Tìm x : a,\(\frac{x-1}{2003}+\frac{x-2}{2002}+\frac{x-3}{2001}-3=0\)
b,\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=-4\)
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)=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
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
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 \(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+x=0\)
b)Tim a,b (2a+5b+1)(2\(^a\)+a\(^2\)+a+b)=105
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
<=>\(\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)
<=>\(\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
<=>\(\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Do \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)
=>416-x=0
<=>x=416
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