Giải pt: x/50+x-1/49+x-2/48+x-3/47+x-150/25=0
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Giải phương trình sau:
\(\dfrac{x}{50}\) +\(\dfrac{x_{ }-1}{49}\)+\(\dfrac{x-2}{48}\)+\(\dfrac{x-3}{47}\)+\(\dfrac{x-150}{25}\)= 0
⇔ \(\dfrac{\left(x-50\right)+50}{50}\)+\(\dfrac{\left(x-50\right)+49}{49}\)+\(\dfrac{\left(x-50\right)+48}{48}\)+\(\dfrac{\left(x-50\right)-100}{25}\)= 0
⇔\(\dfrac{x-50}{50}\)+ 1 + \(\dfrac{x-50}{49}\)+1+\(\dfrac{x-50}{48}\)+1+\(\dfrac{x-50}{47}\)+1+\(\dfrac{x-50}{25}\)-4 = 0
⇔\(\dfrac{x-50}{50}\)+\(\dfrac{x-50}{49}\)+\(\dfrac{x-50}{48}\)+\(\dfrac{x-50}{47}\)+\(\dfrac{x-50}{25}\)= 0
⇔ (x - 50 ) ( \(\dfrac{1}{50}\)+ \(\dfrac{1}{49}\)+\(\dfrac{1}{48}\)+\(\dfrac{1}{47}\)+\(\dfrac{1}{25}\)) = 0
⇔ x-50 =\(\dfrac{0}{\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}}\)
⇔ x- 50 = 0
⇔ x = 50
vậy S = \(\left\{50\right\}\)
Ta có : \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{49}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)
\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{49}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)
\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{49}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)
<=> x - 100 = 0
<=> x = 100
Vậy ..
Ta có: \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{48}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)
\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{48}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)
\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{48}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)
mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}>0\)
nên x-100=0
hay x=100
Vậy: S={100}
\(\frac{x}{50}+\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-150}{25}=0\)
\(\Leftrightarrow\)\(\frac{x}{50}-1+\frac{x-1}{49}-1+\frac{x-2}{48}-1+\frac{x-3}{47}-1+\frac{x-150}{25}+4=0\)
\(\Leftrightarrow\)\(\frac{x-50}{50}+\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\)\(\left(x-50\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{25}\right)=0\)
\(\Leftrightarrow\)\(x-50=0\) (vì 1/50 + 1/49 + 1/48 + 1/47 + 1/25 > 0 )
\(\Leftrightarrow\)\(x=50\)
Vậy....
1) 1996 + 3992 + 5988 + 7984
= 1996 x 1 + 1996 x 2 + 1996 x 3 + 1996 x 4
= 1996 x ( 1 + 2 + 3 + 4)
= 1996 x 10 = 19960
2) 2 x 3 x 4 x 8 x 50 x 25 x 125
= 3 x (2 x 50) x (4 x 25) x (8 x 125)
= 3 x 100 x 100 x 1000
= 30.000.000
3(45 x 46 + 47 x 48) x (51 x 52 - 49 x 48) x (45 x 128 - 90 x 64)
x (1995 x 1996 + 1997 x 1998)
Ta thÊy : 45 x 128 - 90 x 64
= 45 x 2 x 64 - 90 x 64
= 90 x 64 - 90 x 64
= 0
1) 1996 + 3992 + 5988 + 7984
= 1996 x 1 + 1996 x 2 + 1996 x 3 + 1996 x 4
= 1996 x ( 1 + 2 + 3 + 4)
= 1996 x 10 = 19960
2) 2 x 3 x 4 x 8 x 50 x 25 x 125
= 3 x (2 x 50) x (4 x 25) x (8 x 125)
= 3 x 100 x 100 x 1000
= 30.000.000
3(45 x 46 + 47 x 48) x (51 x 52 - 49 x 48) x (45 x 128 - 90 x 64)
x (1995 x 1996 + 1997 x 1998)
Ta thÊy : 45 x 128 - 90 x 64
= 45 x 2 x 64 - 90 x 64
= 90 x 64 - 90 x 64
= 0
L I K E NHÁ BẠN^^
x1+x2+x3+x4+...x49+x50+x51=.
\(\Rightarrow\)(x1+x2)+(x3+x4)+...+(x49+x50)+x51=0
Theo de bai ta co:\(\Rightarrow\)1+1+...+1+x51=0
\(\Rightarrow\)1*25+x51=0
\(\Rightarrow\)x51=-25
Ma x60+x51=0\(\Rightarrow\)-50+(-25)=1\(\Rightarrow\)x50=26
Nho tich nha
x1+x2+x3+...+x49+x50+x51=0
(x1+x2)+(X3+x4)+ ...+ (x49+x50) + x51=0
1.25+x51=0
=> x51= -25 ( vì -25+25=0 nhá ^^)
mà x50+ (-25)(tức là x51 ý ) =1
=> x50=26
Ta có : \(\frac{x}{50}+\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-150}{25}=0\)
=> \(\frac{x}{50}-1+\frac{x-1}{49}-1+\frac{x-2}{48}-1+\frac{x-3}{47}-1+\frac{x-150}{25}+4=0\)
=> \(\frac{x-50}{50}+\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{25}=0\)
=> \(\left(x-50\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{25}\right)=0\)
=> \(x-50=0\)
=> \(x=50\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{50\right\}\)