Ответы
Ответ дал:
0
1.
![sqrt{1frac{24}{25}}-sqrt{0,09}+sqrt{3^2+4^2}=\
=sqrt{frac{1cdot25+24}{25}}-sqrt{0,01cdot9}+sqrt{9+16}=\
=sqrt{frac{25+24}{25}}-sqrt{0,01}cdotsqrt9+sqrt{25}=\
=sqrt{frac{49}{25}}-0,1cdot3+5=frac{7}{5}-0,3+6=1,4-0,3+5=6,1](https://tex.z-dn.net/?f=sqrt%7B1frac%7B24%7D%7B25%7D%7D-sqrt%7B0%2C09%7D%2Bsqrt%7B3%5E2%2B4%5E2%7D%3D%5C%0A%3Dsqrt%7Bfrac%7B1cdot25%2B24%7D%7B25%7D%7D-sqrt%7B0%2C01cdot9%7D%2Bsqrt%7B9%2B16%7D%3D%5C%0A%3Dsqrt%7Bfrac%7B25%2B24%7D%7B25%7D%7D-sqrt%7B0%2C01%7Dcdotsqrt9%2Bsqrt%7B25%7D%3D%5C%0A%3Dsqrt%7Bfrac%7B49%7D%7B25%7D%7D-0%2C1cdot3%2B5%3Dfrac%7B7%7D%7B5%7D-0%2C3%2B6%3D1%2C4-0%2C3%2B5%3D6%2C1 "sqrt{1frac{24}{25}}-sqrt{0,09}+sqrt{3^2+4^2}=\
=sqrt{frac{1cdot25+24}{25}}-sqrt{0,01cdot9}+sqrt{9+16}=\
=sqrt{frac{25+24}{25}}-sqrt{0,01}cdotsqrt9+sqrt{25}=\
=sqrt{frac{49}{25}}-0,1cdot3+5=frac{7}{5}-0,3+6=1,4-0,3+5=6,1")
2.
а)
б)
в)
![left(3sqrt{7,5}right)^2-sqrt3sqrt{0,12}+frac{sqrt2}{sqrt8}=9cdot7,5-sqrt{3cdot0,12}=\
=67,5-sqrt{0,36}+sqrt{frac28}=67,5-0,6+frac14=61,5+0,25=61,75.](https://tex.z-dn.net/?f=left%283sqrt%7B7%2C5%7Dright%29%5E2-sqrt3sqrt%7B0%2C12%7D%2Bfrac%7Bsqrt2%7D%7Bsqrt8%7D%3D9cdot7%2C5-sqrt%7B3cdot0%2C12%7D%3D%5C%0A%3D67%2C5-sqrt%7B0%2C36%7D%2Bsqrt%7Bfrac28%7D%3D67%2C5-0%2C6%2Bfrac14%3D61%2C5%2B0%2C25%3D61%2C75. "left(3sqrt{7,5}right)^2-sqrt3sqrt{0,12}+frac{sqrt2}{sqrt8}=9cdot7,5-sqrt{3cdot0,12}=\
=67,5-sqrt{0,36}+sqrt{frac28}=67,5-0,6+frac14=61,5+0,25=61,75.")
