文章目录
一元二次方程
- a x 2 + b x + c = 0 , a ≠ 0 ax^2+bx+c=0,\ a\neq0 ax2+bx+c=0, a=0
- Δ = b 2 − 4 a c \Delta =b^2-4ac Δ=b2−4ac
- i f Δ ≥ 0 t h e n if \Delta \ge 0\ then ifΔ≥0 then
- x 1 = − b + b 2 − 4 a c 2 a \quad \displaystyle x_1=\frac{-b+\sqrt{b^2-4ac}}{2a} x1=2a−b+b2−4ac
- e l s e else else
- x 2 = − b − b 2 − 4 a c 2 a \quad \displaystyle x_2=\frac{-b-\sqrt{b^2-4ac}}{2a} x2=2a−b−b2−4ac
- s i n ( x + π 4 ) + c o s ( x − π 3 ) + l n ( e x + 1 + 5 x ) = 0 sin(x+\frac{\pi}{4})+cos(x-\frac{\pi}{3})+ln(e^{x+1}+5x)=0 sin(x+4π)+cos(x−3π)+ln(ex+1+5x)=0
