5x+2 - 5x = 6. 102
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Ta có : \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\Leftrightarrow5x-200=0\)
\(\Leftrightarrow x=40\)
Vậy ...
Ta có: \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}>0\)
nên 5x-200=0
\(\Leftrightarrow5x=200\)
hay x=40
Vậy: S={40}
\(pt\Leftrightarrow\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-16}{46}-14=0\)
\(\Leftrightarrow\frac{5x-150}{50}-1+\frac{5x-102}{49}-2+\frac{5x-56}{48}-3+\frac{5x-12}{47}-4+\frac{5x-16}{46}-4=0\)
\(\Leftrightarrow\frac{5x-200}{50}+\frac{5x-200}{49}+\frac{5x-200}{48}+\frac{5x-200}{47}+\frac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\right)=0\)
Do \(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\ne0\) nên \(5x-200=0\Rightarrow x=\frac{200}{5}=40\)
Vậy x= 40
\(\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-660}{46}=0\)
\(\Leftrightarrow\)\(\left(\frac{5x-150}{50}-1\right)+\left(\frac{5x-102}{49}-2\right)+\left(\frac{5x-56}{48}-3\right)+\left(\frac{5x-12}{47}-4\right)+\left(\frac{5x-660}{46}+10\right)=0\)
\(\Leftrightarrow\)\(\frac{5x-200}{50}+\frac{5x-200}{49}+\frac{5x-200}{48}+\frac{5x-200}{47}+\frac{5x-200}{46}=0\)
\(\Leftrightarrow\)\(\left(5x-200\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\right)=0\)
\(\Leftrightarrow\)\(5x-200=0\)
\(\Leftrightarrow\)\(5x=200\)
\(\Leftrightarrow\)\(x=40\)
Vậy x = 40
a, 71.2 – 6.(2x+5) = 10 5 : 10 3
71.2 – 6.(2x+5) = 10 2
6.(2x+5) = 71.2 – 100
6.(2x+5) = 42
x = 1
b, 5 x + 3 4 . 6 8 = 6 9 . 3 4
5 x + 3 4 . 6 8 = 6 8 . 6 . 3 4
5 x + 3 4 = 6 8 . 6 . 3 4 : 6 8 = 6 . 3 4
5x = 6 . 3 4 - 3 4 = 5 . 3 4
x = 3 4
c, 12:{390:[5. 10 2 – ( 5 3 + x . 7 2 )]} = 4
390:[5. 10 2 – ( 5 3 + x . 7 2 )] = 12:4 = 3
5. 10 2 – ( 5 3 + x . 7 2 ) = 390:3 = 130
5 3 + x . 7 2 = 5. 10 2 – 130 = 370
x . 7 2 = 370 – 5 3 = 245
x = 245: 7 2 = 5
d, 5 3 .(3x+2):13 = 10 3 : 13 5 : 13 4
5 3 .(3x+2):13 = 10 3 : 13
3x+2 = 10 3 : 13 : 5 3 .13 = 8
x = 2
b) \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\Leftrightarrow5x-200=0\)
\(\Leftrightarrow x=40\)
b)
\(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Rightarrow\left(\dfrac{5x-150}{50}-1\right)+\left(\dfrac{5x-102}{49}-2\right)+\left(\dfrac{5x-56}{48}-3\right)+\left(\dfrac{5x-12}{47}-4\right)\)
\(+\left(\dfrac{5x-660}{46}+10\right)=0\)
\(\Rightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Rightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\ne0\)
\(\Rightarrow5x-200=0\Rightarrow x=40\)
\(\left|5x+3\right|-9x=102\)( điều kiện: \(x\ge-\frac{3}{2}\))
\(\Leftrightarrow5x+3-9x=102\)
\(\Leftrightarrow-4x=99\)
\(\Leftrightarrow x=-\frac{99}{4}\)(thỏa mãn)
vậy...
a ) 5x - 32 = 48
5x = 48 + 32
5x = 80
x = 80 : 5
x = 16
b ) 215 - ( 2x + 143 ) = 16
( 2x + 143 ) = 215 - 16
( 2x + 143 ) = 199
2x = 199 -143
2x = 56
x = 56 : 2
x = 28
bài c hình như sai đề rồi bạn
\(\left(5x-6\right)^2-\left(3x-7\right)^2=-9x\left(x-8\right)+\left(5x-6\right)^2-13\)
\(\Rightarrow25x^2-60x+36-9x^2+42x-49=-9x^2+72x+25x^2-60x+36-13\)
\(\Rightarrow\left(25x^2-25x^2\right)-\left(60x-60x\right)+\left(36-36\right)-\left(9x^2-9x^2\right)+\left(42x-72x\right)-\left(49-13\right)=0\)
\(\Rightarrow-30x-36=0\)
\(\Rightarrow-30x=36\)
\(\Rightarrow x=-\dfrac{36}{30}\)
\(\Rightarrow x=-\dfrac{6}{5}\)
Vậy: \(x=-\dfrac{6}{5}\)
`5^{x+2}-5^{x}=6.10^2`
`5^{x}(5^{2}-1)=6.100`
`5^{x}.24=600`
`5^{x}=25`
`5^{x}=5^{2}`
`x=2`