\(\sqrt{32}-\sqrt{50}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
tính
1/\(2\sqrt{20}-\sqrt{50}+3\sqrt{80}\)\(-\sqrt{320}\)
2/\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(1,4\sqrt{5}-5\sqrt{2}+12\sqrt{5}-8\sqrt{5}=8\sqrt{5}-5\sqrt{2}\)
\(2,\left(\sqrt{27}+\sqrt{32}-\sqrt{50}\right)\left(\sqrt{27}-\sqrt{32}+\sqrt{50}\right)\)
\(=27-\left(\sqrt{32}-\sqrt{50}\right)^2=27-2=25\)
Rút gọn các biểu thức:
1. A=\(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
2. B= \(\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
3. C= \(\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)\)
4. D= \(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
5. E= \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4\)
6. F= \(\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}\)
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
Bài 1: Tính và rút gọn biểu thức:
\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)\)
\(B=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(C=1-\left(\sqrt{45}-\sqrt{20}-\sqrt{3}\right)\left(\sqrt{20}-\sqrt{45}-\sqrt{3}\right)\)
\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right):\frac{1}{\sqrt{6}}\)
\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)=5\sqrt{5}-5\sqrt{3}+15-3\sqrt{15}\)
Bạn ghi nhầm đề thì phải, ngoặc đầu là \(\sqrt{5}+\sqrt{3}\) mới rút gọn được theo HĐT số 3
\(B=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)
\(C=1-\left(3\sqrt{5}-2\sqrt{5}-\sqrt{3}\right)\left(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{5}-\sqrt{3}\right)\left(-\sqrt{5}-\sqrt{3}\right)=1+\left(5-3\right)=3\)
\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right).\sqrt{6}=\frac{\left(3-2\right)}{\sqrt{6}}.\sqrt{6}=1\)
Tính:
\(\left(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\right)+\left(\dfrac{1}{3+2\sqrt{2}}-\dfrac{1}{3-2\sqrt{2}}\right)\)
\(\left(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\right)+\left(\dfrac{1}{3+2\sqrt{2}}-\dfrac{1}{3-2\sqrt{2}}\right)\)
\(=\left(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\right)+\left(3+2\sqrt{2}-\left(3+2\sqrt{2}\right)\right)\)
\(=\left(9\sqrt{2}\right)+\left(3-2\sqrt{2}-3-2\sqrt{2}\right)\)
\(=9\sqrt{2}+\left(-4\sqrt{2}\right)\)
\(=9\sqrt{2}-4\sqrt{2}\)
\(=5\sqrt{2}\)
\(\left(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\right)+\left(\dfrac{1}{3+2\sqrt{2}}-\dfrac{1}{3-2\sqrt{2}}\right)\)
= \(\left(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\right)+\left(\dfrac{3-2\sqrt{2}-3-2\sqrt{2}}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\right)\)
= \(9\sqrt{2}+\left(\dfrac{-4\sqrt{2}}{1}\right)\) = \(9\sqrt{2}-4\sqrt{2}\) = \(5\sqrt{2}\)
Tính:
\(\left(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\right)+\left(\dfrac{1}{3+2\sqrt{2}}-\dfrac{1}{3-2\sqrt{2}}\right)\)
= \(\left(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\right)\left[3-2\sqrt{2}-\left(3+2\sqrt{2}\right)\right]\)
= \(9\sqrt{2}\left(3-2\sqrt{2}-3-2\sqrt{2}\right)\)
= \(9\sqrt{2}\left(-4\sqrt{2}\right)\)
= \(-9.2.4\)
= \(-72\)
\(\left(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\right)+\left(\dfrac{1}{3+2\sqrt{2}}-\dfrac{1}{3-2\sqrt{2}}\right)\)
\(=\left(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\right)+\left[3-2\sqrt{2}-\left(3+2\sqrt{2}\right)\right]\)
\(=\left(9\sqrt{2}\right)+\left(3-2\sqrt{2}-3-2\sqrt{2}\right)\)
\(=9\sqrt{2}+\left(-4\sqrt{2}\right)\)
\(=9\sqrt{2}-4\sqrt{2}\)
\(=5\sqrt{2}\)
a,(\(\left(\sqrt{50}-\sqrt{32}+\sqrt{8}\right):\sqrt{2}\) b,\(\dfrac{4}{\sqrt{5}-1}-5\sqrt{\dfrac{1}{5}}\)
a: \(\dfrac{\sqrt{50}-\sqrt{32}+\sqrt{8}}{\sqrt{2}}\)
\(=\dfrac{5\sqrt{2}-4\sqrt{2}+2\sqrt{2}}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{\sqrt{2}}=3\)
b: \(\dfrac{4}{\sqrt{5}-1}-5\sqrt{\dfrac{1}{5}}\)
\(=\dfrac{4\left(\sqrt{5}+1\right)}{5-1}-\sqrt{5}\)
\(=\sqrt{5}+1-\sqrt{5}\)
=1
tính
A=\(\left(1-\sqrt{7}\right).\dfrac{\sqrt{7}+7}{2\sqrt{7}}\)
B=\(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
C=\(\sqrt{32}-\sqrt{50}+\sqrt{18}\)
D=\(\sqrt{72}+\sqrt{4\dfrac{1}{2}}-\sqrt{32}-\sqrt{162}\)
E=\(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
a
\(\sqrt{32}\)+\(\sqrt{50}\) - 2\(\sqrt{200}\) + 3\(\sqrt{72}\)
b)\(\dfrac{3}{\sqrt{ }2-1}\) + \(\sqrt{\left(3-\sqrt{2}\right)^{^2}}\) - 2\(\sqrt{2}\)
rút gọn các biểu thức trên
\(a.4\sqrt{2}+5\sqrt{2}-20\sqrt{2}+18\sqrt{2}=7\sqrt{2}\)
\(a,=4\sqrt{2}+5\sqrt{2}-20\sqrt{2}+18\sqrt{2}=7\sqrt{2}\\ b,=\dfrac{3\left(\sqrt{2}+1\right)}{1}+\left|3-\sqrt{2}\right|-2\sqrt{2}\\ =3\sqrt{2}+3+3-\sqrt{2}-2\sqrt{2}=6\)
`a)`
`\sqrt{32} + \sqrt{50} - 2\sqrt{200} + 3\sqrt{72}`
`= 4\sqrt{2} + 5\sqrt{2} - 20\sqrt{2} + 18\sqrt{2}`
`= (4 + 5 - 20 + 18) . \sqrt{2}`
`= 7\sqrt{2}`
`b)`
`3/(\sqrt{2} - 1) + \sqrt{(3 - \sqrt{2})^2} - 2\sqrt{2}`
`= (3 . (\sqrt{2} + 1))/1 + |3 - \sqrt{2}| - 2\sqrt{2}`
`= 3\sqrt{2} + 3 + 3 - \sqrt{2} - 2\sqrt{2}`
`= (3 - 1 - 2) . \sqrt{2} + 6`
`= 6`