\sqrt{5}}{2+\sqrt{5}}}\)
e) E = \(\dfrac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right):\sqrt{\sqrt{13}+2}}\)
Tinh giá trị biểu thuc \(A=x^2+2016-2017\)
Biết \(x=\frac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right):\sqrt{\sqrt{13}+3}}\)
Tính giá trị A = \(x^2+2002x-2003\)
Với x = \(\frac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right):\sqrt{\sqrt{13}+2}}\)
\(x=\frac{\left(5+\sqrt{2}\right)^2\sqrt{\left(5-\sqrt{2}\right)^2}-\left(5-\sqrt{2}\right)^2\sqrt{\left(5+\sqrt{2}\right)^2}}{\frac{\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}-2\right)}+\sqrt{\left(\sqrt{13}+3\right)\left(\sqrt{13}-2\right)}}{\sqrt{13-4}}}\)
\(=\frac{\left(5+\sqrt{2}\right)\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)-\left(5-\sqrt{2}\right)\left(5-\sqrt{2}\right)\left(5+\sqrt{2}\right)}{\frac{\sqrt{19-5\sqrt{13}}+\sqrt{7+\sqrt{13}}}{3}}\)
\(=\frac{69\left(5+\sqrt{2}-5+\sqrt{2}\right)}{\frac{1}{\sqrt{2}}\left(\sqrt{38-10\sqrt{13}}+\sqrt{14+2\sqrt{13}}\right)}=\frac{276}{\sqrt{\left(5-\sqrt{13}\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}}\)
\(=\frac{276}{5-\sqrt{13}+\sqrt{13}+1}=46\)
\(\Rightarrow A=...\)
Tính giá trị biểu thức: A=\(x^2+2002x-2003\) với x=\(\frac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right):\sqrt{\sqrt{13}+2}}\)
Cho \(P=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
a, Rút gọn P
b, Tính P khi \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
a) Ta có: \(P=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=5+\sqrt{2}-4-\sqrt{2}\)
=1
Thay x=1 vào P, ta được:
\(P=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)
Bài 1: Rút gọn biểu thức
1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\) 2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)
3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\) 4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)
5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) 6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)
7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\) 8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)
9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\) 10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\) 12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
13) \(\sqrt{15-6\sqrt{6}}\) 14) \(\sqrt{8-2\sqrt{15}}\) 15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)
Rút gọn:
\(\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}\)
\(\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\sqrt{2}\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3-\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3-\sqrt{3-2\sqrt{3}+1}}}=\dfrac{\left(6+2\sqrt{5}\right)\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{\sqrt{2}.\sqrt{4-\sqrt{3}}}=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{4-\sqrt{3}}}\)
\(=\dfrac{4\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{4-\sqrt{3}}}\)
Bài 1: Rút gọn biểu thức:
a) \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}+\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}+\sqrt{5}}\right)\)
b) \(\dfrac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\dfrac{1}{3}}\)
c) \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
d) \(\sqrt{\dfrac{3}{4}}+\sqrt{\dfrac{1}{3}}+\sqrt{\dfrac{1}{12}}\)
Bài 2: Giải các phương trình sau:
a) \(x^2+4x+5=2\sqrt{2x+3}\)
b) \(x^2+9x+20=2\sqrt{3x+10}\)
c) \(x^2+7x+14=2\sqrt{x+4}\)
d) \(4\sqrt{x+1}=x^2-5x+14\)
e) \(\sqrt{6-x}=3x-4\)
f) \(\sqrt{5x-9}=9-2x\)
Mọi người làm ơn giúp mình với. Mình đang cần gấp ạ. Cảm ơn mọi người rất nhiều
Bài 1:
a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)
\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)
\(=\dfrac{5}{2}\sqrt{5}\)
c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)
\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)
Bài 2:
e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)
Ta có: \(\sqrt{6-x}=3x-4\)
\(\Leftrightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow9x^2-24x+16+6-x=0\)
\(\Leftrightarrow9x^2-25x+22=0\)
\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)
Vậy: Phương trình vô nghiệm