5(x-7)=0

=>x=7

25(x-4)=0

=>x=4

342(2x-6)=0

=>x=3

2016(3x-12)=0

=>x=4

47(5x-15)=0

=5x-15=0

=>5x=15

=>x=3

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)

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

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 : \(A=2^{2017}-1\)

\(x^7\times x^5=3^{12}\)

\(x^{12}=3^{12}\)

\(x=3\)

^^

\(\left(x+1\right)^4=\frac{5^8}{25^4}\)

\(\left(x+1\right)^4=\frac{\left(5^2\right)^4}{\left(5^2\right)^4}\)

\(\left(x+1\right)^4=1\)

\(\left[\begin{array}{nghiempt}x+4=1\\x+4=-1\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1-4\\x=-1-4\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=-3\\x=-5\end{array}\right.\)

^^

\(x^6=x\)

\(\left[\begin{array}{nghiempt}x=1\\x=-1\\x=0\end{array}\right.\)

^^

\(\frac{4^{20}+4^{15}}{4^{10}+4^5}=\frac{4^{15}\times\left(4^5+1\right)}{4^5\times\left(4^5+1\right)}=4^{10}\)

^^

\(A=2^0+2^1+2^2+...+2^{2016}\)

\(2A=2^1+2^2+2^3+...+2^{2017}\)

\(2A-A=\left(2^1+2^2+2^3+...+2^{2017}\right)-\left(2^0+2^1+2^2+...+2^{2016}\right)\)

\(A=2^{2017}-1\)

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 \(x\in\left\{15;-2;-3\right\}\)

k minh minh giai cho

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}

dễ mà

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: \(-x+120=-14\)

\(\Rightarrow-x=-14-120\)

\(\Rightarrow-x=-134\)

\(\Rightarrow x=134\)