1: Đặt A=\(5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x+2\right)\)

\(=20x^3-10x^2+5x-20x^3+10x^2-4x=x\)

Thay x=15 vào A, ta được:

A=x=15

2: Đặt \(B=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)

\(=6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2-5xy^3\)

\(=19x^2y^2-11xy^3-8x^3\)

Thay x=0,5 và y=2 vào B, ta được:

\(B=19\cdot0,5^2\cdot2^2-11\cdot0,5\cdot2^3-8\cdot2^3\)

=19-44-64

=-89

3: x=4 nên x+1=5

\(x^5-5x^4+5x^3-5x^2+5x-1\)

\(=x^5-x^4\left(x+1\right)+x^3\left(x+1\right)-x^2\left(x+1\right)+x\left(x+1\right)-1\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-1\)

=x-1=4-1=3

4: x=7 nên x+1=8

\(x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\)

\(=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-5\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+...+x^2+x-5\)

=x-5=7-5=2

5: \(M=\left(2x-1\right)^2+2\left(2x-1\right)\left(3x+1\right)+\left(3x+1\right)^2\)

\(=\left(2x-1+3x+1\right)^2=\left(5x\right)^2=25x^2\)

6: \(N=\left(3x-1\right)^2-2\left(9x^2-1\right)+\left(3x+1\right)^2\)

\(=\left(3x-1\right)^2-2\left(3x-1\right)\left(3x+1\right)+\left(3x+1\right)^2\)

\(=\left(3x-1-3x-1\right)^2=\left(-2\right)^2=4\)

a:

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

 \(Q=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)

\(=\dfrac{3\left(x-3\right)+x+3+18}{x^2-9}\)

\(=\dfrac{3x-9+x+21}{\left(x-3\right)\left(x+3\right)}=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\)

b: \(R=Q\cdot x=\dfrac{4x}{x-3}=\dfrac{4x-12+12}{x-3}=4+\dfrac{12}{x-3}\)

Để R nguyên thì \(12⋮x-3\)

=>\(x-3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(x\in\left\{4;2;5;1;6;0;7;-1;9;-3;15;-9\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;2;5;1;6;0;7;-1;9;15;-9\right\}\)

`(5x - 1)^6 = 729`

`=> (5x - 1)^6 = 3^6`

`=> 5x - 1 = 3` hoặc `5x - 1 = -3`

`=> 5x = 4` hoặc `5x = -2`

`=> x = 4/5` hoặc `x = -2/5`

-------------------

`(2x + 1)^3 = -0,001`

`=> (2x + 1)^3 = (-0,1)^3`

`=> 2x + 1 = -0,1`

`=> 2x  = -1001/1000`

`=> x = -1001/2000`

???

a: \(2\cdot16>=2^n>4\)

=>\(2^5>=2^n>2^2\)

=>2<n<=5

mà n là số tự nhiên

nên \(n\in\left\{3;4;5\right\}\)

b: \(9\cdot27< =3^n< =243\)

=>\(243< =3^n< =243\)

=>\(3^n=243\)

=>n=5

c: \(27< 3^n< 3\cdot81\)

=>\(3^3< 3^n< 3^5\)

=>3<n<5

mà n là số tự nhiên

nên n=4

d: \(4^{15}\cdot9^{15}< 2^n\cdot3^n< 18^{16}\cdot2^{16}\)

=>\(36^{15}< 6^n< 36^{16}\)

=>\(6^{30}< 6^n< 6^{32}\)

=>30<n<32

mà n là số tự nhiên

nên n=31

\(a.2\cdot16\ge2^n>4\\ 2\cdot2^4\ge2^n>2^2\\ 2^5\ge2^n>2^2\\ 5\ge n>2\\ n\in\left\{3;4;5\right\}\\ b.9\cdot27\le3^n\le243\\ 3^2\cdot3^3\le3^n\le3^5\\ 3^5\le3^n\le3^5\\ n=5\\ c.27< 3^n< 3\cdot81\\ 3^3< 3^n< 3\cdot3^4\\ 3^3< 3^n< 3^5\\ 3< n< 5\\ n=4\\ d.4^{15}\cdot9^{15}< 2^n\cdot3^n< 18^{16}\cdot2^{16}\\ 36^{15}< 6^n< 36^{16}\\ \left(6^2\right)^{15}< 6^n< \left(6^2\right)^{16}\\ 6^{30}< 6^n< 6^{32}\\ n=31\)

a/

Gọi d là ước chung của 2n+1 và 3n+1 nên

\(2n+1⋮d\Rightarrow3\left(2n+1\right)=6n+3⋮d\)

\(3n+1⋮d\Rightarrow2\left(3n+1\right)=6n+2⋮d\)

\(\Rightarrow6n+3-\left(6n+2\right)=1⋮d\Rightarrow d=1\)

Điều đó chứng tỏ rằng 2n+1 và 3n+1 là 2 số nguyên tố sánh đôi

Các câu b;c;d làm tương tự

\(a.\left(\dfrac{2}{33}\right)^n\cdot11^n=\dfrac{4}{9}\\ \left(\dfrac{2}{33}\cdot11\right)^n=\left(\dfrac{2}{3}\right)^2\\ \left(\dfrac{2}{3}\right)^n=\left(\dfrac{2}{3}\right)^2\\ n=2\\ b.\dfrac{125}{5^n}=5\\\dfrac{ 5^3}{5^n}=5\\ 5^{3-n}=5^1\\ 3-n=1\\ n=3-1\\ n=2\\ c.\dfrac{\left(-6\right)^n}{36}=-216\\ \dfrac{\left(-6\right)^n}{\left(-6\right)^2}=\left(-6\right)^3\\ =\left(-6\right)^{n-2}=\left(-6\right)^3\\ n-2=3\\ n=2+3\\ n=5\\ d.20^n:14^n=\dfrac{10}{7}\\ \left(\dfrac{20}{14}\right)^n=\dfrac{10}{7}\\ \left(\dfrac{10}{7}\right)^n=\left(\dfrac{10}{7}\right)^1\\ n=1\)

\(\left\{{}\begin{matrix}0,2x+0,5y=0,7\\4x+10y=9\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}4x+10y=3,5\\4x+10y=9\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}0x=5,5\left(ko\exists\right)\\4x+2y=3,5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}0,2x+0,5y=0,7\\4x+10y=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+10y=14\\4x+10y=9\end{matrix}\right.\)

=> Hpt vô nghiệm 

\(\dfrac{-4}{9}\cdot x=\dfrac{-2}{7}:\dfrac{4}{21}\\ -\dfrac{4}{9}\cdot x=\dfrac{-2}{7}\cdot\dfrac{21}{4}\\ -\dfrac{4}{9}\cdot x=\dfrac{-3}{2}\\ x=-\dfrac{3}{2}:\dfrac{-4}{9}\\ x=\dfrac{3}{2}\cdot\dfrac{9}{4}\\ x=\dfrac{27}{8}\)

Vậy: ... 

\(a.BC\left(3\cdot5^2;5^2\cdot7\right)\\ =B\left(3\cdot5^2\cdot7\right)\\ =B\left(525\right)=\left\{0;525;1050;...\right\}\\ b.ƯC\left(2^2\cdot3\cdot5;3^2\cdot7;3\cdot5\cdot11\right)\\ =Ư\left(3\right)=\left\{1;3\right\}\)

tìm cả bội chung và ức chung của cả hai