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a: =2015+6-5=2016
b: =10căn 2+5căn 2-6căn 2=9căn 2
c: =3căn 3-4căn 3-5căn 3=-6căn 3
d: =2căn 3+3căn 3-4căn 3=căn 3
\(A=2015+6-5==2015+1=2016\)
\(B=5\sqrt{2^3}+\sqrt{5^2.2}-2\sqrt{3^2.2}\\ =10\sqrt{2}+5\sqrt{2}-6\sqrt{2}\\ =\left(10+5-6\right)\sqrt{2}=9\sqrt{2}\)
\(C=\sqrt{3^3}-2\sqrt{2^2.3}-\sqrt{5^2.3}\\ =3\sqrt{3}-4\sqrt{3}-5\sqrt{3}\\ =\left(3-4-5\right)\sqrt{3}=-6\sqrt{3}\)
\(D=\sqrt{2^2.3}+\sqrt{3^3}-\sqrt{4^2.3}\\ =2\sqrt{3}+3\sqrt{3}-4\sqrt{3}\\ =\left(2+3-4\right)\sqrt{3}=\sqrt{3}\)
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\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=2\sqrt{2}+\sqrt{3}\)
c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=3\sqrt{5}-1\)
d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)
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a: \(A=6\sqrt{3}+10\sqrt{3}-12\sqrt{3}=4\sqrt{3}\)
b: \(B=7\sqrt{3}+5\sqrt{3}-12\sqrt{3}=0\)
c: \(=12\sqrt{2}-6+3\left(9-4\sqrt{2}\right)=12\sqrt{2}-6+27-12\sqrt{2}=21\)
d: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
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Bài 1 :
a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)
Mà \(x^2+1\ge1>0\)
\(\Rightarrow2x+1\ge0\)
\(\Rightarrow x\ge-\dfrac{1}{2}\)
Vậy ...
b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)
\(=-3+4-\left(-4\right)=-3+4+4=5\)
Bài 2 :
\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)
\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)
\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)
\(=3\)
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\(a.\sqrt{72}-5\sqrt{2}+3\sqrt{12}\\ =6\sqrt{2}-5\sqrt{2}+6\sqrt{3}\\ =\sqrt{2}+6\sqrt{3}\\ b.6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\\ =3\sqrt{2}-\sqrt{2}-5\sqrt{2}\\ =-3\sqrt{2}\\ c.\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\\ =2+1+\sqrt{3}-\sqrt{3}\\ =3\\ d.\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\\ =4+3+4\\ =11\)
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a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=-1+3\sqrt{5}\)
b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}=2-\sqrt{3}+1+\sqrt{3}=3\)
a: \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)
\(=3\sqrt{5}-1\)
b: \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}+1\)
=3
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\(E=2\sqrt{3}+3\sqrt{3^3}-\sqrt{100.3}\\ =2\sqrt{3}+9\sqrt{3}-10\sqrt{3}\\ =\left(2+9-10\right)\sqrt{3}=\sqrt{3}\)
\(F=\sqrt{3^2.2}+4\sqrt{18}=\sqrt{18}+4\sqrt{18}=\left(1+4\right)\sqrt{18}=5\sqrt{18}\)
\(G=2\sqrt{3}-4\sqrt{3^3}+5\sqrt{4^2.3}=2\sqrt{3}-12\sqrt{3}+20\sqrt{3}=\left(2-12+20\right)\sqrt{3}=10\sqrt{3}\)
\(H=\left(3\sqrt{25.2}-5\sqrt{9.2}+3\sqrt{2^3}\right)\sqrt{2}\\ =\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\sqrt{2}\\ =6\sqrt{2}.\sqrt{2}=6\)