a) \(2x-5-3+x=1\)

\(3x=9\)

\(x=3\)

b) \(9-3x-2-5x+1=x-3\)

\(12-8x=x-3\)

\(-9x=-15\)

\(x=\dfrac{15}{9}=\dfrac{5}{3}\)

Trong tam giác cân đường trung tuyến đồng thời là đường cao

\(BM=\dfrac{1}{2}BC=5cm\)

Áp dụng định lí Pytago vào tam giác vuông AMB, ta có:

\(AM=\sqrt{AB-BM}=\sqrt{13^2-5^2}=12cm\)

BN KO TRẢ LỜI THÌ THÔI BN ĐỪNG CÓ BL LINH TINH

\(f\left(2\right)=4a-14+2=8\)

\(\Rightarrow a=5\)

\(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{1}{4}\)

\(x=\left(\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{2}{3}=-\dfrac{3}{8}\)

\(\frac{1}{2}+\frac{2}{3}x=\frac{1}{4}\)

             \(\frac{2}{3}x=\frac{1}{4}-\frac{1}{2}\)

             \(\frac{2}{3}x=\frac{1}{4}+\left(\frac{-1}{2}\right)=\frac{1}{4}+\frac{-2}{4}=\frac{-1}{4}\)

                  \(x=\left(\frac{-1}{4}\right)\div\frac{2}{3}=\left(\frac{-1}{4}\right)\times\frac{3}{2}=\frac{-3}{8}\)

\(\dfrac{10}{n-1}\in Z\Rightarrow n-1=Ư\left(10\right)\)

\(\Rightarrow n-1=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

\(\Rightarrow n=\left\{-9;-4;-1;0;2;3;6;11\right\}\)

Để \(\frac{10}{n-1}\)nguyên thì \(n-1\inƯ\left(10\right)\)

\(\Rightarrow n-1\in\left\{-1,1,10,-10\right\}\)

\(\Rightarrow n-1\in\left\{0,2,-9,11\right\}\)

Vậy \(n\in\left\{0;2;-9;11\right\}\)thì \(\frac{10}{n-1}\)nguyên

\(f\left(x\right)=ax^2+bx+c\)

\(f\left(4\right)=a.4^2+b.4+c=16a+4b+c\)

\(f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=4a-2b+c\)

\(f\left(4\right)-f\left(-2\right)=\left(16a+4b+c\right)-\left(4a-2b+c\right)=12a+6b\)

\(=6\left(2a+b\right)=0\)

\(\Leftrightarrow f\left(4\right)=f\left(-2\right)\)

\(f\left(4\right)+2f\left(-2\right)=\left(16a+4b+c\right)+2\left(4a-2b+c\right)=24a+3c=3\left(8a+c\right)\ne0\)

Suy ra \(f\left(4\right)=f\left(-2\right)\ne0\)suy ra đpcm.