г)
![sqrt{48}+sqrt{75}-sqrt{108}=sqrt{16cdot3}+sqrt{25cdot3}-sqrt{36cdot3}=\
=sqrt{16}cdotsqrt3+sqrt{25}cdotsqrt3-sqrt{36}cdotsqrt3=\
=left(sqrt{16}+sqrt{25}-sqrt{36}right)cdotsqrt3=left(4+5-6right)sqrt3=3sqrt3](https://tex.z-dn.net/?f=sqrt%7B48%7D%2Bsqrt%7B75%7D-sqrt%7B108%7D%3Dsqrt%7B16cdot3%7D%2Bsqrt%7B25cdot3%7D-sqrt%7B36cdot3%7D%3D%5C%0A%3Dsqrt%7B16%7Dcdotsqrt3%2Bsqrt%7B25%7Dcdotsqrt3-sqrt%7B36%7Dcdotsqrt3%3D%5C%0A%3Dleft%28sqrt%7B16%7D%2Bsqrt%7B25%7D-sqrt%7B36%7Dright%29cdotsqrt3%3Dleft%284%2B5-6right%29sqrt3%3D3sqrt3 "sqrt{48}+sqrt{75}-sqrt{108}=sqrt{16cdot3}+sqrt{25cdot3}-sqrt{36cdot3}=\
=sqrt{16}cdotsqrt3+sqrt{25}cdotsqrt3-sqrt{36}cdotsqrt3=\
=left(sqrt{16}+sqrt{25}-sqrt{36}right)cdotsqrt3=left(4+5-6right)sqrt3=3sqrt3")
д)
![left(4-sqrt3right)^2-left(2sqrt5-1right)cdotleft(2sqrt5+1right)=\
=4^2-2cdot4cdotsqrt3+left(sqrt3right)^2-left(left(2sqrt5right)^2-left(1right)^2right)=\
=16-8sqrt3+9-left(4cdot5-1right)=25-8sqrt3-20+1=6-8sqrt3](https://tex.z-dn.net/?f=left%284-sqrt3right%29%5E2-left%282sqrt5-1right%29cdotleft%282sqrt5%2B1right%29%3D%5C%0A%3D4%5E2-2cdot4cdotsqrt3%2Bleft%28sqrt3right%29%5E2-left%28left%282sqrt5right%29%5E2-left%281right%29%5E2right%29%3D%5C%0A%3D16-8sqrt3%2B9-left%284cdot5-1right%29%3D25-8sqrt3-20%2B1%3D6-8sqrt3 "left(4-sqrt3right)^2-left(2sqrt5-1right)cdotleft(2sqrt5+1right)=\
=4^2-2cdot4cdotsqrt3+left(sqrt3right)^2-left(left(2sqrt5right)^2-left(1right)^2right)=\
=16-8sqrt3+9-left(4cdot5-1right)=25-8sqrt3-20+1=6-8sqrt3")
е)
![frac{sqrt2+1}{sqrt3+1}-frac{sqrt2-1}{sqrt3-1}=frac{(sqrt2+1)cdot(sqrt3-1)-(sqrt2-1)cdot(sqrt3+1)}{(sqrt3-1)cdot(sqrt3+1)}=\
=frac{sqrt6+sqrt3-sqrt2-1-sqrt6+sqrt3-sqrt2-1}{3-1}=frac{2sqrt3-2sqrt2}{2}=sqrt3-sqrt2](https://tex.z-dn.net/?f=frac%7Bsqrt2%2B1%7D%7Bsqrt3%2B1%7D-frac%7Bsqrt2-1%7D%7Bsqrt3-1%7D%3Dfrac%7B%28sqrt2%2B1%29cdot%28sqrt3-1%29-%28sqrt2-1%29cdot%28sqrt3%2B1%29%7D%7B%28sqrt3-1%29cdot%28sqrt3%2B1%29%7D%3D%5C%0A%3Dfrac%7Bsqrt6%2Bsqrt3-sqrt2-1-sqrt6%2Bsqrt3-sqrt2-1%7D%7B3-1%7D%3Dfrac%7B2sqrt3-2sqrt2%7D%7B2%7D%3Dsqrt3-sqrt2 "frac{sqrt2+1}{sqrt3+1}-frac{sqrt2-1}{sqrt3-1}=frac{(sqrt2+1)cdot(sqrt3-1)-(sqrt2-1)cdot(sqrt3+1)}{(sqrt3-1)cdot(sqrt3+1)}=\
=frac{sqrt6+sqrt3-sqrt2-1-sqrt6+sqrt3-sqrt2-1}{3-1}=frac{2sqrt3-2sqrt2}{2}=sqrt3-sqrt2")
2.
а)
б)
в)
г)
д)
е)
Ответ дал:
0
Давайте еще!
Ответ дал:
0
ты сказал, что любые
Ответ дал:
0
Давай второе плииииз
Ответ дал:
0
надо очень
Похожие вопросы
2 года назад
7 лет назад
7 лет назад
10 лет назад
10 лет назад
10 лет назад
10 лет назад