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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
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
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=-4\)
\(\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=0\)
\(\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\left(416-x\right).\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)nên 416 - x = 0
\(\Rightarrow x=416\)
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)
\(\frac{315-x}{101}+\frac{101}{101}+\frac{313-x}{103}+\frac{103}{103}+\frac{311-x}{105}+\frac{105}{105}+\frac{309-x}{107}+\frac{107}{107}=0\)
\(\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)nên
416-x=0
x=416
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
<=>\(\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)
<=>\(\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
<=>\(\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Do \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)
=>416-x=0
<=>x=416
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}+4=0\\ \Leftrightarrow\dfrac{315-x}{101}+1+\dfrac{313-x}{103}+1+\dfrac{311-x}{105}+1+\dfrac{309-x}{107}+1=0\\ \Leftrightarrow\dfrac{416-x}{101}+\dfrac{416-x}{103}+\dfrac{416-x}{105}+\dfrac{416-x}{107}=0\\ \Leftrightarrow\left(416-x\right)\left(\dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}\right)=0\\ \dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}>0\\ \Rightarrow416-x=0\\ \Leftrightarrow x=416\)
Lời giải:
PT \(\Leftrightarrow \frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=4\)
\(\Leftrightarrow \frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=4\)
\(\Leftrightarrow (416-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=4\)
\(\Rightarrow 416-x=\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
\(\Rightarrow x=416-\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\left(1+\frac{315-x}{101}\right)+\left(1+\frac{313-x}{103}\right)+\left(1+\frac{311-x}{105}\right)+\left(1+\frac{309-x}{107}\right)=0\)
\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\Leftrightarrow416-x=0\)
\(\Leftrightarrow x=416\)
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=0\)
\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)\ne0\)
\(\Rightarrow416-x=0\Leftrightarrow x=416\)