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`(x+1)/99+(x+2)/98+(x+3)/97+(x+4)/96=-4`
`=>(x+1)/99+1+(x+2)/98+1+(x+3)/97+1+(x+4)/96+1=-4+4`
`=>(x+100)/99+(x+100)/98+(x+100)/97+(x+100)/96=0`
`=>(x+100)(1/99+1/98+1/97+1/96)=0`
`=>x+100=0` (Vì `1/99+1/98+1/97+1/96\ne0`)
`=>x=-100`
Vậy ...
`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\\ \dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}+4=0\\ \left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\\ \dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\\ \left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)
mà `1/99+1/98+1/97+1/96 \ne 0`
nên `x+100=0`
`x=-100`
a, \(15:\left(x+2\right)=3\Leftrightarrow x+2=3\Leftrightarrow x=1\)
b, \(20:\left(x+1\right)=2\Leftrightarrow x+1=10\Leftrightarrow x=9\)
c, \(240:\left(x-5\right)=2^2.5^2-20=80\Leftrightarrow x-5=3\Leftrightarrow x=8\)
d, \(96-3\left(x+1\right)=42\Leftrightarrow3\left(x+1\right)=54\Leftrightarrow x+1=18\Leftrightarrow x=17\)
a) 15 : (x + 2) = 3
x + 2 = 15 : 3
x + 2 = 5
x = 5 – 2 = 3
b) 20 : (1 + x) = 2
1 + x = 20 : 2
1 + x = 10
x = 10 – 1 = 9
c) 240 : (x – 5) = 22.52 – 20
240 : (x – 5) = 4.25 – 20
240 : (x - 5) = 100 – 20
240 : (x - 5) = 80
x – 5 = 240 : 80
x – 5 = 3
x = 3 + 5 = 8
d) 96 - 3(x + 1) = 42
3(x + 1) = 96 – 42
3(x + 1) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 – 1
x = 17
c)96 - 3.(x + 1) = 42
3.(x + 1)= 96 - 42
3.(x + 1) = 54
x + 1= 54:3
x + 1= 18
x = 18 - 1
x = 17
\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+4}{96}+\frac{x+1}{99}=-4\)
\(\Rightarrow\frac{x+3}{97}+1+\frac{x+5}{95}+1+\frac{x+4}{96}+1+\frac{x+1}{99}+1=0\)
\(\Rightarrow\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{96}+\frac{x+100}{99}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\right)=0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
Vậy x = -100
\(2^{x+2}-2^x=96\)
\(\Rightarrow2^x\cdot2^2-2^x=96\)
\(\Rightarrow2^x\left(2^2-1\right)=96\)
\(\Rightarrow2^x\left(4-1\right)=96\)
\(\Rightarrow2^x\cdot3=96\)
\(\Rightarrow2^x=96:3\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(5^x+5^{x+1}=750\)
\(\Rightarrow5^x+5^x\cdot5=750\)
\(\Rightarrow5^x\left(1+5\right)=750\)
\(\Rightarrow5^x\cdot6=750\)
\(\Rightarrow5^x=750:6\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
\(2^{x+3}+2^x=144\)
\(\Rightarrow2^x\cdot2^3+2^x=144\)
\(\Rightarrow2^x\left(2^3+1\right)=144\)
\(\Rightarrow2^x\cdot9=144\)
\(\Rightarrow2^x=144:9\)
\(\Rightarrow2^x=16\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
a. 96 - 3(x + 1) = 42
<=> 96 - 3x - 3 = 42
<=> 96 - 3 - 42 = 3x
<=> 3x = 51
<=> x = 17
b. (x - 37) : 12 = 9
<=> \(\dfrac{x}{12}-\dfrac{37}{12}=9\)
<=> \(\dfrac{x}{12}-\dfrac{37}{12}=\dfrac{108}{12}\)
<=> x - 37 = 108
<=> x = 108 + 37
<=> x = 145
c. 125 : (x - 3) = 5
<=> \(\dfrac{125}{x}-\dfrac{125}{3}=5\)
<=> \(\dfrac{375}{3x}-\dfrac{125x}{3x}=\dfrac{625x}{3x}\)
<=> 375 - 125x = 625x
<=> 375 = 625x + 125x
<=> 500x = 375
<=> x = \(\dfrac{375}{500}=\dfrac{3}{4}\)
96-3(x+1)=42
3(x+1)=96-42
3(x+1)=54
x+1 = 54 : 3
x+1 = 18
x = 18 - 1
x = 17
96 - 3(x + 1) = 42
\(\Rightarrow\)3x + 3 = 54
\(\Rightarrow\)3x = 51
\(\Rightarrow\)x = 17
96 – 3(x + 1) = 42
3(x + 1) = 96 - 42
3(x + 1) = 54
x + 1 = 54:3
x + 1 = 18
x = 18 - 1
x = 17
Vậy x = 17
=> 3(x+1)=96-42
=> x+1=54:3
=> x=18-1
=> x=17
X=17 k đi