giải các phương trình sau
x+2/50+x+4/49=x+6/48+x+8/47
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9 tháng 2 2021
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 ..
9 tháng 2 2021
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}
KT
10 tháng 2 2018
10) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/86 + 1/85 + 1/84 + 1/83 + 1/4 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy....
CM
25 tháng 2 2019
a, Ta có :
1996 + 3992 + 5988 + 7984
= 1 x 1996 + 2 x 1996 + 3 x 1996 + 4 x 1996
= (1 + 2 + 3 + 4) x 1996
= 10 x 1996
= 19960
b, 2 x 3 x 4 x 8 x 50 x 25 x 125
= 3 x 2 x 4 x 50 x 8 x 25 x 125
= 3 x (2 x 50) x (4 x 25) x (8 x 125)
= 30 000 000.
c, Ta nhận thấy :
45 x 128 – 90 x 64 = 45 x (2 x 64) – 90 x 64
= (45 x 2) x 64 – 90 x 64
= 90 x 64 – 90 = 0
Trong 1 tích có 1 thừa số bằng 0. Vậy tích đó bằng 0, tức là :
(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) = 0
KT
5 tháng 2 2018
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy.....
KT
7 tháng 2 2018
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy...
26 tháng 3 2020
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 \(S=\left\{50\right\}\)
24 tháng 3 2020
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
HT
13 tháng 1 2018
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 = \(\left\{50\right\}\)
KN
10 tháng 3 2020
\(ĐKXĐ:x\ne49;x\ne50\)
Đặt \(x-49=u;x-50=v\)
Phương trình trở thành \(\frac{50}{u}+\frac{49}{v}=\frac{u}{50}+\frac{v}{49}\)
\(\Rightarrow\frac{50v+49u}{uv}=\frac{49u+50v}{2450}\)
\(\Rightarrow\orbr{\begin{cases}50v+49u=0\\uv=2450\end{cases}}\)
+) \(50v+49u=0\)
\(\Rightarrow50v=-49u\)
\(\Rightarrow\frac{v}{-49}=\frac{u}{50}=\frac{\left(x-50\right)-\left(x-49\right)}{-49-50}\)
\(=\frac{-1}{-99}=\frac{1}{99}\)
\(\Rightarrow\hept{\begin{cases}v=\frac{-49}{99}\\u=\frac{50}{99}\end{cases}}\Rightarrow x=\frac{4901}{99}\)(tm)
+) \(uv=2450\)
hay \(\left(x-49\right)\left(x-50\right)=2450\)
\(\Leftrightarrow x^2-99x+2450=2450\)
\(\Leftrightarrow x^2-99x=0\Leftrightarrow x\left(x-99\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=99\end{cases}}\left(tm\right)\)
Vậy phương trình có 3 nghiệm \(S=\left\{0;\frac{4901}{99};99\right\}\)
PT <=> \(\left(\frac{x+2}{50}+2\right)+\left(\frac{x+4}{49}+2\right)=\left(\frac{x+6}{48}+2\right)+\left(\frac{x+8}{47}+2\right)\)
<=> \(\frac{x+102}{50}+\frac{x+102}{49}=\frac{x+102}{48}+\frac{x+102}{47}\)
<=> \(\left(x+102\right)\left(\frac{1}{50}+\frac{1}{49}-\frac{1}{48}-\frac{1}{47}\right)=0\)
Mà \(\frac{1}{50}+\frac{1}{49}-\frac{1}{48}-\frac{1}{47}\ne0\)
<=> x + 102 = 0
<=> x = -102
oki cảm ơn bạn nhó <3