ПОМОГИТЕ !!!!!!!!!!!!!!!!!! 32 / 33 / 34 / 35 С РЕШЕНИЕМ. Created by Kristina7942401. matematika-ru

32–ой номер: 33–ий номер: 34–ый номер: 35–ый номер:

35 С РЕШЕНИЕМ...

July 2022 1 20 Report

ПОМОГИТЕ !!!!!!!!!!!!!!!!!! 32 / 33 / 34 / 35 С РЕШЕНИЕМ

Answers & Comments

skvrttt 32–ой номер:

$(\frac{a^{-1,25}-a^{-0,75}}{a^{-1}-a^{-0,5}})^{-2}=(\frac{a^{-\frac{5}{4}}-a^{-\frac{3}{4}}}{a^{-1}-a^{-\frac{1}{2}}})^{-2}=(\frac{(a^{-\frac{1}{4}})^5-(a^{-\frac{1}{4}})^3}{(a^{-\frac{1}{2}})^2-(a^{-\frac{1}{2}})^1})^{-2}=\\(\frac{(a^{-\frac{1}{4}})^3[(a^{-\frac{1}{4}})^2-1]}{a^{-\frac{1}{2}}(a^{-\frac{1}{2}}-1)})^{-2}=(\frac{a^{-\frac{3}{4}}}{a^{-\frac{1}{2}}})^{-2}=(a^{-\frac{3}{4}+\frac{2}{4}})^{-2}=(a^{-\frac{1}{4}})^{-2}=\sqrt{a};\\\sqrt{1,6*10^{-3}}=\sqrt{0,0016}=0,04=4*10^{-2}$

33–ий номер:

$\sqrt[6]{213,1^3+12*2131*1869+186,9^3}=\\\sqrt[6]{213,1^3+186,9^3+12*2131*1869}=\\\sqrt[6]{400(213,1^2-213,1*186,9+186,9^2)+1200*213,1*186,9}=\\\sqrt[6]{400(213,1^2-213,1*186,9+186,9^2+3*213,1*186,9)}=\\\sqrt[6]{400(213,1^2+2*213,1*186,9+186,9^2)}=\\\sqrt[6]{400(213,1+186,9)^2}=\sqrt[6]{400^3}=400^{\frac{3}{6}}=400^{\frac{1}{2}}=20$

34–ый номер:

$\frac{212,2^3-111,2^3}{101}-212,2*333,6-100^2-10^2-1=\\\frac{(212,2-111,2)(212,2^2+212,2*111,2+111,2^2)}{101}-212,2*333,6-10^4-10^2-1=\\212,2^2+212,2*111,2+111,2^2-3*212,2*111,2-10^4-10^2-1=\\212,2^2+111,2^2-2*212,2*111,2-10^4-10^2-1=\\(212,2^2-2*212,2*111,2+111,2^2)-10^2(10^2+1)-1=\\(212,2-111,2)^2-10^2(100+1)-1=101^2-10^2*101-1=\\101(101-10^2)-1=100$

35–ый номер:

$(\frac{199*201+299*301+2}{1999*2001+2999*3001+2})^{-0,5}=\\(\frac{(200-1)(200+1)+(300-1)(300+1)+2}{(2000-1)(2000+1)+(3000-1)(3000+1)+2})^{-0,5}=(\frac{200^2-1+300^2-1+2}{2000^2-1+3000^2-1+2})^{-0,5}=\\(\frac{200^2+300^2}{2000^2+3000^2})^{-0,5}=(\frac{(2*10^2)^2+(3*10^2)^2}{(2*10^3)^2+(3*10^3)^2})^{-0,5}=(\frac{2^2*10^4+3^2*10^4}{2^2*10^6+3^2*10^6})^{-0,5}=\\(\frac{10^4(2^2+3^2)}{10^6(2^2+3^2)})^{-0,5}=(\frac{1}{10^2})^{-0,5}=(10^2)^{0,5}=10$

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