1. \(\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\frac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\frac{3^{10}\cdot2^4}{3^9\cdot2^4}=3\)

2. \(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\frac{2^{10}\cdot\left(13+65\right)}{2^8\cdot104}=\frac{2^{10}\cdot78}{2^8\cdot104}=\frac{2^8\cdot2^2\cdot2\cdot3\cdot13}{2^8\cdot2^3\cdot13}=\frac{2^8\cdot2^3\cdot3\cdot13}{2^8\cdot2^3\cdot13}=3\)

3. \(\frac{72^2\cdot54^2}{108^4}=\frac{\left(2^3\cdot3^2\right)^2\cdot\left(2\cdot3^3\right)^2}{\left(2^2\cdot3^3\right)^4}\)

\(=\frac{2^6\cdot3^4\cdot2^2\cdot3^6}{2^8\cdot3^{12}}=\frac{2^8\cdot3^{10}}{2^8\cdot3^{12}}=\frac{3^{10}}{3^{12}}=3^{-2}=\frac{1}{9}\)

4. \(\frac{21^2\cdot14\cdot125}{35^5\cdot6}=\frac{\left(3\cdot7\right)^2\cdot2\cdot7\cdot5^3}{\left(5\cdot7\right)^5\cdot2\cdot3}=\frac{3^2\cdot7^2\cdot2\cdot7\cdot5^3}{5^5\cdot7^5\cdot2\cdot3}=\frac{3^2\cdot7^3\cdot2\cdot5^3}{5^3\cdot5^2\cdot7^2\cdot7^3\cdot2\cdot3}=\frac{3^2}{5^2\cdot3\cdot7^2}=\frac{3}{1225}\)

1. \(\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^4=\left(-\frac{1}{2}\right)^5=-\frac{1}{32}\)

2. \(6.3^2-24:2^3=6.9-24:8=54-3=51\)

3. \(\left(\frac{1}{4}\right)^2.\left(\frac{1}{4}\right)^3=\left(\frac{1}{4}\right)^5=\frac{1}{1024}\)

1) (-1/2).(-1/2)^4 = ( -1/2)^ 5 = -1/32

2) 6.3^2 - 24:2^3 = 6,9 - 24 : 8 = 54 - 3 = 51

3) (1/4)^2 . (1/4)^3 = ( 1/4)^5 = 1/1024

A = -a² + 3a + 4

A = -( a² - 3a + 9/4 ) + 25/4

A = -( a - 3/2 )² + 25/4

-( a - 3/2 )² ≤ 0 ∀ x => -( a - 3/2 )² + 25/4 ≤ 25/4

Đẳng thức xảy ra <=> a - 3/2 = 0 => a = 3/2

=> MaxA = 25/4 <=> a = 3/2

\(A=-a^2+3a+4\)

\(\Rightarrow A=-a^2+3a-\frac{9}{4}+\frac{25}{4}\)

\(\Rightarrow A=-\left(a-\frac{3}{2}\right)^2+\frac{25}{4}\)

Vì \(\left(a-\frac{3}{2}\right)^2\ge0\forall a\)\(\Rightarrow-\left(a-\frac{3}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(a-\frac{3}{2}\right)^2=0\Leftrightarrow a-\frac{3}{2}=0\Leftrightarrow a=\frac{3}{2}\)

Vậy maxA = 25/4 <=> a = 3/2

\(\left(9x^2-12x+4\right)-\left(y+2\right)^2\)

\(=\left[\left(3x^2\right)-2.3x.2+2^2\right]-\left(y+2\right)^2\)

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

\(\left(9x^2-12x+4\right)-\left(y+2\right)^2\)   

= \(9x^2-12x+4-\left(y^2+4y+4\right)\)     

=\(9x^2-12x+4-y^2-4y-4\)    

=\(9x^2-y^2-12x-4y\) 

=\(\left(3x-y\right)\left(3x+y\right)-4\left(3x+y\right)\) 

=\(\left(3x+y\right)\left(3x-y-4\right)\)

\(8\frac{2}{7}-\left(1\frac{1}{6}+25\%\right)=\frac{58}{7}-\left(\frac{7}{6}+\frac{1}{4}\right)=\frac{58}{7}-\frac{17}{12}=\frac{577}{84}\)

\(4\frac{3}{4}+\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)

\(=\frac{19}{4}+\left(-\frac{83}{20}\right)+\frac{37}{12}=\frac{3}{5}+\frac{37}{12}=\frac{221}{60}\)

\(8\frac{2}{7}-\left(1\frac{1}{6}+25\%\right)=\frac{58}{7}-\left(\frac{7}{6}+\frac{1}{4}\right)=\frac{58}{7}-\frac{17}{12}=\frac{577}{84}\)

