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2 hình nha hok tốt

\(a)13-\left(40-X\right)=35\)

\(\Leftrightarrow40-X=13-35\)

\(\Leftrightarrow40-X=-22\)

\(\Leftrightarrow X=40-\left(-22\right)\)

\(\Leftrightarrow X=62\)

Vậy \(X=62\)

\(b)14-3x\left(5-X\right)=8\)

\(\Leftrightarrow3x\left(5-X\right)=14-8\)

\(\Leftrightarrow3x\left(5-X\right)=6\)

\(\Leftrightarrow5-X=6:3\)

\(\Leftrightarrow5-X=3\)

\(\Leftrightarrow X=5-3\)

\(\Leftrightarrow X=2\)

Vậy \(X=2\)

\(c)\left(3X-2\right)x\left(5+X\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3X-2=0\\5+X=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3X=0+2\\X=0-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3X=2\\X=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}X=2:3\\X=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}X=\frac{2}{3}\\X=-5\end{cases}}\)

Vậy \(X\in\left\{-5;\frac{2}{3}\right\}\)

\(d)\left(3X-6\right)x3=3^4\)

\(\Leftrightarrow\left(3X-6\right)x3=81\)

\(\Leftrightarrow3X-6=81:3\)

\(\Leftrightarrow3X-6=27\)

\(\Leftrightarrow3X=27+6\)

\(\Leftrightarrow3X=33\)

\(\Leftrightarrow X=33:3\)

\(\Leftrightarrow X=11\)

Vậy\(X=11\)

\(\left|\overrightarrow{MA}+\overrightarrow{MC}-\overrightarrow{MN}\right|=\left|\overrightarrow{MA}+\overrightarrow{MD}+\overrightarrow{DC}-\overrightarrow{MN}\right|\)\(=\left|\overrightarrow{DC}-\frac{1}{2}\overrightarrow{DC}-\frac{1}{2}\overrightarrow{AB}\right|=\left|\overrightarrow{DC}-\frac{3}{4}\overrightarrow{DC}\right|=\frac{1}{A}DC=\frac{a}{2}\)

mk bận đi ch nên chỉ tạm câu a nha 

vẽ 3 đường trung tuyến AD ; BE ; CF 

VT = 

\(GA+GB+GC\)   ( nhớ thêm dấu vec tơ nha ) 

\(=-\frac{2}{3}AD-\frac{2}{3}BE-\frac{2}{3}CF\)  

\(=-\frac{2}{3}\cdot\frac{1}{2}\left(AB+BC\right)-\frac{2}{3}\cdot\frac{1}{2}\left(BA+BC\right)-\frac{2}{3}\cdot\frac{1}{2}\left(CA+CB\right)\)     ( quy tắc hình bình hành ) 

\(=-\frac{1}{3}\left(AB+AC\right)-\frac{1}{3}\left(BA+BC\right)-\frac{1}{3}\left(CA+CB\right)\) 

\(=-\frac{1}{3}AB-\frac{1}{3}AC-\frac{1}{3}BA-\frac{1}{3}BC-\frac{1}{3}CA-\frac{1}{3}CB\)    

\(=0=VP\)

.... chua hoc

Giả sử \(x\ge y\ge z\)cũng được mà.

a) \(A=\left\{0;1;2;3;...;13\right\}\)

b) Ta có: \(x^2+3x-9=0\)

\(\Leftrightarrow\left(x-\frac{-3+3\sqrt{5}}{2}\right)\left(x+\frac{3+3\sqrt{5}}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-3+3\sqrt{5}}{2}\\x=\frac{-3-3\sqrt{5}}{2}\end{cases}}\)

c) \(C=\left\{-7;-6;-5;...;5;6;7\right\}\)