k,\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)\) l,\(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)
thực hiện phép tính
a)\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\frac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
b)\(2\sqrt{2}\left(2-3\sqrt{3}\right)+\left(1-2\sqrt{2}\right)^2+6\sqrt{6}\)
c)\(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)
Thực hiện phép tính a)\(2\sqrt{2}.\left(2-2\sqrt{ }2\right)+1-\left(2\sqrt{ }2\right)^3+6\sqrt{3}\) b) \(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}:\sqrt{ }15\) c). \(\left(\right)\)
thực hiện phép tính
a)(\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot\sqrt{8}\right)3\sqrt{6}\)
b)\(2\sqrt{2}\left(2-3\sqrt{3}\right)+\left(1-2\sqrt{2}\right)+6\sqrt{67}\)
c)\(\left(20\sqrt{30}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)
a,
\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)
\(=6\sqrt{6}^2-12\sqrt{9.2}+15\sqrt{4.3}-\dfrac{3}{4}\sqrt{16.3}\)
= \(6.6-12.3\sqrt{2}+15.2\sqrt{3}-\dfrac{3}{4}.4\sqrt{3}\)
\(=36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}\)
\(=36-36\sqrt{2}+27\sqrt{3}\)
Thực hiện phép tính sau : \(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)
Ta có : \(\frac{20\sqrt{300}-15\sqrt{675}+5\sqrt{75}}{\sqrt{15}}=\frac{\sqrt{15}\left(20\sqrt{20}-15\sqrt{45}+5\sqrt{5}\right)}{\sqrt{15}}\)
\(=20\sqrt{20}-15\sqrt{45}+5\sqrt{5}=40\sqrt{5}-45\sqrt{5}+5\sqrt{5}=0\)
a \(\left(2\sqrt{6-}4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)5\sqrt{6}\)
b \(\left(\dfrac{1}{\sqrt{5-}\sqrt{2}}+\dfrac{1}{\sqrt{5}+\sqrt{2}}+1\right).\dfrac{1}{\sqrt{2}+1^2}\)
c \(\dfrac{3}{\sqrt{2+}\sqrt{7}}\dfrac{1}{\sqrt{2-}\sqrt{7}}-1\)
d \(2x\sqrt{300}-15\sqrt{75}+5\sqrt{75}.15\)
a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)
\(=60-60\sqrt{2}+45\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)
Rút gọn \(A=\left(\sqrt{6+\sqrt{20}}-2\sqrt{3-\sqrt{5}}+\sqrt{15-10\sqrt{2}}\right):\left(2+\sqrt{8}\right)\)
\(A=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{2}.\sqrt{6-2\sqrt{5}}+\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{2}\left(\sqrt{5}-1\right)+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{10}+\sqrt{2}+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}=\dfrac{1}{2}\)
Bài 1: Tính
a) \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
b) \(1\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
c) \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
d) \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
Bài 2: Cho (d₁): y = \(\dfrac{1}{2}x-4\) và (d₂): y = \(-3x+3\) . Vẽ (d₁) và (d₂) trên cùng một hệ trục tọa độ. Tìm tọa độ giao điểm A của 2 đường thẳng trên.
Helpp!!
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
Tính (rút gọn)
a) \(\left(\sqrt{3+\sqrt{5}}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-}\sqrt{15}\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3+2}}\)
d) \(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
f)\(\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\)
g)\(\dfrac{\sqrt{2+\sqrt{3}}}{3+\sqrt{3}}\)
h)\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
\(d,\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\\ =\left(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{4+\sqrt{5}-1}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\sqrt{24+8\sqrt{5}}\left(\sqrt{5}-1\right)\\ =\sqrt{\left(2\sqrt{5}+2\right)^2}\left(\sqrt{5}-1\right)\\ =2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\\ =2\left(5-1\right)\\ =8\)
tính
1.\(\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}\)
2.\(3\sqrt{2}\left(4-\sqrt{2}\right)+3\left(1-2\sqrt{2}\right)^2\)
3.\(\dfrac{1}{2}\left(\sqrt{6}+\sqrt{5}\right)^2-\dfrac{1}{4}\sqrt{120}-\sqrt{\dfrac{15}{2}}\)
4.\(\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
5.\(\left(\sqrt{\sqrt{14}+\sqrt{5}}+\sqrt{\sqrt{14}-\sqrt{5}}\right)^2\)
6.\(\left(\sqrt{3}+1\right)^3-\left(\sqrt{3}-1\right)^3\)
7.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
8.\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
9.\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)