Question

As dori si eu sa imi rezolvati exercitile de la 3 la 6!! Va rog frumoss!

Answer 1

3. a) $7^{2}- 3^{2}+(- \sqrt{5}) ^{2}=49-9+ \sqrt{5} ^{2} =40+5=45$
b) $0^{100}+(- \sqrt{7}) ^{3}+ \sqrt{243}=0+(- \sqrt{7}) ^{3}+9 \sqrt{3} =- \sqrt{7} ^{3}+9 \sqrt{3} =$$- \sqrt{7 ^{3}}+9 \sqrt{3} =-7 \sqrt{7}+9 \sqrt{3}$
c) $\sqrt{1.44}-(1.2) ^{2}= \sqrt{ \frac{36}{25} }-( \frac{6}{5}) ^{2} = \frac{6}{5}- \frac{36}{25}= \frac{30-36}{25}= \frac{-6}{25}=-\frac{6}{25}$
d) $\sqrt{3 ^{2} +4 ^{2} } - \sqrt{5 ^{2} }= \sqrt{9+16}-5= \sqrt{25}-5=5-5=0$

4. a) $[ \frac{3}{ \sqrt{3} }-(0.4+ \sqrt{3})]*10=[ \sqrt{3}-0.4- \sqrt{3}]*10=[-0.4]*10=$$-[0.4*10]=-4$
b) $[-3 \sqrt{5}-( \frac{1}{2}-2 \sqrt{5})]*2= [-3 \sqrt{5}- \frac{1}{2}-2 \sqrt{5}]*2=[- \sqrt{5}- \frac{1}{2}]*2=$$-2 \sqrt{5}- \frac{1}{2}*2=-2 \sqrt{5}-1$
c) $\sqrt{( \sqrt{3}- \sqrt{2}) ^{2} } -(\sqrt{3}- \sqrt{2})=\sqrt{3}- \sqrt{2}-\sqrt{3}+ \sqrt{2}=0$

5. a) $2^{-1} + 3^{-1}= \frac{1}{2}+ \frac{1}{3}= \frac{3+2}{6}= \frac{5}{6}$
b) $(- \sqrt{2}) ^{-2}+( \sqrt{3}) ^{-2}=- \sqrt{2} ^{-2}+3 ^{-1}=2 ^{-1}+ \frac{1}{3}= \frac{1}{2}+ \frac{1}{3}= \frac{3+2}{6}= \frac{5}{6}$
c) $\sqrt{27}*( \sqrt{3}) ^{-3}+1= \sqrt{27}\sqrt{3 ^{-3}}+1= \sqrt{27*3 ^{-3} }+1= \sqrt{ 3^{3} *3 ^{-3} }+1=$$\sqrt{1}+1=1+1=2$
d) $[(2+3) ^{-1}+ \frac{1}{ \sqrt{5} }]:(1+ \sqrt{5})=[5 ^{-1}+ \frac{1}{ \sqrt{5} }]* \frac{1}{1+ \sqrt{5} }=[ \frac{1}{5}+\frac{1}{ \sqrt{5} }]* \frac{1}{1+ \sqrt{5} }=$$\frac{ \sqrt{5}+5 }{5 \sqrt{5} }* \frac{1}{1+ \sqrt{5} }= \frac{( \sqrt{5}+5)*1 }{5 \sqrt{5}*(1+ \sqrt{5}) }= \frac{( \sqrt{5}+5) }{5 \sqrt{5}*(1+ \sqrt{5}) }= \frac{( \sqrt{5}+5) \sqrt{5} }{5 \sqrt{5}*(1+ \sqrt{5}) \sqrt{5} }=$$\frac{( \sqrt{5}+5) \sqrt{5} }{5 *5(1+ \sqrt{5}) }=\frac{( \sqrt{5}+5) \sqrt{5} }{25(1+ \sqrt{5}) }=\frac{5+5 \sqrt{5} }{25(1+ \sqrt{5}) }= \frac{5(1+ \sqrt{5}) }{{25(1+ \sqrt{5}) }}= \frac{1}{5}$

6. a) $4-4(2 \sqrt{2}- \frac{4}{ \sqrt{2} })=4-4(2 \sqrt{2}- \frac{4 \sqrt{2} }{ 2})=4-4(2 \sqrt{2}-2 \sqrt{2})=$$4-4*0=4-0=4$
b) $[- \sqrt{75}-4( \sqrt{3}-2 \sqrt{3})]:1 ^{50}=[-5 \sqrt{3}-4*(- \sqrt{3})]:1 =$$-5 \sqrt{3}+4 \sqrt{3}=(-5+4) \sqrt3}=-1 \sqrt{3}=- \sqrt{3}$
c) $5-5*[( \sqrt{6}+ \sqrt{3}): \sqrt{3}- \sqrt{2}]=5-5[ \frac{ \sqrt{6}+ \sqrt{3} }{ \sqrt{3} }- \sqrt{2} ]=$$5- \frac{5( \sqrt{6}+ \sqrt{3}) }{ \sqrt{3} } +5 \sqrt{2}=5- \frac{5 \sqrt{6}+5 \sqrt{3} }{ \sqrt{3} } +5 \sqrt{2}=5- \frac{(5 \sqrt{6}+5 \sqrt{3}) \sqrt{3} }{3}+ 5\sqrt{2}=$$5- \frac{5 \sqrt{18} +15 }{3}+5 \sqrt{2}=5- \frac{15 \sqrt{2} +15 }{3}+5 \sqrt{2}=5- \frac{3(5 \sqrt{2} +5) }{3}+5 \sqrt{2}=$$5-(5 \sqrt{2}+5)+5 \sqrt{2}=5-5 \sqrt{2}-5+5 \sqrt{2}=0$