\(\left(5-x\right)^2+\left(5-x\right)^5=0\)

\(\Leftrightarrow\left(5-x\right)^2.\left[1+\left(5-x\right)^3\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(5-x\right)^2=0\\1+\left(5-x\right)^3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5-x=0\\\left(5-x\right)^3=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\5-x=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)

Vậy \(x\in\left\{5,6\right\}\)

a. \(\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}=\frac{3^3.5^3.5^4}{\left(-3\right)^5.5^6}\)

\(=\frac{3^3.5^7}{\left(-3\right)^5.5^6}=\frac{5}{-9}\)

b. \(\frac{6^3.2^5.\left(-3\right)^2}{\left(-2\right)^9.3^7}=\frac{2^3.3^3.2^5.3^2}{\left(-2\right)^9.3^7}\)

\(=\frac{2^8.3^5}{\left(-2\right)^9.3^7}=\frac{1}{\left(-2\right).3^2}=-\frac{1}{18}\)

\(6-\left|\frac{1}{2}-x\right|=\frac{2}{5}\)

\(\Leftrightarrow\left|\frac{1}{2}-x\right|=6-\frac{2}{5}\)

\(\Leftrightarrow\left|\frac{1}{2}-x\right|=\frac{28}{5}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}-x=\frac{28}{5}\\\frac{1}{2}-x=-\frac{28}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}-\frac{28}{5}\\x=\frac{1}{2}-\left(-\frac{28}{5}\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{51}{10}\\x=\frac{61}{10}\end{cases}}\)

Vậy : \(x\in\left\{-\frac{51}{10},\frac{61}{10}\right\}\)

<=>|1/2-x|=6-2/5

<=>|1/2-x|=30/5-2/5

<=>|1/2-x|=28/5

=>1/2-x=28/5  hoặc 1/2-x=-28/5

=>x=5/10-56/10 hoặc x=5/10+56/10

=>x=-51/10 hoặc x=61/10

Bn tự vẽ hình nha

a)Ta có:\(\widehat{bOm}+\widehat{aOb}=90^o\left(Oa\perp Om\right)\)

\(\widehat{aOn}+\widehat{aOb}=90^o\left(Ob\perp On\right)\)

\(\Rightarrow\widehat{bOm}=\widehat{aOn}\)

b)Ta có:\(\widehat{bOm}+\widehat{bOn}=150^o\)hay\(\widehat{bOm}+90^o=150^o\)

\(\Rightarrow\widehat{bOm}=150^o-90^o=60^o\)mà\(\widehat{bOm}=\widehat{aOn}\)

\(\Rightarrow\widehat{bOm}=\widehat{aOn}=60^o\)

Ta lại có:\(\widehat{bOm}=\widehat{bOy}+\widehat{yOm}\)mà\(\widehat{bOy}=\widehat{yOm}\)(Oy là phân giác của\(\widehat{bOm}\))

\(\Rightarrow\widehat{bOy}=\widehat{yOm}=\frac{\widehat{bOm}}{2}=\frac{60^o}{2}=30^o\)

\(\widehat{aOn}=\widehat{aOx}+\widehat{xOn}\)mà\(\widehat{aOx}=\widehat{xOn}\)(Ox là phân giác của\(\widehat{aOn}\))

\(\Rightarrow\widehat{aOx}=\widehat{xOn}=\frac{\widehat{aOn}}{2}=\frac{60^o}{2}=30^o\)

\(\widehat{aOn}+\widehat{aOb}+\widehat{bOm}=150^o\)hay\(60^o+\widehat{aOb}+60^o=150^o\)

\(\Rightarrow\widehat{aOb}=150^o-60^o-60^o=30^o\)

Ta lại có:\(\widehat{xOy}=\widehat{bOy}+\widehat{aOb}+\widehat{aOx}=30^o+30^o+30^o=90^o\)

\(a,\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)\(=\frac{3^3.5^3}{\left(-3\right)^5.5^2}\)\(=-\frac{5}{\left(3\right)^2}=-\frac{5}{9}\)

\(b,\frac{6^3.2.\left(-3\right)^2}{\left(-2\right)^9.3^7}\)\(=-\frac{6^3}{2^8.3^5}\)\(=-\frac{2^3.3^3}{2^8.3^5}\)\(=-\frac{1}{2^5.3^2}=-\frac{1}{288}\)

\(c,\frac{3^6.7^2-3^7.7}{3^7.21}\)\(=\frac{3^6.7\left(7-3\right)}{3^7.21}\)\(=\frac{3^6.7.4}{3^7.7.3}\)\(=\frac{4}{3.3}=\frac{4}{9}\)

\(a,\left(x-1,2\right)^2=4\)

\(\Rightarrow x-1,2=2\)

\(\Rightarrow x=3,2\)

\(b,\left(x+1\right)^3=-125\)

\(\Rightarrow\left(x+1\right)^3=\left(-5\right)^3\)

\(\Rightarrow x+1=-5\Rightarrow x=-6\)

\(c,\left(x-5\right)^3=2^6\)

\(\Rightarrow\left(x-5\right)^3=4^3\)

\(\Rightarrow x-5=4\Rightarrow x=9\)

\(d,\left(2x+1\right)^{x+1}=5^{x+1}\)

\(\Rightarrow2x+1=5\Rightarrow x=2\)

                                                   Bài giải

\(x=-\frac{1}{7}\)

\(\Rightarrow\text{ }\left|x\right|=\left|-\frac{1}{7}\right|=\frac{1}{7}\)

\(x=-3\frac{1}{5}\)

\(\Rightarrow\text{ }\left|x\right|=\left|-3\frac{1}{5}\right|=\left|\frac{-16}{5}\right|=\frac{16}{5}\)

                                                   Bài giải

\(x=-\frac{1}{7}\)

\(\Rightarrow\text{ }\left|x\right|=\left|-\frac{1}{7}\right|=\frac{1}{7}\)

\(x=-3\frac{1}{5}\)

\(\Rightarrow\text{ }\left|x\right|=\left|-3\frac{1}{5}\right|=\left|\frac{-16}{5}\right|=\frac{16}{5}\)

\(\left|3+x\right|\left|x-1\right|=0\)

\(\Leftrightarrow\orbr{\begin{cases}3+x=0\\x-1=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=0-3\\x=0-1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-1\end{cases}}\)

vi |3+x| va |x-1| luon luon lon hon hoac bang 0

vi de bai cho |3+x|+|x-1|=0

=>khong ton tai x sao cho thoa man de bai

vi /3-x/ lớn hơn hoặc bằng 0  mà để̉̀̀̀̀̃̃̃ min A nhỏ̉̉ nhất => /3-x/ phả̉̉̉̀̉̀i nhỏ̉ nhất   

1.                                                                                             =>/3-x/=0 => 3-x=0=>x=3 =>min A=2

câu B bạ̣̣n tự lam nha