Giải phương trình sau 1909-x/91+1907-x/93+1905-x/95+1903-x/91+4=0 18/07/2021 Bởi Samantha Giải phương trình sau 1909-x/91+1907-x/93+1905-x/95+1903-x/91+4=0
$\frac{1909-x}{91}+$ $\frac{1907-x}{93}+$ $\frac{1905-x}{95}+$ $\frac{1903-x}{97}+4=0$ $<=>$ $(\frac{1909-x}{91}+1)$ $(\frac{1907-x}{93}+1)+$ $(\frac{1905-x}{95}+1)+($ $\frac{1903-x}{97}+1)=0$ $<=>$$\frac{2000-x}{91}+$ $\frac{2000-x}{93}+$ $\frac{2000-x}{95}+$ $\frac{2000-x}{97}$ $<=>(2000-x)($$\frac{1}{91}+$ $\frac{1}{93}+$ $\frac{1}{95}+$ $\frac{1}{97})=0$ $<=>2000-x=0.(Vì: $ $\frac{1}{91}+$ $\frac{1}{93}+$ $\frac{1}{95}+$ $\frac{1}{97})$$\neq0)$ $<=>x=2000$ Bình luận
$\frac{1909-x}{91}$+$\frac{1907-x}{93}$+$\frac{1905-x}{95}$+$\frac{1903-x}{97}$ +4= 0 ⇔ $\frac{1909-x}{91}$+1+$\frac{1907-x}{93}$+1+$\frac{1905-x}{95}$+1+$\frac{1903-x}{97}$+1= 0 ⇔ $\frac{2000-x}{91}$+$\frac{2000-x}{93}$+$\frac{2000-x}{95}$+$\frac{2000-x}{97}$= 0 ⇔ $( 2000-x)$.( $\frac{1}{91}$+$\frac{1}{93}$+$\frac{1}{95}$+$\frac{1}{97}$)= 0 ⇔ $2000-x= 0$ ⇔ $x=2000$ Bình luận
$\frac{1909-x}{91}+$ $\frac{1907-x}{93}+$ $\frac{1905-x}{95}+$ $\frac{1903-x}{97}+4=0$
$<=>$ $(\frac{1909-x}{91}+1)$ $(\frac{1907-x}{93}+1)+$ $(\frac{1905-x}{95}+1)+($ $\frac{1903-x}{97}+1)=0$
$<=>$$\frac{2000-x}{91}+$ $\frac{2000-x}{93}+$ $\frac{2000-x}{95}+$ $\frac{2000-x}{97}$
$<=>(2000-x)($$\frac{1}{91}+$ $\frac{1}{93}+$ $\frac{1}{95}+$ $\frac{1}{97})=0$
$<=>2000-x=0.(Vì: $ $\frac{1}{91}+$ $\frac{1}{93}+$ $\frac{1}{95}+$ $\frac{1}{97})$$\neq0)$
$<=>x=2000$
$\frac{1909-x}{91}$+$\frac{1907-x}{93}$+$\frac{1905-x}{95}$+$\frac{1903-x}{97}$ +4= 0
⇔ $\frac{1909-x}{91}$+1+$\frac{1907-x}{93}$+1+$\frac{1905-x}{95}$+1+$\frac{1903-x}{97}$+1= 0
⇔ $\frac{2000-x}{91}$+$\frac{2000-x}{93}$+$\frac{2000-x}{95}$+$\frac{2000-x}{97}$= 0
⇔ $( 2000-x)$.( $\frac{1}{91}$+$\frac{1}{93}$+$\frac{1}{95}$+$\frac{1}{97}$)= 0
⇔ $2000-x= 0$
⇔ $x=2000$