5 ( x - 7 ) = 0
25 ( x - 4 ) =0
34 ( 2 x - 6 ) =0
2016 ( 3 x - 12 ) = 0
47 ( 5 x - 15 ) = 0
Giúp mk ik T^T chi tiết hộ mình
Tìm x
5 ( x - 7 ) = 0
25 ( x - 4 ) = 0
342 ( 2 x - 6 = 0
2016 ( 3 x - 12 ) = 0
47 ( 5 x - 15 ) = 0
Cứu tick cho
342(2x-6)=0
=>x=3
2016(3x-12)=0
=>x=4
47(5x-15)=0
=5x-15=0
=>5x=15
=>x=3
Tìm x:
1) x.(x+7) = 0
2) (x + 12) . (x-3) = 0
3) (-x + 5) . (3-x) = 0
4) x .(2 + x) . (7-x) = 0
5) ( x-1) . (x+2). (-x - 3) = 0
(Mk cần gấp , ai trả lời nhanh , đầy đủ và chi tiết nhất thì mk sẽ tick)
1) \(x.\left(x+7\right)=0\)
\(=>\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.=>\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
2) \(\left(x+12\right).\left(x-3\right)=0\)
\(=>\left[\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.=>\left[\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
3) \(\left(-x+5\right).\left(3-x\right)=0\)
\(=>\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.=>\left[\begin{matrix}x=5\\x=3\end{matrix}\right.\)
4) \(x.\left(2+x\right).\left(7-x\right)=0\)
\(=>\left[\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.=>\left[\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
5) \(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)
\(=>\left[\begin{matrix}x-1=0\\x+2=0\\-x-3=0\end{matrix}\right.=>\left[\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
Tổng quát: tích ab = 0 thì a=0 hoặc b=0
\(a\cdot b\cdot c=0\Leftrightarrow\left[\begin{matrix}a=0\\b=0\\c=0\end{matrix}\right.\)(nếu có 3 thừa số)
Tìm x:
7.x-x=5^21.5^19+3.2^2-7^0
Giải chi tiết hộ mình nha
bài 19: tìm x
a) 5 . ( x - 7 ) = 0
b) 25 ( x - 4 ) = 0
c) ( 34 - 2x ) . ( 2x - 6 ) = 0
d) ( 2019 - x ) . ( 3x - 12 ) 0
e) 57 . ( 9x - 27 ) = 0
f) 25 + ( 15 - x ) = 30
g) 43 - ( 24 - x ) = 20
h) 2 . ( x - 5 ) - 17 = 25
i) 3 . ( x + 7 ) - 15 = 27
j) 15 + 4 . ( x - 2 ) = 95
k) 20 - ( x + 14 ) = 5
l) 14 + 3 . ( 5 - x ) = 27
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
Các bạn giúp mình trình bày 2 bài này với
1) tìm x
x^7 X x^5= 3^12
(x+1)^4=5^8 : 25^4
x^6= x
2)Tính
(4^20 + 4^15) : (4^10 + 4^5)
A= 2^0 + 2^1 + 2^2 +........+ 2^2016
1, Tìm x :
a, \(x^7.x^5=3^{12}\)
\(\Rightarrow x^{12}=3^{12}\)
\(\Rightarrow x=3\)
Vậy x = 3
b, \(\left(x+1\right)^4=5^8\div25^4\)
\(\left(x+1\right)^4=5^8\div\left(5^2\right)^4\)
\(\left(x+1\right)^4=5^8\div5^8\)
\(\left(x+1\right)^4=1\)
\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
c, \(x^6=x\)
\(\Rightarrow x^6-x=0\)
\(\Rightarrow x.x^5-x.1=0\)
\(\Rightarrow x\left(x^5-1\right)=0\)
x = 0 hoặc x5 - 1 = 0
x = 0 hoặc x5= 1
x = 0 hoặc x5 = 1
\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
2, Tính :
\(\left(4^{20}+4^{15}\right)\div\left(4^{10}+4^5\right)\)
\(=4^{15}.\left(4^5+1\right)\div4^5.\left(4^5+1\right)\)
