Ta có: \(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(\Rightarrow P=\dfrac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{5^{32}-1}{2}\)

Ta có:

( 5 2 - 1).P = ( 5 2  – 1).12.( 5 2  + 1)( 5 4  + 1)( 5 8  + 1)( 5 16  + 1)

= 12.(  5 2  – 1).( 5 2  + 1)( 5 4 + 1)( 5 8  + 1)( 5 16 + 1)

= 12.(  5 4  - 1)(  5 4  + 1)(  5 8  + 1)( 5 16  + 1)

= 12.(  5 8  - 1)(  5 8  + 1)( 5 16  + 1)

= 12.(  5 16  - 1)( 5 16  + 1)

= 12.(  5 32  - 1)

\(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\left(5^{128}-1\right)=2.5^{128}-2\)

c: Ta có: \(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{64}-1\right)\left(5^{64}+1\right)\)

\(=2\cdot\left(5^{128}-1\right)\)

\(=2\cdot5^{128}-2\)

d: Ta có: \(E=\left(x-2\right)^3-\left(x+1\right)\left(x^2-x+1\right)+6\left(x-1\right)^2\)

\(=x^3-6x^2+12x-8-x^3-1+6x^2-12x+6\)

\(=-3\)

Bài4:

=>x(x^2+1)=0

>x=0

Bài 5: 

=>\(3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)

Bài 4:

x^3+x=0

=>x(x^2+1)=0

=>x=0

Bài 5:

\(3n^3+10n^2-5⋮3n+1\)

\(\Leftrightarrow3n^3+n^2+9n^2-1-4⋮3n+1\)

=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)

                                     Bài giải:

Theo bài ra ta có:

Tổ 1 và tổ 2: 52 cây

Tổ 2 và tổ 3: 54 cây

Tổ 3 và tổ 4: 52 cây

Tổ 1 và tổ 4: 58 cây

Nhìn vào sơ đồ ta thấy mỗi tổ đều xuất hiện 2 lần vậy:

Tổng số các tổ trồng được là:

( 52 + 54 + 52 + 58 ) : 2 = 108 ( cây )

Trung bình mỗi tổ trồng được số cây là:

108 : 4 = 27 ( cây )

Đáp số: 27 cây

MÌNH KO CHẮC LẮM ĐÂU NHƯNG NẾU ĐÚNG THÌ TICK GIÚP MÌNH NHA

trung bình mỗi tổ sẽ lấy tổng số cây của 4 tổ cộng lại chia

                   tổng số cây của 4 tổ là:

                           52+52=104[cây]

                   trung bình mỗi tổ trồng được số cây là:

                           104:4=26[cây]

                                     đáp số:26 cây

12 + 14 + 16 + 18 + 20 + ... + 50 + 52 + 54 + 56 + 58

= ( 12 + 58 ) + ( 14 + 56 ) + ... + ( 34 + 36 ) 

= 70 + 70 + ... + 70 

= 70 x 12 

= 840

~ Hok tốt ~

=(12+58)+(14+56)+(16+54)+....+(30+30)

=    60     +   60     +      60  +.....+  60

=      60  x   12

=       720

Hok tốt nha

12+14+16+....+58

=(12+58)×12

=70×12

=840

Bài 1: 

1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)

\(=-12-18\)

=-30

2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)

\(=36-2020+2019-136-27\)

\(=1-100-27\)

\(=-126\)

3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)

\(=144-97-244+197\)

\(=-100+100=0\)

4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)

\(=-24\cdot13+24\cdot3\)

\(=24\cdot\left(-13+3\right)\)

\(=24\cdot\left(-10\right)=-240\)

5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)

\(=54+55+56+57+58-64-65-66-67-68\)

\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)

\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)

=-50

6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)

\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)

\(=-24\cdot5+16\cdot5\)

\(=5\cdot\left(-24+16\right)\)

\(=-5\cdot8=-40\)

7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)

\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)

\(=47\cdot50-23\cdot50\)

\(=50\cdot\left(47-23\right)\)

\(=50\cdot24=1200\)

8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)

\(=-31\cdot\left(47+52+1\right)\)

\(=-31\cdot100=-3100\)

Bài 2: 

1) Ta có: \(-17-\left(2x-5\right)=-6\)

\(\Leftrightarrow-17-2x+5+6=0\)

\(\Leftrightarrow-2x-6=0\)

\(\Leftrightarrow-2x=6\)

hay x=-3

Vậy: x=-3

2) Ta có: \(10-2\left(4-3x\right)=-4\)

\(\Leftrightarrow10-8+6x+4=0\)

\(\Leftrightarrow6x+6=0\)

\(\Leftrightarrow6x=-6\)

hay x=-1

Vậy: x=-1

3) Ta có: \(-12+3\left(-x+7\right)=-18\)

\(\Leftrightarrow-12-3x+21+18=0\)

\(\Leftrightarrow-3x+27=0\)

\(\Leftrightarrow-3x=-27\)

hay x=9

Vậy: x=9

4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)

\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)

\(\Leftrightarrow-2x-3=-3\)

\(\Leftrightarrow-2x=0\)

hay x=0

Vậy: x=0

5) Ta có: x(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3\right\}\)

6) Ta có: (x-2)(x+4)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-4\right\}\)

7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-1;3\right\}\)

Bài 1: 

1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)

=−12−18=−12−18

=-30

2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27

=36−2020+2019−136−27=36−2020+2019−136−27

=1−100−27=1−100−27

=−126

Tớ chcs cậu học thật giỏi nha !