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\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\left(1+\frac{315-x}{101}\right)+\left(1+\frac{313-x}{103}\right)+\left(1+\frac{311-x}{105}\right)+\left(1+\frac{309-x}{107}\right)=0\)
\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\Leftrightarrow416-x=0\)
\(\Leftrightarrow x=416\)
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=0\)
\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)\ne0\)
\(\Rightarrow416-x=0\Leftrightarrow x=416\)
Tâ có \(\frac{315-x}{101}+\frac{313-x}{103}-\frac{x-311}{105}-\frac{x-309}{107}=-4\)
\(\Leftrightarrow\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{x-309}{107}+1=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}-\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}-\frac{1}{107}\right)=0\)
\(\Rightarrow416-x=0\Leftrightarrow x=416\)
#học tốt
chia đều 4 vào 4 số hạng => tử số đều là (101+315)
=> pt có nghiệm (x=(101+315)
\(\Leftrightarrow\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Mà \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)
\(\Rightarrow416-x=0\Rightarrow x=416\)
Vậy x = 416
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)
\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)
\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)
Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)
\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy \(x=100\)