Tìm x để M = |x-33|+|x-43|+|x-53|
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(Lần sau, "bé" nhớ sử dụng công cụ công thức trực quan nhé, dịch của "bé" mình mệt ghê :V)
a,
\(\frac{x+1}{55}+\frac{x+3}{53}+\frac{x+5}{51}=\frac{x+7}{49}+\frac{x+9}{47}+\frac{x+11}{45}\\ \Leftrightarrow\frac{x+1}{55}+1+\frac{x+3}{53}+1+\frac{x+5}{51}+1=\frac{x+7}{49}+1+\frac{x+9}{47}+1+\frac{x+11}{45}+1\\ \Leftrightarrow\frac{x+56}{55}+\frac{x+56}{53}+\frac{x+56}{51}=\frac{x+56}{49}+\frac{x+56}{47}+\frac{x+56}{45}\\ \Leftrightarrow\left(x+56\right)\left(\frac{1}{55}+\frac{1}{53}+\frac{1}{51}-\frac{1}{49}-\frac{1}{47}-\frac{1}{45}\right)=0\)
And... u know dat right? ( ͡~ ͜ʖ ͡°)
b,
\(\frac{x+29}{31}-\frac{x+27}{33}=\frac{x+17}{43}-\frac{x+15}{45}\\ \Leftrightarrow\frac{x+29}{31}+1-\frac{x+27}{33}+1=\frac{x+17}{43}+1-\frac{x+15}{45}+1\\ \Leftrightarrow\frac{x+60}{31}-\frac{x+60}{33}=\frac{x+60}{43}-\frac{x+60}{45}\\ \Leftrightarrow\left(x+60\right)\left(\frac{1}{31}-\frac{1}{33}-\frac{1}{43}+\frac{1}{45}\right)=0\)
Ok, phần còn lại là của bạn nhé. :)
Chúc bạn học tốt nha.
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) x = 19 20 x = 19 20 b ) x = 83 12 x = − 37 12
c ) x = 31 12 x = − 25 12 d ) x = 13 4 x = − 7 12
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x − 2 = − 3 5 + 2 3 ⇔ x − 2 = 19 15 ⇔ x − 2 = 19 15 x − 2 = − 19 15 ⇔ x = 49 15 x = 11 15
b) x − 4 3 = 1 6 + 5 3 − 1 2 + 7 12 ⇔ x − 4 3 = 23 12 ⇔ x − 4 3 = 23 12 x − 4 3 = − 23 12 ⇔ x = 13 4 x = − 7 12
![](https://rs.olm.vn/images/avt/0.png?1311)
x-27+15=53
x-27=53-15
x-27=38
x=38+27
x=65
x-35=43-14
x-35=29
x=29+35
x=64
x-27+15=53
<=> x-27=53-15
<=>x-27=38
<=>x=38+27
<=>65
x-35=43-14
<=>x-35=29
<=>x=29+35
<=>x=64
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\)
\(\Leftrightarrow\dfrac{x+35}{65}+1+\dfrac{x+39}{61}+1=\dfrac{x+43}{57}+1+\dfrac{x+47}{53}+1\)
\(\Leftrightarrow\dfrac{x+100}{65}+\dfrac{x+100}{61}-\dfrac{x+100}{57}-\dfrac{x+100}{53}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\right)=0\Leftrightarrow x=-100\)
Ta có:
\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\\ \Rightarrow\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+39}{61}+1\right)=\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+47}{53}+1\right)\\ \Rightarrow\dfrac{x+100}{53}+\dfrac{x+100}{61}=\dfrac{x+100}{57}+\dfrac{x+100}{53}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\right)=0\)
Ta thấy:
\(\dfrac{1}{65}< \dfrac{1}{57}\\ \dfrac{1}{61}< \dfrac{1}{53}\\ \Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{62}\right)-\left(\dfrac{1}{57}+\dfrac{1}{53}\right)< 0\)
Hay \(\dfrac{1}{65}+\dfrac{1}{62}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\)
\(\Rightarrow x+100=0\\ \Rightarrow x=0-100\\ \Rightarrow x=-100\)
Vậy \(x=-100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a./ x-15=53
x=53+15
x=68
b./43-x=23
x=43-23
x=20
c./ 23+x=56
x=56-23
x=33
a) x - 15 = 53
=> x = 53 + 15
=> x = 68
b) 43 - x = 23
=> x = 43 - 23
=> x = 20
c) 23 + x = 56
=> x = 56 - 23
=> x = 33