a,Tính giá trị của biểu thức:
N=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
a) Tính giá trị biểu thức:
N=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
b)Rút gọn biểu thức:
A=\(\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}+2}\),trị x>2
\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3-2\sqrt{2}}+\sqrt{9+4\sqrt{2}}+\sqrt{12-8\sqrt{2}}}\)
CÁC BẠN GIÚP MÌNH TÌM GIÁ TRỊ NHÉ
cho P = \(\frac{\sqrt{x}+2}{\sqrt{x}+1}\) , Tìm GTLN của P
ĐKXĐ: \(x\ge0\)
Ta có: \(P=\frac{\sqrt{x}+2}{\sqrt{x}+1}=1+\frac{1}{\sqrt{x}+1}\)
Để P lớn nhất thì: \(\frac{1}{\sqrt{x}+1}\)phải lớn nhất.Hay: \(\sqrt{x}+1\)nhỏ nhất
Theo ĐKXĐ,lại có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\)
=>Min \(\sqrt{x}+1\)là 1 tại \(\sqrt{x}=0\Rightarrow x=0\)
=>Max P = \(1+\frac{1}{0+1}=2\)tại x=0
=.= hk tốt!!
Tính \(E=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
Tính: \(N=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
Trục căn thức ở mẫu và rút gọn:
a) \(\frac{20}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
b) \(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2.\sqrt{3+2\sqrt{5}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
\(C=\dfrac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
\(B=\dfrac{20}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
\(C=\dfrac{\sqrt{10}-\sqrt{5}+2\sqrt{2}+\sqrt{5}-\sqrt{10}-1}{2\sqrt{2}+2+2\sqrt{2}-1+2\sqrt{2}+2}\)
\(=\dfrac{2\sqrt{2}-1}{6\sqrt{2}+3}=\dfrac{9-4\sqrt{2}}{21}\)
\(B=\dfrac{40}{6+2\sqrt{5}+\sqrt{4\sqrt{5}+4}}\)
\(=\dfrac{40}{\left(\sqrt{5}+1\right)^2+2\sqrt{\sqrt{5}+1}}\)
\(=\dfrac{40}{\sqrt{\sqrt{5}+1}\left(\sqrt{\sqrt{5}+1}+2\right)}\)
\(=\dfrac{40\sqrt{\sqrt{5}-1}}{2\left(\sqrt{\sqrt{5}+1}+2\right)}\)
\(=\dfrac{20\left(\sqrt{\sqrt{5}-1}\right)\left(\sqrt{\sqrt{5}+1}-2\right)}{\sqrt{5}+1-4}\)
\(=\dfrac{20\left(\sqrt{\sqrt{5}-1}\right)\left(\sqrt{\sqrt{5}+1}-2\right)}{-3+\sqrt{5}}\)
\(=-5\left(3+\sqrt{5}\right)\left(\sqrt{\sqrt{5}-1}\right)\left(\sqrt{\sqrt{5}+1}-2\right)\)
Tính
A=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
B=\(\left(3-\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\cdot\sqrt{3-\sqrt{5}}\)
C=\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{ }}3}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
E=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{5}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
Tính giá trị của biểu thức:
\(\sqrt{2+\sqrt{3+\sqrt[3]{4+5\sqrt[5]{6+7\sqrt[7]{8+9\sqrt[9]{10+11\sqrt[11]{12}}}}}}}\)
Tính giá trị các biểu thức:
a.\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\sqrt{3}\)
b.\(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
c.\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)
d.\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)
\(=19\sqrt{3\sqrt{2}}\)