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\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Leftrightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Leftrightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-5\right)^4\left(x-6\right)\left(x-4\right)=0\)
\(\Leftrightarrow x=5;x=6;x=4\)
\(x^{2006}=x^2\)
\(\Leftrightarrow x^{2006}-x^2=0\)
\(\Leftrightarrow x^2\left(x^{2004}-1\right)=0\)
\(\Leftrightarrow x^2\left(x^{1002}-1\right)\left(x^{1002}+1\right)=0\)
\(\Leftrightarrow x^2\left(x^{1001}+1\right)\left(x^{1001}-1\right)\left(x^{1002}+1\right)=0\)
\(\Rightarrow x=0;x=-1;x=1\)
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a)(x-6)2 . (x-6) =64
(x-6)3=64
(x-6)3=43
=>x-6=3
x=9
b)(x+5)3 : (x+5) =144
(x+5)2=144
(x+5)2=122
=>x+5\(\in\left\{12;-12\right\}\)
\(\Rightarrow x\in\left\{7;-17\right\}\)
c)3x+42=196 : (194 . 19)-2.12036
3x+42=196 :195-2
3x+42=19-2
3x+42=17
3x+16=17
3x=1
=>x=0
d)(19x + 2.52) =52 - 42
(19x + 2.52) =25-16
(19x + 2.52) =9
19x + 50 =9
19x = -41
x=\(\frac{-41}{19}\)
tính hợp lí(nếu có thể)
(52004 - 52006):(52005.5)
=(52004 - 52006):52006
=(52004 - 52006).\(\frac{1}{5^{2006}}\)
=\(\frac{5^{2004}}{5^{2006}}\) - 1
=\(\frac{1}{5^2}\) -1
=\(\frac{1}{25}-1=\frac{-24}{25}\)
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a)Ta có:04=06
14=16
(-1)4=(-1)6
=>x-5 E {-1;0;1}
=>x E {4;5;6}
b)Ta có: 02006=02
12006=12
(-1)2006=(-1)2
=>x E {-1;'0;1}
a) (x-5)4 = (x-5)6
=>(x-5)4 - (x-5)6=0
=>(x-5)4[1-(x-5)2]=0
=>(x-5)4[1-x2+10x-25]=0
=>(x-5)4[6x-24+4x-x2]=0
=>(x-5)4[6(x-4)-x(x-4)]=0
=>(x-5)4(6-x)(x-4)=0
=>(x-5)4=0 hoặc 6-x=0 hoặc x-=0
=>x=5 hoặc x=6 hoặc x=4
b)x2006=x2
=>x2006-x2=0
=>x2(x2004-1)=0
=>x2=0 hoặc x2004-1=0
=>x=0 hoặc x=1 hoặc -1
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a. 2x . 4 = 128
2x = 32
2x = 25
=> x = 5
b. (2x+1)3 = 125
(2x+1)3 = 53
=> 2x+1 = 5
2x = 5 - 1
2x = 4
x = 4:2
x = 2
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a,X^2006-X^2=0
X^2(X^2014-1)=0
X^2=0;x^2014=1
X=-1;0;1
b,2^x+2-2^x=96
2^x.4-2^x=96
2^x(4-1)=96
2^x=96:3
2^x=32
2^x=2^5
x=5
c,x.(x^2)^3=x^5
x.x^6=x^5
x^7-x-5=0
...
x=-1;0;1
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23x + 20060 x = 95:81
8x +x = 95: 92
9x=93
x=93:9
x=92
x=81
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Bài 3:
a. 5x.5x.5x = 5x3
b.x1.x2......x2006 = x1+2+..+2006
c.x.x4.x7....x100 = x1+4+7+...+100
d.x2.x5.x8....x2003 = x2+5+8+..+2003
Bài 3 :
a) 5x . 5x . 5x = ( 5x )3
b) x1 . x2 . ... . x2006
= x1+2+3+...+2006
= x(2006+1).2006:2
= x2013021
c) x . x4 . x7 . ... . x100
= x1+4+7+...+100
= x(100+1).34:2
= x1717
d) x2 . x5 . x8 . ... . x2003
= x2+5+8+...+2003
= x(2003+2).668:2
= x669670
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g) C = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + ... + 2002 - 2003 - 2004 + 2005 + 2006
Số số hạng của C từ 1 đến 2004 là : ( 2004 - 1 ) : 1 + 1 = 2004 số
Nhóm 4 số thành 1 cặp ta được : 2004 : 4 = 501 cặp
=> C = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + ... + ( 2001 + 2002 - 2003 - 2004 ) + ( 2005 + 2006 )
C = -4 + ( -4 ) + ... + ( -4 ) + 4011
C = -4 . 501 + 4011
C = -2004 + 4011
C = 2007
h) D = 12 - 22 + 32 - 42 + ... + 992 - 1002 + 1012
=> D = ( 12 - 22 ) + ( 32 - 42 ) + ... + ( 992 - 1002 ) + 1012
=> D = (−1)(1 + 2) + (−1)(3 + 4)+. . . . +(−1)(99 + 100) + 1012
=> D = −(1 + 2+. . . . . +99 + 100) + 1012
=> D = \(-\frac{100\left(100+1\right)}{2}+101^2=101^2-50\cdot101=101\cdot51=5151\)
Vậy D = 5151
\(\frac{-4}{8}=\frac{x}{-10}=\frac{-7}{y}=\frac{z}{-24}\)* \(\frac{-4}{8}=\frac{x}{-10}=\frac{-7}{y}=\frac{z}{-24}\)
Ta có : \(-4\cdot\left(-10\right)=8\cdot x\Rightarrow x=5\)
\(5y=-7\cdot\left(-10\right)\Rightarrow y=14\)
\(\frac{-7}{14}=\frac{z}{-24}\Rightarrow-7\cdot\left(-24\right)=14\cdot z\Rightarrow z=12\)
* \(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
=> x thuộc { -3 ; -2 ; -1 ; 0 ; 1 ; 2 ; 3 ; 4 }
~ Xin lỗi . Mình chỉ giúp được đến đây thôi :( ~
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a) Có : \(6^{x+2}-6^{x+1}=1080\)
=> \(6^{x+1}.\left(6-1\right)=6^{x+1}.5=1080\)
=> \(6^{x+1}=1080:5=216\)
=> \(6^{x+1}=6^3=216\)
=> x+1 = 3
=> x = 3 - 1 = 2
Vậy x = 2
Ủng hộ mik nhá
Câu x thứ 1: x = 4
Câu x thứ 2 : x = 1 hoặc x = -1
ý a)\(\left(x-5\right)^4=\)\(\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4=\)\(\left(x-5\right)^4\left(x-5\right)\)
\(\Rightarrow1=x-5\)
\(\Rightarrow x=4\)
ý b)\(x^{2006}=x^2\)
vì 2006 ; 2 là số chẵn
\(\Rightarrow x=1;-1\)