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Bài 2:
a)|x| < 3
x\(\in\){-2;-1;0;1;2}
b)|x - 4 | < 3
x\(\in\){ 6 ; 5 ; 4 ; 3 ; 2 }
c) | x + 10 | < 2
x\(\in\){ -2 ; -10 }
Bài 1:
A = 1 + 2 - 3 + 4 + 5 - 6 +...+98 - 99
A = (1 + 4 + 7 +...+97) + [(2-3)+(5-6)+...+(98-99)]
A = 1617 + [(-1)+(-1)+...+(-1)]
A = 1617 + (-49)
A = +(1617-49) = A = 1568
B = - 2 - 4 + 6 - 8 + 10 + 12 - .... + 60
B =
2)
a) \(x\in\left\{2;1;0;-1;-2\right\}\)
b) \(x\in\left\{6;-6;5;-5;4\right\}\)
c) \(x\in\left\{-9;-11;-10\right\}\)
3)
\(\left(a;b\right)\in\left\{\left(0;1\right);\left(0;-1\right);\left(1;0\right);\left(-1;0\right)\right\}\)
(2x + 1)(y-5) = 12
Ta có: 12 = 1.12 = 2.6 = 3.4 = (-1)(-12) = (-2) (-6) = (-3) (-4)
Bận liệt ke ra nhé
( 3x - 6 ) . 3 = 18
3x - 6 = 18 : 3 = 6
3x = 6 + 6 = 12
x= 4
6x - 5 = 613
6x = 613 + 5 =618
x=103
x-36:18=12
x - 2 = 12
x= 12 + 2
x= 14
\(\left(3x-6\right).3=18\)
\(\Rightarrow3x-6=19:3=6\)
\(\Rightarrow3x=6+6=12\)
\(\Rightarrow x=12:3=4\)
\(6.x-5=613\)
\(\Rightarrow6x=613+ 5=618\)
\(\Rightarrow x=618:6=103\)
\(x-36:18=12\)
\(\Rightarrow x-2=12\)
\(\Rightarrow x=12+2=14\)
Ủng hộ nha m.n ^_^
Bài 1 :
A = 12 + 22 + 32 +....+n2
A = 12 + 2.(1+1) + 3.(2 +1) + 4.( 3 +1) +.....+n(n-1 + 1)
A = 1 + 1.2 + 2 + 2.3 + 3 + 3.4 + 4 +.....+ n.(n-1) + n
A = ( 1 + 2 + 3 + 4 +....+n) + ( 1.2 + 2.3 + 3.4 +....+(n-1).n
A = (n+1).{(n-1):n+1)/2 +1/3.[1.2.3 +2.3.3 +.....+(n-1)n.3]
A = (n+1).n/2+1/3.[1.2.3 +2.3.(4-1)+ ...+(n-1).n [(n+1) - (n -2)]
A = (n+1)n/2+1/3.( 1.2.3 + 2.3.4 -1.2.3 +..+ (n-1)n(n+1)- (n-2)(n-1)n)
A =(n+1)n/2 + 1/3.(n-1)n(n+1)
A = n(n+1)[1/2 + 1/3 .(n-1)]
A = n.(n+1) \(\dfrac{3+2n-2}{6}\)
A= n.(n+1)(2n+1)/6
Bài 2 :
a, (x+1) +(x+2) + (x+3)+...+(x+10) = 5070
(x+10 +x+1).{( x+10 - x -1): 1 +1):2 = 5070
(2x + 11)10 : 2 = 5070
( 2x + 11)5 = 5070
2x+ 11 = 5070:5
2x = 1014 - 11
2x = 1003
x = 1003 :2
x = 501,5
b, 1 + 2 + 3 +...+x = 820
( x + 1)[ (x-1):1 +1] : 2 = 820
(x +1).x = 820 x 2
(x +1).x = 1640
(x +1) .x = 40 x 41
x = 40
1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
\(\dfrac{x-3}{10}-\dfrac{x-3}{15}=\dfrac{x-3}{12}-\dfrac{x-3}{18}\)
\(\Rightarrow\dfrac{x-3}{1}-\dfrac{x-3}{15}-\dfrac{x-3}{12}+\dfrac{x-3}{18}=0\)
\(\Rightarrow\left(x-3\right)\left(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\right)=0\)
Mà \(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\ne0\)
\(\Rightarrow x-3=0\Leftrightarrow x=3\)
(3x-12).(x-3)=0
=> 3x-12=0 hoặc x-3=0
bạn tự giải 2 trường hợp đó ra
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
\(\left(x-7\right)^{2025}=125\left(x-7\right)^{2020}\\ =>\left(x-7\right)^{2020}.\left[\left(x-7\right)^5-125\right]=0\)
\(=>\left[{}\begin{matrix}\left(x-7\right)^{2020}=0\\\left(x-7\right)^5=125\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x-7=0\\x-7=\sqrt[5]{125}\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=7\\x=7+\sqrt[5]{125}\end{matrix}\right.\)
Lời giải:
$(x-7)^{2025}=125(x-7)^{2020}$
$\Rightarrow (x-7)^{2025}-125(x-7)^{2020}=0$
$\Rightarrow (x-7)^{2020}[(x-7)^5-125]=0$
$\Rightarrow (x-7)^{2020}=0$ hoặc $(x-7)^5=125$
$\Rightarrow x-7=0$ hoặc $x-7=sqrt[5]{125}+7$
\(\dfrac{\left(x+3\right)}{6}=\dfrac{10}{12}\)
\(=>\dfrac{\left(x+3\right)}{6}=\dfrac{5}{6}\)
\(=>x+3=5\)
\(=>x=2\)