\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)=0
Tìm x nhé
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
\(\frac{315-x}{101}+\frac{313-x}{103}-\frac{x-311}{105}-\frac{x-309}{107}=-4\)
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
Tìm x thuộc Q biết:
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
giúp mình nha!
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\left(1+\frac{315-x}{101}\right)+\left(1+\frac{313-x}{103}\right)+\left(1+\frac{311-x}{105}\right)+\left(1+\frac{309-x}{107}\right)=0\)
\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\Leftrightarrow416-x=0\)
\(\Leftrightarrow x=416\)
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=0\)
\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)\ne0\)
\(\Rightarrow416-x=0\Leftrightarrow x=416\)
Bài làm :
Ta có :
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\left(1+\frac{315-x}{101}\right)+\left(1+\frac{313-x}{103}\right)+\left(1+\frac{311-x}{105}\right)+\left(1+\frac{309-x}{107}\right)=0\)
\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\text{Vì : }\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)
\(\Rightarrow416-x=0\)
\(\Leftrightarrow x=416\)
Vậy x=416
Tìm x : a,\(\frac{x-1}{2003}+\frac{x-2}{2002}+\frac{x-3}{2001}-3=0\)
b,\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=-4\)
Bạn tham khảo nhé
\(a)\) \(\frac{x-1}{2003}+\frac{x-2}{2002}+\frac{x-3}{2001}-3=0\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2003}-1\right)+\left(\frac{x-2}{2002}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(-3+3\right)=0\)
\(\Leftrightarrow\)\(\frac{x-2004}{2003}+\frac{x-2004}{2002}+\frac{x-2004}{2001}=0\)
\(\Leftrightarrow\)\(\left(x-2004\right)\left(\frac{1}{2003}+\frac{1}{2002}+\frac{1}{2001}\right)=0\)
Vì \(\frac{1}{2003}+\frac{1}{2002}+\frac{1}{2001}\ne0\)
\(\Rightarrow\)\(x-2004=0\)
\(\Rightarrow\)\(x=2004\)
Vậy \(x=2004\)
Chúc bạn học tốt ~
\(b)\) \(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=-4\)
\(\Leftrightarrow\)\(\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=-4+4\)
\(\Leftrightarrow\)\(\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\)\(\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)
\(\Rightarrow\)\(416-x=0\)
\(\Rightarrow\)\(x=416\)
Vậy \(x=416\)
Chúc bạn học tốt ~
tìm x
\(\frac{315-x}{101}\)+\(\frac{313-x}{103}\)+\(\frac{311-x}{105}\)+\(\frac{309-x}{107}\)=\(-4\)
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=-4\)
\(\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=0\)
\(\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\left(416-x\right).\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)nên 416 - x = 0
\(\Rightarrow x=416\)
a)Tim x biet \(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+x\)=0
b) Tìm a,b thuộc Z biết (2a + 5b +1) . 2|3a +b| = 105
cho một lời khuyên nhá, bạn nên nhờ trần như hoặc trieu dang
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 ;
\(\frac{315-x}{101}\)+\(\frac{313-x}{103}\)+ \(\frac{311-x}{105}\)+\(\frac{309-x}{107}\)+4 = 0
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)
\(\frac{315-x}{101}+\frac{101}{101}+\frac{313-x}{103}+\frac{103}{103}+\frac{311-x}{105}+\frac{105}{105}+\frac{309-x}{107}+\frac{107}{107}=0\)
\(\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)nên
416-x=0
x=416
bạn triều dâng làm đúng đấy .tớ cũng làm thế