Question

  giải chi tiết giúp 

Answer 1

Ta có: \(4\left(m_{b}^2+m_{c}^2\right)=5a^2\)

=>\(4\cdot\left(\frac{BA^2+BC^2}{2}-\frac{AC^2}{4}+\frac{CA^2+CB^2}{2}-\frac{AB^2}{4}\right)=5\cdot BC^2\)

=>\(4\cdot\frac{2BA^2+2BC^2-AC^2+2CA^2+2CB^2-AB^2}{4}=5BC^2\)

=>\(BA^2+4BC^2+AC^2=5BC^2\)

=>\(BA^2+AC^2=BC^2\)

=>ΔABC vuông tại A

\(\frac{1+cosB}{\sin^2\left(A+C\right)}=\frac{2c}{2c-a}\)

=>\(\frac{1+cosB}{\sin^2\left(180^0-B\right)}=\frac{2\cdot AB}{2\cdot AB-BC}\)

=\(\frac{1+cosB}{\sin^2B}=2\cdot\frac{AB}{2AB-BC}\)

=>\(\frac{1+cosB}{cos^2C}=\frac{2AB}{2AB-BC}\)

=>\(\frac{1+\frac{BA}{BC}}{\left(\frac{CA}{CB}\right)^2}=\frac{2AB}{2AB-BC}\)

=>\(\frac{BC+BA}{BC}:\frac{CA^2}{CB^2}=\frac{2AB}{2AB-BC}\)

=>\(\frac{BC+AB}{BC}\cdot\frac{BC^2}{BC^2-AB^2}=\frac{2AB}{2AB-BC}\)

=>\(BC\cdot\frac{BC+AB}{\left(BC+BA\right)\left(BC-BA\right)}=\frac{2AB}{2AB-BC}\)

=>\(\frac{BC}{BC-BA}=\frac{2AB}{2AB-BC}\)

=>\(BC\left(2AB-BC\right)=2AB\left(BC-BA\right)\)

=>\(2\cdot AB\cdot BC-BC^2=2AB\cdot BC-2AB^2\)

=>\(BC^2=2AB^2\)

=>\(BC=AB\cdot\sqrt2\)

Xét ΔABC vuông tại A có \(\sin C=\frac{AB}{BC}=\frac{1}{\sqrt2}\)

nên \(\hat{C}=45^0\)

=>m=45