tìm x, biết x+2x+3x+....+1000x =5005:104
Những câu hỏi liên quan
a) tính :
1+2+3+...+1000
b) Tìm x biết
x+2x+3x+....+1000x = 5005:104
a) tù 1-1000 có:1000 số
1+2+3+..1000=(1000+1).1000:2=500500
b)x+2x+3x+...+1000x=385/8
x(1+2+3+...+1000)=385/8
x.500500=385/8
x=1/10400
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Tìm x, biết:
x + 2x + 3x + ... + 1000x = 5005 . 104
x+2x+3x+...+1000x=5005x\(10^4\)
ta có:x+2x+3x+...+1000x=x(1+2+3+...+1000)
=x((1000+1):2x1000)
=x500500
\(\Rightarrow\) x500500=5005x\(10^4\)
x5005.\(100\)=5005x\(10^4\)
x5005.\(10^2\)=5005x\(10^4\)
\(\Rightarrow x.10^2=10^4\)
x=\(10^4:10^2\)
x=\(10^2\)
x=100
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27x^3 - 27 x^2 +3x - 1
1/27 + x^3
x^3- 3x^2+3x-1
0,001-1000x^3
12/5 x^2y^2-9x^4 - 4/25y^4
a^2y^2+b^2x^2-2axby
100-(3x-y)^2
64x^2-(8a+b)^2
27x^3-a^3b^3
b: \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
c: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
e: \(a^2y^2-2axby+b^2x^2\)
\(=\left(ay\right)^2-2\cdot ay\cdot bx+\left(bx\right)^2\)
\(=\left(ay-bx\right)^2\)
f: \(100-\left(3x-y\right)^2\)
\(=\left(10-3x+y\right)\left(10+3x-y\right)\)
g: \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
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Tìm x
a. 3x(4x-2) - 4x (3x-1) = 6
b. (3x-2)(4x-1) - (6x+1)(2x-3) = -7+x
c. (x+2)(x+4) - x(x+2) = 104
d. (x-1)(x-4) - x(x-1) = 308
a) 3x ( 4x - 2 ) - 4x ( 3x - 1 ) = 6
12 x2 - 6x - 12 x2+ 4x = 6
( 12 x2 - 12 x2 ) - ( 6x - 4x ) = 6
0 - 2x = 6
2x = 6
x = 3
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a) 3x ( 4x - 2 ) - 4x ( 3x - 1 ) = 6
12 x2 - 6x - 12 x2+ 4x = 6
( 12 x2 - 12 x2 ) - ( 6x - 4x ) = 6
0 - 2x = 6
2x = 6
x = 3
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Tìm x
a, (x+2)^3 - x(x^2+6x-3)=0
b,(x+4)^3 - x(x+6)^2=7
c, (x-1)^3 - x(x^2-3x-2)=0
d,1000x^3+90x(10x+3)+27=0
e,125x^3+15x(5x+1)= -1
f, 27x^3+45x(3x+5)+133=0
a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)
\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)
b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P
Tự làm nốt nhé
Cho P(x) = x^2019- 1000x^2018 + 1000x^2017- 1000x^2016 +...+ 1000 x - 1.Tính P(999).
P(x) = x2019 - 1000x2018 + 1000x2017 - 1000x2016 + ... + 1000x - 1
Với x = 999 => 1000 = x + 1
=> P(999) = x2019 - ( x + 1 )x2018 + ( x + 1 )x2017 - ( x + 1 )x2016 + ... + ( x + 1 )x - 1
= x2019 - x2019 - x2018 + x2018 + x2017 - x2017 - x2016 + ... + x2 + x - 1
= x - 1 = 999 - 1 = 998
Vậy ...
a , ( x\(^2\)- 3x - 5 ) ⋮ ( x - 3 )
b , ( 5x + 2 ) ⋮ ( x + 1 )
c , ( 2x\(^2\)+ 3x + 2 ) ⋮ ( x + 1 )
a) \(x^2-3x-5=x\left(x-3\right)-5\)
Để \(^2-3x-5\)chia hết cho x-3 thì x(x-3) -5 phải chia hết cho x-3
mà x(x-3) chia hết cho x-3 => -5 phải chia hết cho x-3
=> x-3\(\inƯ\left(-5\right)=\left\{-1;-5;1;5\right\}\)
Lập bảng giải tiếp
\(5x+2=5\left(x+1\right)-3\)
Để 5x+2 chia hết cho x+1 thì 5(x+1)-3 phải chia hết cho x+1
mà 5(x+1) chia hết cho x+1
=> -3 phải chia hết cho x+1
=> x+1\(\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\)
Lập bảng giải tiếp nhé! :3
Tìm x
a) (x+8)(x+6) -x^2 =104
b) 6x^2 -(2x-3)(3x+2) -1=0
c) (x+4)^2 +(x+3)^2 -5(x+7)=0
d) (10x+9)x -(5x -1)(2x+3) =8
\(a,\left(x+8\right)\left(x+6\right)-x^2=104\)
\(\Rightarrow x^2+14x+48-x^2=104\)
\(\Rightarrow14x=56\)
\(\Rightarrow x=4\)
Vậy x=4
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cho A(x)=x^99-109x^98+x^98-109x^97+x^97-109x^96+...+x-109
Tính A(109)
cho B(x)=x^49-10x^48+10x^47-10x^46+...+10x-1
Tính B(9)
cho C(x)=x^999-1000x^998+1000x^997-1000x^996+...+1000x-1
Tính C(999)
Giúp mình với!mai thi rồi
a: \(A\left(109\right)=x^{99}-x^{98}\cdot x+x^{98}-x^{97}\cdot x+...+x-x\)
\(=x^{99}-x^{99}+x^{98}-x^{98}+...+0\)
=0
b: x=9 nên x+1=10
\(B\left(x\right)=x^{49}-x^{48}\left(x+1\right)+x^{47}\left(x+1\right)-...+x\left(x+1\right)-1\)
\(=x^{49}-x^{49}-x^{48}+x^{48}+...+x^2+x-1\)
=x-1=8
c: x=999 nên x+1=1000
\(C\left(x\right)=x^{999}-x^{998}\left(x+1\right)+x^{997}\left(x+1\right)-...+x\left(x+1\right)-1\)
\(=x^{999}-x^{999}-x^{998}+x^{998}+...+x^2+x-1\)
=x-1=998
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