\(100+23+x=469\)                \(567-x=278-35\)                \(x-120=403+60\)

\(123+x=469\)                         \(567-x=243\)                            \(x-120=463\)

\(x=469-123\)                        \(x=567-243\)                             \(x=463+120\)

\(x=346\)                                       \(x=324\)                                           \(x=583\)

Vậy x = 346                                         Vậy x = 324                                             Vậy : x = 583

a) 100+23+x=469

\(\Leftrightarrow123+x=469\)

\(\Leftrightarrow x=469-123\)

\(\Leftrightarrow x=346\)

b) 567-x = 278-35

\(\Leftrightarrow567-x=243\)

\(\Leftrightarrow x=567-243\)

\(\Leftrightarrow x=324\)

Đề bài là gì bạn , chẳng nhẽ tính ?

a) (9x + 1) - (4x + 2) = 9x + 1 - 4x - 2 = (9x - 4x) + (1 - 2) = 5x - 1

b) (3x³ + 1) - (3x² - 4x + 5) = 3x³ + 1 - 3x² + 4x - 5 = 3x³ - 3x² + 4x + (1 - 5) = 3x³ - 3x² + 4x  - 4

a) \(\left(9x+1\right)-\left(4x+2\right)\)

\(=9x+1-4x-2\)

\(=5x-1\)

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

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

\(=4x-4\)

a. \(\frac{n^2+1}{n+1}\in Z\)

Ta có : \(\frac{n^2+1}{n+1}=\frac{n\left(n+1\right)-n+1}{n+1}=n-1=0\)

\(\Leftrightarrow n=1\)

b. \(\frac{n^2-3}{n+2}\in Z\)

Ta có : \(\frac{n^2-3}{n+2}=\frac{n\left(n+2\right)-2n-3}{n+2}=n-\frac{2n+4-7}{n+2}=n-2-\frac{7}{n+2}\)

Để n^2 - 3 / n + 2 thuộc Z thì 7 / n + 2 thuộc Z, n thuộc Z

=> n + 2 thuộc { - 7 ; - 1 ; 1 ; 7 }

=> n thuộc { - 9 ; - 3 ; - 1 ; 5 }

a ) Để \(n^2+1⋮n+1\)

mà \(n\left(n+1\right)⋮n+1\)

\(\Rightarrow n\left(n+1\right)-n^2-1⋮n+1\)

\(\Rightarrow n^2+n-n^2-1⋮n+1\)

\(\Rightarrow n-1⋮n+1\)

\(\Rightarrow n+1-2⋮n+1\)

mà \(n+1⋮n+1\)

\(\Rightarrow2⋮n+1\left(n\inℤ\right)\)

\(\Rightarrow n+1\inƯ\left(2\right)=\left\{1;-1;2-2\right\}\)

\(\Rightarrow n\in\left\{0;-2;1;-3\right\}\)

b ) \(n^2-3⋮n+2\)

mà \(n\left(n+2\right)⋮n+2\)

\(\Rightarrow n\left(n+2\right)-n^2+3⋮n+2\)

\(\Rightarrow n^2+2n-n^2+3⋮n+2\)

\(\Rightarrow2n+3⋮n+2\)

\(\Rightarrow2n+4-1⋮n+2\)

\(\Rightarrow2\left(n+2\right)-1⋮n+2\)

mà \(2\left(n+2\right)⋮n+2\)

\(\Rightarrow1⋮n+2\)

\(\Rightarrow n+2\in\left\{1;-1\right\}\)

\(\Rightarrow n\in\left\{-1;-3\right\}\)

c ) \(n+3⋮n^2+2\)

\(\Rightarrow n\left(n+3\right)⋮n^2+2\)

mà \(n^2+2⋮n^2+2\)

\(\Rightarrow n\left(n+3\right)-n^2-2⋮n^2+2\)

\(\Rightarrow n^2+3n-n^2-2⋮n^2+2\)

\(\Rightarrow3n-2⋮n^2+2\)

mà \(3\left(n+3\right)⋮n^2+2\left(n+3⋮n^2+2\right)\)

\(\Rightarrow3\left(n+3\right)-3n+2⋮n^2+2\)

\(\Rightarrow3n+9-3n+2⋮n^2+2\)

\(\Rightarrow11⋮n^2+2\left(n\in Z\right)\)

\(\Rightarrow n^2+2\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow n^2=9\)

\(\Rightarrow\orbr{\begin{cases}n=3\\n=-3\end{cases}}\)

Đối chiều đề bài , ta có \(n=-3\) thỏa mãn .