x+315=311+5
315-x /101 +313-x /103 +311-x /105 +309-x /107 =-4. Tim x
làm như thế này đứng chưa:
315-x/101+313-x/103+311-x/105+309-x/107=-4
<=>(315-x/101+1)+(313-x/103+1)+(311-x/105+1)+(309-x/107+1)=-4+4
<=>x+416/101+x+416/103+x+416/105+x+416/107=0
<=>(x+416)(1/101+1/103+1/105+1/107)=0
<=>x+416=0
=>x=-416
tại s cái bước 2 lại là x + 416/101 chứ k pải là -x+416/101
Giải phương trình:
1)(315-x)/101+(313-x)/103+(311-x)/105+(309-x)/107+1=0
2)(x+2)/327+(x+3)/326+(x+4)/325+(x+5)/324+(x+349)/5=0
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}+4=0\)
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}+4=0\\ \Leftrightarrow\dfrac{315-x}{101}+1+\dfrac{313-x}{103}+1+\dfrac{311-x}{105}+1+\dfrac{309-x}{107}+1=0\\ \Leftrightarrow\dfrac{416-x}{101}+\dfrac{416-x}{103}+\dfrac{416-x}{105}+\dfrac{416-x}{107}=0\\ \Leftrightarrow\left(416-x\right)\left(\dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}\right)=0\\ \dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}>0\\ \Rightarrow416-x=0\\ \Leftrightarrow x=416\)
\(\frac{315-x}{101}+\frac{313-x}{103}-\frac{x-311}{105}-\frac{x-309}{107}=-4\)
tìm x
à mk bt làm r :v thôi nhé :))
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
dòng 3,4 đánh sai kìa :D
Rút gọn biểu thức
a) (-15) + y +(- y ) + 315
b) 500 + (- x) + (-200) + x
c)567 + x + /311+ (-456)
làm như thế này đứng chưa:
315-x/101+313-x/103+311-x/105+309-x/107=-4
<=>(315-x/101+1)+(313-x/103+1)+(311-x/105+1)+(309-x/107+1)=-4+4
<=>x+416/101+x+416/103+x+416/105+x+416/107=0
<=>(x+416)(1/101+1/103+1/105+1/107)=0
<=>x+416=0
=>x={-416}
tìm x biết
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
Tìm x
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
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
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)=0
Tìm x nhé