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

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

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

\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)

\(=\frac{4.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)

\(=\frac{4}{7}+\frac{3}{7}=1\)

\(a)\)\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)

\(=\)\(\frac{4.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)

\(=\)\(\frac{4}{7}+\frac{3}{7}\)

\(=\)\(1\)