\(=4^{15}\div4^5\)
\(=4^{10}\)
Vậy giá trị biểu thức trên bằng 410
\(A=2^0+2^1+2^2+...+2^{2016}\)
\(2A=2+2^2+2^3+...+2^{2017}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2017}\right)-\left(2^0+2^1+2^2+...+2^{2016}\right)\)
\(\Rightarrow A=2^{2017}-1\)
Vậy : \(A=2^{2017}-1\)
\(x^7\times x^5=3^{12}\)
\(x^{12}=3^{12}\)
\(x=3\)
^^
\(\left(x+1\right)^4=\frac{5^8}{25^4}\)
\(\left(x+1\right)^4=\frac{\left(5^2\right)^4}{\left(5^2\right)^4}\)
\(\left(x+1\right)^4=1\)
\(\left[\begin{array}{nghiempt}x+4=1\\x+4=-1\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1-4\\x=-1-4\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=-3\\x=-5\end{array}\right.\)
^^
\(x^6=x\)
\(\left[\begin{array}{nghiempt}x=1\\x=-1\\x=0\end{array}\right.\)
^^
\(\frac{4^{20}+4^{15}}{4^{10}+4^5}=\frac{4^{15}\times\left(4^5+1\right)}{4^5\times\left(4^5+1\right)}=4^{10}\)
^^
\(A=2^0+2^1+2^2+...+2^{2016}\)
\(2A=2^1+2^2+2^3+...+2^{2017}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2017}\right)-\left(2^0+2^1+2^2+...+2^{2016}\right)\)
\(A=2^{2017}-1\)
Bài 14: Tìm x
1/ x.(x + 17) = 0
2/ (x + 1112).(x-3) = 0
3/ (-x + 25).(3 – x ) = 0
4/ x.(12 + x).( 7 – x) = 0
5/ (x - 15).(x +2).(-x -3) = 0
1/ x(x+17)=0
⇒ \(\left[{}\begin{matrix}x=0\\x+17=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-17\end{matrix}\right.\)
2/ (x+1112)(x-3)=0
⇒\(\left[{}\begin{matrix}x+1112=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1112\\x=3\end{matrix}\right.\)
3/ (-x+25)(3-x)=0
⇒\(\left[{}\begin{matrix}-x+25=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=25\\x=3\end{matrix}\right.\)
4/ x(12+x)(7-x)=0
⇒ \(\left[{}\begin{matrix}x=0\\12+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-12\\x=7\end{matrix}\right.\)
5/ (x-15)(x+2)(-x-3)=0
⇒\(\left[{}\begin{matrix}x-15=0\\x+2=0\\-x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\\x=-2\\x=-3\end{matrix}\right.\)
\(x\left(x+17\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+17=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-17\end{matrix}\right.\)
Vậy \(x\in\left\{0;-17\right\}\)
\(\left(x+1112\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1112=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1112\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-1112;3\right\}\)
\(\left(-x+25\right)\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-x+25=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=25\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{25;3\right\}\)
\(x\left(12+x\right)\left(7-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\12+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-12\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{0;-12;7\right\}\)
