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(4x+5):3-121:11=4
(4x+5):3-11=4
(4x+5):3=15
4x+5=45
4x=40
x=10.
1+3+5+...+x=1600
Gọi x=2k+1
Ta có 1+3+...+x=1600
1+3+..+2k+1=1600
(2k+2)[(2k+1-1)/2+1]/2=1600
(2k+2)(k+1)=3200
2(k+1)(k+1)=3200
(k+1)^2=1600
=>k+1=40
k=39. Vậy x=2k+1=2.39+1=79.
2^x+2^x+3=144
2^x(1+2^3)=144
2^x.9=144
2^x=16
2^x=2^4
=>x=4
(x-5)^4=(x-5)^6
(x-5)^6-(x-5)^4=0
(x-5)^4[(x-5)^2-1]=0
(x-5)^4=0, (x-5)^2-1=0
x-5=0, (x-5)^2=1
x=5, x-5=1, x-5=-1
x=5, x=6, x=4.
1+3+5+...+x=1600
=(x+1).[(x-1):2+1] /2 =1600
=(x+1).(x+1) /2 =1600
=(x+1)^2:2=40^2
=(x+1):2=40
=x+1=80
=x=79
1 + 3 + 5 + ... + x = 1600
{(n - 1)/(2) + 1} ^ 2 = 1600
(n-1)/(2)+1 = 40^2
(n-1)/(2)+1 = 40
(n-1)/(2) = 40-1
(n-1)/(2) = 39
n-1 = 39*2
n-1 = 78
n = 78 +1
n = 79
a ) \(\left(4x+5\right)\div3-121\div11=4\)
\(\left(4x+5\right)\div3-11=4\)
\(\left(4x+5\right)\div3=4+11\)
\(\left(4x+5\right)\div3=15\)
\(\left(4x+5\right)=15\cdot3\)
\(4x+5=45\)
\(4x=45-5\)
\(4x=40\)
\(x=10\)
(4x + 5) : 3 - 121 : 11 = 4
=> (4x + 5) : 3 - 11 = 4
=> (4x + 5) : 3 = 15
=> 4x + 5 = 45
=> 4x = 40
=> x = 10
b) 1 + 3 + 5 + ... + x = 1600
=>[(x - 1) : 2 + 1] . (x + 1) : 2 = 1600
=> \(\left(\frac{x}{2}-\frac{1}{2}+1\right).\frac{x+1}{2}=1600\)
=> \(\frac{x+1}{2}.\frac{x+1}{2}=1600\)
=> \(\left(\frac{x+1}{2}\right)^2=1600\)
=> \(\frac{x+1}{2}=40\)
=> x + 1 = 80
=> x = 79