a) \(5,4+\left(-23,2\right)+4,6+\left(-0,8\right)\)

\(=\left(5,4+4,6\right)+\left(-23,2-0,8\right)=10-24=-14\)

b) \(\left(-14,3\right)+5,1+4,9+\left(-15,7\right)\)

\(=\left(5,1+4,9\right)+\left(-14,3-15,7\right)=10-30=-20\)

c) \(12,3+3,7+\left(-24,2\right)+\left(-12,3\right)+24,2\)

\(=\left(12,3-12,3\right)+\left(-24,2+24,2\right)+3,7=3,7\)

d) \(\left(-16,5\right).3,4+3,4+\left(-3,5\right)\)

\(=3,4\left(-16,5-3,5\right)=3,4\left(-20\right)=-68\)

\(c,TH1:\frac{2}{3}x-\frac{4}{9}=0=>x=\frac{2}{3}\)

\(TH2:\frac{1}{2}+-\frac{3}{7}:x=0=>x=\frac{6}{7}\)

Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)

c) \(\left(\frac{2}{3}.x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x=\frac{4}{9}\\\frac{-3}{7}:x=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)

\(10a+b+10b+c+10c+a=100a+10b+c\)

\(89a-b=10c\)

\(c\le9\Rightarrow10c\le90\Rightarrow89a-b\le90\Rightarrow a=1\)

\(\Rightarrow89-b=10c\)

\(10c⋮10\Rightarrow89-b⋮10\Rightarrow b=9\Rightarrow89-9=10c\Rightarrow c=8\)

Ta có \(19+98+81=198\)

\(\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)

\(\left(x+2\right).\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)

Vậy \(x\in\left\{-2;-3\right\}\)

làm giá trị lớn nhất nhá

`A =- |x+5| + 10`

Vì `-|x+5| =< 0`

`-> - |x+5|+10 =< 10`

`->A =< 10`

Dấu "`=`" xảy ra khi : `<=> |x+5|=0 <=> x=-5`

Vậy `max A=10 <=>x=-5`

k mik

đúng

hi

Xét ∆ABC và ∆ADE : 
AB = AD(gt) 
Góc BAC = góc EAD (cùng phụ với gócCAD ) 
AC = AE (gt) 
=>∆ABC = ∆ADE (c - g - c) 
=> BC = DE 
=> AM = BC/2 = DE/2

k cho mk nha

chúc bn trung thu vui vẻ

HT