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
Những câu hỏi liên quan
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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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)Ta có: 109=x =>x99-109x98+109x97-...+x-109
A(109)=(x99-xx98)+(xx97-xx96 )+...+(x-x)=(x99-x99)+(x97-x97)+...+0=0
Vậy A(109)=0
b) Ta có :x=9 => x49-(x+1)x48+(x+1)x47+...+(x+1)x-1
B(9)=x49-x49-x48+x48+x47-...+x2+x-1=x-1=9-1=8
Vậy :B(9)=8
c) Làm tương tự câu b => C(999)=998
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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 ...
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
Bài 1
a,(2x+1)2/5 - (x-1)2/37x2-14x-5/15
b,(x+10)(x+4)/12 - (x-4)(2-x)/4(x+10)(x-2)/3
c,(x-2)2/3 - (2x-3)(2x+3)/8 + (x-4)2/60
Bài 2
a,x-23/24 + x-23/25x-23/26 + x-23/27
b,(x+2/98 + 1)+(x+3/97 + 1)(x+4/96 + 1)+(x+5/95 + 1)
c,x+1/2004 + x+2/2003x+3/2002 + x+4/2001
d,201-x/99 + 203-x/97205-x/95 + 30
e,x-45/55 + x-47/53x-55/45 + x-53/47
f,x+1/9 + x+2/8x+3/7 + x+4/6
g,x+2/98 +x+4/96x+6/94 + x+8/92
h,2-x/2002 - 11-x/2003 - x/2004
i,x2-10x-29/1971...
Đọc tiếp
Bài 1
a,(2x+1)2/5 - (x-1)2/3=7x2-14x-5/15
b,(x+10)(x+4)/12 - (x-4)(2-x)/4=(x+10)(x-2)/3
c,(x-2)2/3 - (2x-3)(2x+3)/8 + (x-4)2/6=0
Bài 2
a,x-23/24 + x-23/25=x-23/26 + x-23/27
b,(x+2/98 + 1)+(x+3/97 + 1)=(x+4/96 + 1)+(x+5/95 + 1)
c,x+1/2004 + x+2/2003=x+3/2002 + x+4/2001
d,201-x/99 + 203-x/97=205-x/95 + 3=0
e,x-45/55 + x-47/53=x-55/45 + x-53/47
f,x+1/9 + x+2/8=x+3/7 + x+4/6
g,x+2/98 +x+4/96=x+6/94 + x+8/92
h,2-x/2002 - 1=1-x/2003 - x/2004
i,x2-10x-29/1971 + x2-10x -27/1973=x2-10x-1971/29 + x2-10x-1973/27
Hơi khó hiểu thông cảm ạ
giải các phương trình
a)\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
b) \(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
c) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
d)\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)
Các câu na ná chắc nên mk làm mẫu 2 bài thui nha !
a, pt <=> x-23/24 + x-23/25 - x-23/26 - x-23/27 = 0
<=> (x-23).(1/24+1/25-1/26-1/27) = 0
<=> x-23=0 ( vì 1/24+1/25-1/26-1/27 > 0 )
<=> x=23
b, pt <=> (201-x/99 + 1)+(203-x/97 + 1)+(205-x/95 + 1) = 0
<=> 300-x/99 + 300-x/97 + 300-x/95 = 0
<=> (300-x).(1/99+1/97+1/95) = 0
<=> 300-x = 0 ( vì 1/99+1/97+1/95 > 0 )
<=> x=300
Tk mk nha
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I : Giải các phương trình sau :
a) dfrac{x+1}{2004}+dfrac{x+2}{2003}dfrac{x+3}{2002}+dfrac{x+4}{2001}
b) dfrac{x-45}{55}+dfrac{x-47}{53}dfrac{x-55}{45}+dfrac{x-53}{47}
c) dfrac{x+2}{98}+dfrac{x+4}{96}dfrac{x+6}{94}+dfrac{x+8}{92}
d) dfrac{201-x}{99}+dfrac{203-x}{97}+dfrac{205-x}{95}+30
e) dfrac{x+1}{9}+dfrac{x+2}{8}dfrac{x+3}{7}+dfrac{x+4}{6}
f) dfrac{2-x}{2002}-1dfrac{1-x}{2003}-dfrac{x}{2004}
g) dfrac{x^2-10x-29}{1971}+dfrac{x^2-10x-27}{1973}dfrac{x^2-10x-1971}{29}+dfrac{x^2-10x-197...
Đọc tiếp
I : Giải các phương trình sau :
a) \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
b) \(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
c) \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
d) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
e) \(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)
f) \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
g) \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
các bạn ơi giúp mình với
mình cảm ơn
b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100
Vậy x = 100
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\(a,\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}-\dfrac{x+3+2002}{2002}-\dfrac{x+4+2001}{2001}=0\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
Vậy pt có tập nghiệm S = { 2005 }
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Xem thêm câu trả lời
1.\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
2.\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
Cho hàm số
1000
x
-
1
+
x
-
2
x
2
-
1...
Đọc tiếp
Cho hàm số 1000 x - 1 + x - 2 x 2 - 1 k h i x > 1 2 a x k h i x ≤ 1 .Tìm a để hàm số liên tục tại x=1?
A. 3 log 10 2
B. 3 ln 10 2
C.Đáp án khác
D. 3 ln 10 + 1 4
