Tìm x biết :
a) x + ( x + 1 ) + ( x + 2) + ........+ ( x + 2006 ) +2007 =2007
b) 2000+ ( 199 +x ) + ( 198 + x ) + ( x + 1 ) + x = 200
Bài 1: Tìm x biết:
1/ x + (x + 1) + (x + 2) + (x + 3) +.....+ (x + 2006) + 2007 = 2007
2/ x + (x + 1) + (x + 2) +...+ 199 + 200 + 201 = 401
3/ x + (x + 1) + (x + 2) +....+ 2009 + 2010 = 2010
Bn nào nhanh mik tick nha!!!
1/x+x+1+x+2+x+3+...+x+2006+2007=2007
------------------------------------------=2007-2007
------------------------------------------=0
x+x+x+...+x+1+2+3+...+2006=0
2007.x+(1+2+...+2006)=0
2007.x+(2006+1).[(2006-1)+1]:2=0
2007.x+2013021=0
2007.x=0-2013021
x=-2013021:2007
x=-1003
2/x+x+1+x+2+...+x+198=401-201-200-199
199.x+(1+2+...+198)=-199
199.x+(1+198).[(198-1)+1]:2=-199
199.x+19701=-199
199.x=-199-19701
x=-19900:199
x=-100
3/x+x+1+x+2+...+x+2008=2010-2010-2009
2009.x+(2008+1).[(2008-1)+1]:2=-2009
2009.x+2017036=-2009
2009.x=-2009-2017036
x=-2019045:2009
x=-1005
Tìm các số nguyên x biết:
a,x+(x+1)+(x+2)+...+2006+2007=2007
b,2000+1999+...+x+1+x=2000
Lời gải:
a. Số số hạng:
$(2007-x):1+x=2008-x$
Suy ra:
$x+(x+1)+(x+2)+....+2006+2007=2007$
$\frac{(x+2007)(2008-x)}{2}=2007$
$(x+2007)(2008-x)=4014=$
$\Rightarrow x=2007$ hoặc $x=-2006$
b.
Số số hạng: $(2000-x):1+1=2001-x$
Suy ra:
$2000+1999+...+(x+1)+x=2000$
$\frac{(2000+x)(2001-x)}{2}=2000$
$(2000+x)(2001-x)=4000$
$\Rightarrow x=2000$ hoặc $x=-1999$
Cho A = 2007 x 2007 x ... x 2007 x 2007 ( có 2006 thừa số 2007 )
Cho B = 2006 x 2006 x … x 2006 x2006 ( có 2007 thừa số 2006 )
Hãy cho biết hiệu A và B có chia hết cho 5 không ? Tại sao ?
Ta co
A=2007^2006( lên lơp 6 e se hoc)
=>A=2007^2 x 2007^2004
=>(...9)x(...1)=(...9) (1)
Ta co:
B=2006^2007=(...6)
A=....9
B=....6
A-B= ....9- ....6= ....3 không chia hết cho 5
=> A-B=(...9)-(...6)=(...3) khong chi het cho 5
tìm giá trị các đa thức sau
\(A=x^{15}+3x^{14}+5\) biết x+3=0
\(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\) biết x= -3
a) \(A=x^{15}+3x^{14}+5\)
\(=x^{14}\left(x+3\right)+5\)
\(=x^{14}.0+5\)
= 5
b) x = -3 => x + 3 = 0
\(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)
\(=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)
\(=\left(x^{2006}.0+1\right)^{2007}\)
\(=1^{2007}=1\)
\(A=x^{15}+3.x^{14}+5\text{ biết x+3=0}\)
\(A=x^{14}.\left(x+3\right)+5\)
\(\text{Do x+3=0}\Rightarrow A=x^{14}.0+5\)
\(A=0+5\)
\(A=5\) \(\text{Vậy }A=5\text{ với x+3=0}\)
\(B=\left(x^{2007}+3.x^{2006}+1\right)^{2007}\text{ biết x=-3}\)
\(B=\left[x^{2006}.\left(x+3\right)+1\right]^{2007}\)
\(\text{Do x=-3}\Rightarrow B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)
\(B=\left(x^{2006}.0+1\right)^{2007}\)
\(B=\left(0+1\right)^{2007}\)
\(B=1^{2007}\)
\(B=1\) \(\text{Vậy }B=1\text{ với x=-3}\)
Tìm x biết : \(\frac{\left(2006-x\right)^2+\left(2006-x\right)\left(x-2007\right)+\left(x-2007\right)^2}{\left(2006-x\right)^2-\left(2006-x\right)\left(x-2007\right)+\left(x-2007\right)^2}=\frac{19}{49}\)
Đặt x -2006 = y
pt <=> \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)
<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)
<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)
<=> \(49y^2-49y+49=57y^2-57y+19\)
<=> \(8y^2-8y-30=0\)
<=> \(4y^2-4y+15=0\)
Giải tiếp nha
Tìm x biết
a) (x + 2006/2007) ^2
b)(x - 1/5 ) ^3 = 8/125
b)
(x-1/5)3=8/125
(x-1/5)3=(2/5)3
=>x-1/5=2/5
x=2/5+1/5
x=3/5
Tìm x biết : x - 1 / 2009 + x - 2 / 2008 = x - 3 / 2007 + x- 4 / 2006
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\right)=0\)
\(\Leftrightarrow x=2010\)
1/ tìm x biết |2x+3|=x+2
2/ tìm gtnn của A=|x-2006|+|2007-x|
1/
l2x+3l=x+2(1)
ta co l2x+3l=\(\hept{\begin{cases}2x+3voix\ge\frac{-3}{2}\\-2x-3voix< \frac{-3}{2}\end{cases}}\)
TH1: neu x>= -3/2 thi (1) <=>2x+3=x+2=>x=-1(chon)
TH2: neu x<= -3/2 thi (1) <=> -2x-3=x+2=>-3x=5=>x=-5/3(chon)
2/
de A dat gtnn thi lx-2006l va l2007l dat gtnn
ma lx-2006l va l2007-xl >=0
=> gtnn cua lx-2006l=0;l2007-xl=0
=> x=2006 hoac 2007
=> gtnn A=1
Tìm x thuộc Z, biết:
x + (x+1) + (x+2) +...+ 2006 + 2007 = 2007