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\(a,=3\sqrt{5}-2\sqrt{5}-\sqrt{5}+5\sqrt{5}=5\sqrt{5}\\ b,=9\sqrt{a}-6\sqrt{a}-\sqrt{a}=2\sqrt{a}\\ c,Sửa:3\sqrt[3]{27}-3\sqrt[3]{-8}-3\sqrt[3]{-125}\\ =3\cdot3-3\left(-2\right)-3\left(-5\right)\\ =9+6+15=30\)
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 :
\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)
\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)
\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)
\(=3\)
\(A=\sqrt{5}.\left(\sqrt{20}-3\right)+\sqrt{45}.\)
\(=\sqrt{5}.\left(\sqrt{4.5}-3\right)+\sqrt{9.5}\)
\(=\sqrt{5}.\left(2\sqrt{5}-3\right)+3\sqrt{5}\)
\(=\sqrt{5}.2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\)
\(=2.\sqrt{5}^2=2.5=10\)
\(a,=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)
\(b,=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}=-6\sqrt{3}\)
\(c,=3\sqrt{3}+7\sqrt{3}-9\sqrt{3}+11\sqrt{3}=12\sqrt{3}\)
a) Ta có: \(-\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{5}\sqrt{125}\)
\(=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{1}{5}\cdot5\sqrt{5}\)
\(=-17\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)
b) Ta có: \(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\)
\(=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\)
\(=-6\sqrt{3}\)
Lời giải:
a.
$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$
b.
$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$
\(I=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-5\sqrt{3}.\sqrt{3^2}+2\sqrt{2^2}.\sqrt{3}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)
\(=-5\sqrt{3}.\dfrac{1}{\sqrt{3}}\)
\(=-5\)
\(K=\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)
\(=\sqrt{5^2.5}-4\sqrt{3^2.5}+3\sqrt{2^2.5}-\sqrt{4^2.5}\)
\(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)
\(=\sqrt{5}.\left(5-12+6-4\right)\)
\(=-5\sqrt{5}\)
\(L=2\sqrt{9}+\sqrt{25}-5\sqrt{4}\)
\(=2\sqrt{3^2}+\sqrt{5^2}-5\sqrt{2^2}\)
\(=2.3+5-5.2\)
\(=1\)
\(N=2\sqrt{32}-5\sqrt{27}-4\sqrt{8}+3\sqrt{75}\)
\(=2.4\sqrt{2}-5.3\sqrt{3}-4.2\sqrt{2}+3.5\sqrt{3}\)
\(=8\sqrt{2}-8\sqrt{2}-15\sqrt{3}+15\sqrt{3}\)
\(=0\)
\(O=2\sqrt{3.5^2}-3\sqrt{3.2^2}+\sqrt{3.3^2}\)
\(=2.5\sqrt{3}-3.2\sqrt{3}+3\sqrt{3}\)
\(=10\sqrt{3}-6\sqrt{3}+3\sqrt{3}\)
\(=7\sqrt{3}\)
\(L=\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}=2-15+8=-5\)
\(K=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)
L=2*3+5-5*2=5-4=1
N=8căn 2-8căn2-15căn3+15căn 3=0
O=10căn 3-6căn3+3căn3=7căn 3
\(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=\sqrt{4.5}-\sqrt{9.5}+3\sqrt{18}+\sqrt{4.18}\)
\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+2\sqrt{18}\)
\(=-\sqrt{5}+5\sqrt{18}\)
a)
\(7\sqrt{12}+\frac{1}{3}\sqrt{27}-\sqrt{75}\)
\(=14\sqrt{3}+\sqrt{3}-5\sqrt{3}\)
\(=10\sqrt{3}\)
b)
\(\left(2\sqrt{20}+\sqrt{125}-3\sqrt{80}\right):5\)
\(=\left(4\sqrt{5}+5\sqrt{5}-12\sqrt{5}\right):5\)
\(=-3\sqrt{5}:5\)
\(=\frac{-3\sqrt{5}}{5}\)
c)
\(3\sqrt{12a}-5\sqrt{3a}+\sqrt{48a}\)
\(=6\sqrt{3a}-5\sqrt{3a}+4\sqrt{3a}\)
\(=5\sqrt{3a}\)