\(124-x+14=87\\ =>124-x=87-14\\ =>124-x=73\\ =>x=124-73\\ =>x=51\)

\(25+x+43=165\\ =>x=165-25-43\\ =>x=97\)

\(c,x\left(3x-16\right)=-5\\ =>3x^2-16x+5=0\\ =>3x^2-15x-x+5=0\\ =>3x\left(x-5\right)-\left(x-5\right)=0\\ =>\left(3x-5\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}3x-5=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=5\end{matrix}\right.\)

\(x^2-9-\left(x-3\right)\left(2x-5\right)=0\\ =>\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(2x-5\right)=0\\ =>\left(x-3\right)\left(x+3-2x+5\right)=0\\ =>\left(x-3\right)\left(8-x\right)=0\\ =>\left[{}\begin{matrix}x-3=0\\8-x=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

\(a,=>x=\left\{13;14;15;16\right\}\)

\(b,=>x=\left\{4\right\}\)

\(AB=3+3=6\\ =>A\)

\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{2}=\dfrac{9}{2}\\ =>\left(x-\dfrac{3}{2}\right)^2=\dfrac{9}{2}-\dfrac{1}{2}\\ =>\left(x-\dfrac{3}{2}\right)^2=4\\ =>\left[{}\begin{matrix}x-\dfrac{3}{2}=2\\x-\dfrac{3}{2}=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

ngủm đi Sun :v 

đề bài là gì nhỉ 

A