`d, 145 - 2x^2 = 136 : 8`

`=> 145 - 2x^2 = 17`

`=> 2x^2 = 145 - 17`

`=> 2x^2  =128`

`=>x^2=128:2`

`=> x^2=64`

`=>x^2=8^2`

`=>x=8`

Vậy: `x=8`

Đọc kĩ đề nha, số tự nhiên

\(\left(5x+1\right)^2=\dfrac{36}{49}\)

=>\(\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{6}{7}-1=-\dfrac{1}{7}\\5x=-\dfrac{6}{7}-1=-\dfrac{13}{7}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{1}{7}:5=-\dfrac{1}{35}\\x=-\dfrac{13}{7}:5=-\dfrac{13}{35}\end{matrix}\right.\)

\(\left(8x\right)^{2x+1}=5^{2x+1}\)

=>8x=5

=>\(x=\dfrac{5}{8}\)

\(x-\left(\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)

=>\(x-\dfrac{8}{729}=\dfrac{64}{729}\)

=>\(x=\dfrac{64}{729}+\dfrac{8}{729}=\dfrac{72}{729}=\dfrac{8}{81}\)

Sửa đề: \(\left(x-2,5\right)^2+\left(y-\dfrac{1}{10}\right)^2< =0\)

mà \(\left(x-2,5\right)^2+\left(y-\dfrac{1}{10}\right)^2>=0\forall x,y\)

nên \(\left\{{}\begin{matrix}x-2,5=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2,5\\y=\dfrac{1}{10}\end{matrix}\right.\)

`(5x + 1)^2 = 36/49`

TH1:

`(5x + 1)^2 = (6/7)^2`

`=> 5x + 1 = 6/7`

`=> 5x      = 6/7 - 1`

`=>  5x    = -1/7`

`=>    x = -1/7 : 5`

`=> x = -1/35`

TH2:

`(5x + 1)^2 = (-6/7)^2`

`=> 5x + 1 = -6/7`

`=> 5x       = -6/7 - 1`

`=>  5x      = -13/7`

`=>   x      = - 13/7 :5`

`=> x        = -13/35`

Vậy `x = -1/35 ; x = -13/35`

`b)(x - 2/9)^3 = (2/3)^6

`=> (x - 2/9)^3 = (8/27)^3`

`=> x - 2/9 = 8/27`

`=> x = 8/27 + 2/9`

`=> x = 14/27`

Vậy `x = 14/27`

`c)(8x)^(2x + 1) = 5^(2x + 1)`

`=> 8x = 5`

`=>  x = 5 : 8`

`=> x = 5/8`

Vậy `x = 5/8`

`d)(x - 2,5)^2 + (y - 1/10)^2 ≥0`

TH1: 

`x - 2.5 = 0`

=> x = 0 + 2,5`

`=> x = 2,5 = 5/2`

TH2:

`y  - 1/10 = 0`

`=> y = 0 + 1/10`

`=> y = 1/10`

Vậy `x = 5/2 ` ; `y = 1/10`

1+1=2

              Giải:

\(x.x\)  = 1 + 3 + 5  +7  + 9 + ...+ 2499

xét vế trái ta có:

VT = 1 + 3 + 5  +7 + 9 + ... + 2499

Xét dãy số 1; 3; 5; 7; 9;...;2499

Dãy số trên là dãy số cách đều với khoảng cách là: 3 - 1  = 2 

Số số hạng của dãy số trên là: (2499  - 1) : 2  + 1 = 1250 

Tổng các số hạng trên là: (2499 + 1) x 1250 : 2  = 1562500

Khi đó ta có: \(x^2\) = 1562500

                    \(x^2\)  = (1250)²

                    \(\left[{}\begin{matrix}x=-12500\\x=12500\end{matrix}\right.\)

Vậy \(x\) \(\in\) { -12500; 12500}

a) Ta có: 

`84 = 2^2 . 3 . 7`

`108 = 2^2 . 3^3`

`=> UCLN(84;108) = 2^2 . 3 = 12`

`=> UC(84;108) = Ư(12) = {-12;-6;-4;-3;-2;-1;1;2;3;4;6;12}`

Do `35 vdots x; 105 vdots x`

`=> x in UC{35;105)`

Mà `105 vdots 35`

`=> x in Ư(35) = {1;5;7;35}`

Mà `x > 5 -> x in {7;35}`

Vậy ...

`b) x vdots 10; x vdots 15`

`=> x in BC(10;15)`

Ta có: 

`10 = 2 . 5`

`15 = 3.5`

`=> BCN``N(10;15) = 2.3.5 = 30`

`=> x in B(30) = {0;30;60;90;120;...}`

Mà `x < 100 -> x in {0;30;60;90}`

Lời giải:

a. $140=2^2.5.7$

$168=2^3.3.7$

$\Rightarrow ƯCLN(140, 168) = 2^2.7=28$

b. 

$525=3.5^2.7$

$375=3.5^3$

$\Rightarrow ƯCLN(525, 375)=3.5^2=75$

`Ư(16) = {-16;-8;-4;-2;-1;1;2;4;8;16}`

`Ư(24) = {-24;-12;-8;-6;-4;-3;-2;-1;1;2;3;4;6;8;12;24}`

`=> ƯC(16;24) = {-8;-4;-2;-1;1;2;4;8}`

`Ư(30) = {-30;-15;-10;-6;-5;-3-2;-1;1;2;3;5;6;10;15;30}`

`Ư(45) = {-45;-15;-9;-5;-3;-1;1;3;5;9;15;45}`

`=> ƯC(30;45) = {-15;-5;-3;-1;1;3;5;15}`

\(A=2+2^2+2^3+...+2^{2023}\)

\(2A=2^2+2^3+2^4+...+2^{2024}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2024}\right)-\left(2+2^2+2^3+...+2^{2023}\right)\)

\(A=2^{2024}-2\)

\(B=3^0+3^1+3^2+...+3^{100}\)

\(3B=3^1+3^2+3^3+...+3^{101}\)

\(3B-B=\left(3^1+3^2+3^3+...+3^{101}\right)-\left(3^0+3^1+3^2+...+3^{100}\right)\)

\(2B=3^{101}-3\)

\(B=\dfrac{3^{101}-3}{2}\)

\(C=4^0+4^2+...+4^{100}\)

=>\(16C=4^2+4^4+...+4^{102}\)

=>\(16C-C=4^2+4^4+...+4^{102}-4^0-4^2-...-4^{100}\)

=>\(15C=4^{102}-1\)

=>\(C=\dfrac{4^{102}-1}{15}\)

\(D=1+5^2+5^4+...+5^{2022}\)

=>\(25D=5^2+5^4+...+5^{2024}\)

=>\(25D-D=5^2+5^4+...+5^{2024}-1-5^2-...-5^{2022}\)

=>\(24D=5^{2024}-1\)

=>\(D=\dfrac{5^{2024}-1}{24}\)

\(S=a^0+a^1+...+a^n\)

=>\(S\cdot a=a^1+a^2+...+a^{n+1}\)

=>\(S\cdot a-S=a^1+a^2+...+a^{n+1}-a^0-a^1-...-a^n\)

=>\(S\left(a-1\right)=a^{n+1}-1\)

=>\(S=\dfrac{a^{n+1}-1}{a-1}\)