5x+1+5x=150
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LT
0
CN
1
NT
0
TT
1
10 tháng 8 2016
a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\Leftrightarrow\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right).\left(4x+5\right)\)
\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)
\(\Leftrightarrow4x+30x-25x-8x=10-6\)
\(\Leftrightarrow1x=4\)
\(\Leftrightarrow x=4:1\)
\(\Leftrightarrow x=4\)
10 tháng 8 2016
b ) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)
\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)
\(\Leftrightarrow138x=966\)
\(\Leftrightarrow x=966:138\)
\(\Leftrightarrow x=7\)
9 tháng 2 2021
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 ...
9 tháng 2 2021
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}
NV
5 tháng 8 2015
a/ \(=5x^2\left(x-y\right)-10x\left(x-y\right)=\left(x-y\right)\left(5x^2-10x\right)=5x\left(x-2\right)\left(x-y\right)\)
b/ \(=2x^2-7x+2x-7=x\left(2x-7\right)+\left(2x-7\right)=\left(x+1\right)\left(2x-7\right)\)
NN
0
TM
1
Trường hợp 1 : 5x+1 + 5x = 150
=> 5x . 5 + 5x = 150
=> x không thỏa mãn
Trường hợp 2 : 5x+1 + 5x = 150
=> 5x . 5 + 5x = 150
=> 5x (5 + 1) = 150
=> 5x . 6 = 150
=> 5x =150 : 6 = 25
=> x = 2
Vậy x = 2