\(\left(x-15\right)\left(x+2\right)\left(-x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-15=0\\x+2=0\\-x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\\x=-2\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{15;-2;-3\right\}\)
tìm số nguyên x biết:
a)3x-17=x+3
b)|x-3|-12=|-5|
c)25-(x-5)=-415-(15-415)
d)2x-15=-47
e)(-5)^2-(5x-3)=43
g)(x-3)(2x+6)=0
giải nhanh hộ mìh nhé
mình đang cần gấp
Tìm x thuộc Z:
1) ( x - 2 ) . ( x + 15 ) = 0
2) ( 7 - x ) . ( x + 19 ) = 0
3) -5 < x < 1
4) |x| < 3
5) ( x - 3 )( x - 5 ) < 0
6) 2x2 - 3 = 29
7) -6x - (-7) = 25
8) 46 - ( x - 11 ) = -48
( Mk cần gấp)
1) ( x - 2 ) . ( x + 15 ) = 0
\(\Rightarrow\) x - 2 = 0 hoặc x + 15 = 0
* Nếu x - 2 = 0
x = 0 + 2
x = 2
* Nếu x + 15 = 0
x = 0 - 15
x = -15
Vậy x \(\in\) { 2 ; -15 }
2) ( 7 - x ) . ( x + 19 ) = 0
\(\Rightarrow\) 7 - x = 0 hoặc x + 19 = 0
* Nếu 7 - x = 0
x = 7 - 0
x = 7
* Nếu x + 19 = 0
x = 0 - 19
x = -19
Vậy x \(\in\) {7 ; -19}
Tìm x ϵ Z biết:
a) |x| = 1005
b) |x| + 15 = 22; x>0
c) |x| + 12 = 25; x<0
d) |3x-15| = 0
e) |-x + 2| = 4
f) |-x + |31-(-89)|| = 14
g) |x+1| + |x-1| = 2
h) |x| < 5
i) 5< |x| \(\le\) 7
Giải chi tiết giúp mình nhé
a) |x|=2005
=> x=2005 hoặc -2005
vì giá trị tuyệt đối của 1 số nguyên dương là chính số đó nên x=2005
vì giá trị tuyệt đối của 1 số nguyên âm là số đối của số đó
=> -2005 có số đối là 2005 nên x cũng có thể bằng -2005
a) x = 1005 hoặc x = -1005
b) x + 15 = 22
x = 22 - 15
x = 7
Vì x là |x| nen cung co the = -7
Nhung vi theo de bai thi x>0 nen x = 7
c) x + 12 = 25
x = 25 - 12
x = 13
Vi x la |x| nen cung co the = -13
Vi theo de thi x<0 nen x = -13
a) \(\left|x\right|=1005\)
\(\Rightarrow\) x = 1005 hoặc x = -1005
b) \(\left|x\right|\) + 15 = 22 ; x > 0
x > 0 \(\Rightarrow\)\(\left|x\right|=x\)
\(\Rightarrow\left|x\right|+15=22\)
\(\Rightarrow x+15=22\)
\(\Rightarrow x=22-15\)
\(\Rightarrow x=7\)
c) \(\left|x\right|+12=25;x< 0\)
\(x< 0\Rightarrow\left|x\right|=-x\)
\(\Rightarrow\left|x\right|+12=25\)
\(\Rightarrow-x+12=25\)
\(\Rightarrow-x=25-12\)
\(\Rightarrow-x=13\)
\(\Rightarrow x=-13\)
d) \(\left|3x-15\right|=0\)
\(\Rightarrow3x-15=0\)
\(\Rightarrow3x=0+15\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=15:3\)
\(\Rightarrow x=5\)
e) \(\left|-x+2\right|=4\)
\(\Rightarrow-x+2=4\) hoặc \(-x+2=-4\)
TH1: \(-x+2=4\)
\(-x=4-2\)
\(-x=2\)
\(\Rightarrow x=-2\)
TH2: \(-x+2=-4\)
\(-x=-4-2\)
\(-x=-6\)
\(\Rightarrow x=6\)
f) \(\left|-x+\left|31-\left(-89\right)\right|\right|=14\)
\(\Rightarrow\left|-x+\left|120\right|\right|=14\)
\(\Rightarrow\left|-x+120\right|=14\)
\(\Rightarrow-x+120=14\) hoặc \(-x+120=-14\)
TH1: \(-x+120=14\)
\(\Rightarrow-x=14-120\)
\(\Rightarrow-x=-106\)
\(\Rightarrow x=106\)
TH2: \(-x+120=-14\)
\(\Rightarrow-x=-14-120\)
\(\Rightarrow-x=-134\)
\(\Rightarrow x=134\)