Các biểu thức dưới dấu căn đều dương

Đat  \(\sqrt{x^2-6x+19}=a\ge0,\sqrt{x^2-6x+10}=b\ge0\)

Ta có  \(a-b=3\)và \(a^2-b^2=9\)

\(\Rightarrow a+b=9\)

Do \(a+b>a-b\) nên  \(b>0\)\(\Leftrightarrow a>0\)

Vậy giá trị của biểu thức A  = 9

1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)

\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)

\(=6\sqrt{3}+2\sqrt{3}-3\sqrt{3}\)

\(=5\sqrt{3}\)

2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)

\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)

\(=\dfrac{-\sqrt{3}-5}{11}\)

3) Ta có: \(\sqrt{\dfrac{2}{5}}\)

\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)

\(=\dfrac{\sqrt{10}}{5}\)

a)\(\frac{21}{\sqrt{14}}\)=\(\frac{21.\sqrt{14}}{14}\)=\(\frac{3\sqrt{14}}{2}\)

b)\(\frac{3}{\sqrt{2}}+\frac{\sqrt{2}}{3}=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{3}=\frac{9\sqrt{2}}{6}+\frac{2\sqrt{2}}{6}=\frac{11\sqrt{2}}{6}\)

c)=\(-46\sqrt{5}\)

a) Ta có:

5√15+12√20+√5515+1220+5

=√52.15+√(12)2.20+√5=√25.15+√14.20+√5=√255+√204+√5=√5+√5+√5=(1+1+1)√5=3√5=52.15+(12)2.20+5=25.15+14.20+5=255+204+5=5+5+5=(1+1+1)5=35

b)  Ta có: 

√12+√4,5+√12,512+4,5+12,5

=√12+√92+√252=√12+√9.12+√25.12=√12+√32.12+√52.12=√12+3√12+5√12=(1+3+5).√12=9√12=91√2=9.√22=9√22=12+92+252=12+9.12+25.12=12+32.12+52.12=12+312+512=(1+3+5).12=912=912=9.22=922

c) Ta có:

√20−√45+3√18+√72=√4.5−√9.5+3√9.2+√36.2=√22.5−√32.5+3√32.2+√62.2=2√5−3√5+3.3√2+6√2=2√5−3√5+9√2+6√2=(2√5−3√5)+(9√2+6√2)=(2−3)√5+(9+6)√2=−√5+15√2=15√2−√520−45+318+72=4.5−9.5+39.2+36.2=22.5−32.5+332.2+62.2=25−35+3.32+62=25−35+92+62=(25−35)+(92+62)=(2−3)5+(9+6)2=−5+152=152−5

d) Ta có:

0,1√200+2√0,08+0,4.√50=0,1√100.2+2√0,04.2+0,4√25.2=0,1√102.2+2√0,22.2+0,4√52.2=0,1.10√2+2.0,2√2+0,4.5√2=1√2+0,4√2+2√2=(1+0,4+2)√2=3,4√2

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1) \(=\frac{2\sqrt{3}}{\sqrt{20}}+\frac{1}{\sqrt{60}}-\frac{1}{\sqrt{15}}=\frac{6\sqrt{60}+\sqrt{60}-4\sqrt{15}}{60}=\frac{\sqrt{15}\left(12+2-4\right)}{60}=\frac{\sqrt{15}}{6}\)

a) \(=\frac{9}{\sqrt{3}}=\frac{9\sqrt{3}}{3}\)

b) \(=\frac{12\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\frac{36+12\sqrt{3}}{9-3}=6+2\sqrt{3}\)

c) \(=\frac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{2+2\sqrt{2}+1}{2-1}=3+2\sqrt{2}\)

d) \(=\frac{\left(7\sqrt{3}-5\sqrt{11}\right)\left(8\sqrt{3}+7\sqrt{11}\right)}{\left(8\sqrt{3}-7\sqrt{11}\right)\left(8\sqrt{3}+7\sqrt{11}\right)}=\frac{217-9\sqrt{11}}{347}\)

e) \(=\frac{\left(1-a\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\frac{1+\sqrt{a}-a\sqrt{a}-a^2}{1-a}=a+\sqrt{a}+1\)

f) \(=\frac{1}{3\sqrt{2}-2\sqrt{2}+\sqrt{8}}=\frac{\sqrt{2}-\sqrt{8}}{\left(\sqrt{2}+\sqrt{8}\right)\left(\sqrt{2}-\sqrt{8}\right)}=\frac{\sqrt{2}}{6}\)

g) \(=\frac{1-\sqrt{2}+\sqrt{3}}{1-\left(\sqrt{2}-\sqrt{3}\right)^2}=\frac{1-\sqrt{2}+\sqrt{3}}{2\sqrt{6}-4}=\frac{\left(1-\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{6}+4\right)}{\left(2\sqrt{6}-4\right)\left(2\sqrt{6}+4\right)}\)

\(=\frac{2\sqrt{6}+4-4\sqrt{3}-4\sqrt{2}+6\sqrt{2}+4\sqrt{3}}{24-16}=\frac{\sqrt{2}+\sqrt{6}+2}{4}\)

f) \(=\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)}=\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{2\sqrt{15}-6}\)

\(=\frac{\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{15}+6\right)}{\left(2\sqrt{15}-6\right)\left(2\sqrt{15}+6\right)}=\frac{2\sqrt{30}+6\sqrt{2}-6\sqrt{5}-6\sqrt{3}+10\sqrt{3}+6\sqrt{5}}{60-36}\)

\(=\frac{\sqrt{30}+3\sqrt{2}+2\sqrt{3}}{12}\)

1) Để căn thức đã cho có nghĩa \(\Leftrightarrow2x+1< 0\) \(\Leftrightarrow x< -\frac{1}{2}\)

2)

a) \(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{2\left(-5\right)^2}\) \(=3-\sqrt{2}+5\sqrt{2}=4+4\sqrt{2}\)

b) \(\frac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}-\frac{2}{\sqrt{3}-1}=\sqrt{3}-1-\sqrt{3}=-1\)

c) \(\frac{\sqrt{8}-2}{\sqrt{2}-1}+\frac{2}{\sqrt{3}-1}-\frac{3}{\sqrt{3}}\) \(=2+1+\sqrt{3}-\sqrt{3}=3\)

Ta có:

\(B=\frac{\sqrt{2+\sqrt{3}}}{2}\div\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)

\(B=\frac{\sqrt{4+2\sqrt{3}}}{2}\div\left(\frac{\sqrt{4+2\sqrt{3}}}{2}-\frac{2\sqrt{3}}{3}+\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{3}}\right)\)

\(B=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}\div\left(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}-\frac{2\sqrt{3}}{3}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{3}}\right)\)

\(B=\frac{\sqrt{3}+1}{2}\div\left(\frac{\sqrt{3}+1}{2}-\frac{2\sqrt{3}}{2}+\frac{\left(\sqrt{3}+1\right)\sqrt{3}}{6}\right)\)

\(B=\frac{\sqrt{3}+1}{2}\div\left[\frac{3\left(\sqrt{3}+1\right)-6\sqrt{3}+3+\sqrt{3}}{6}\right]\)

\(B=\frac{\sqrt{3}+1}{2}\div\frac{6-2\sqrt{3}}{6}\)

\(B=\frac{\sqrt{3}+1}{2}.\frac{6}{6-2\sqrt{3}}\)

\(B=\frac{3+2\sqrt{3}}{2}\)