=(2^3)^3-1=(2^3-1)(2^6+2^3+1)=7(64+8+1)=7.73

b) Câu hỏi của Nguyễn Hoàng Thanh Trúc - Toán lớp 8 - Học toán với OnlineMath tham khảo

1.=(x-2)(x 2+2x+7)+2(x-2)(x+2)-5(x-2) = 0
=>(x-2)(x 2+2x+7+2x+4-5) = 0
=>(x-2)(x 2+4x+6) = 0
Mà x 2+4x+6 (E Z)
=> x 2+4x+6 > 0
Vậy (x-2)=0 => x = 2
 

Bài 1 : Tìm x, biết :

\(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\) \(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+\left(x-2\right)\left(2\left(x+2\right)-5\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2\left(x+2\right)-5\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\\left(x+2\right)^2+2>0\end{matrix}\right.\Rightarrow x=2\)

a: \(=\left(2^3-1\right)\left(2^6+2^3+1\right)=73\cdot7⋮73\)

b: Đề sai rồi bạn

\(a;43^2+43.17=43\left(43+17\right)=43.60⋮60\left(đpcm\right)\)

\(b;27^5-3^{11}=3^{15}-3^{11}=3^{11}\left(3^4-1\right)=3^{11}.80⋮80\left(đpcm\right)\)

a)43^2+43.17=43(43+17)=43.60 chia het cho 60

b)27^5-3^11=3^15-3^11=3^11.(3^4-1)=3^11.80 chia het cho 80

nho k ung ho mik nhe

a/ 2⁹ - 1 = \(\left(2^3\right)^3\) - 1 = 8³ - 1 = (8-1)( 8² +8.1 + 1) = (8-1).73 \(⋮\) 73

b/ 5⁶ - 10⁴ = 5⁴(5² - 2⁴) = 5⁴ (25 - 16) = 5⁴ .9 chia hết cho 9

c, (n+6)²-(n-6)²=(n+6-n+6)(n+6+n-6)(hđt số 3

=12 .2n=24n

1)\(\overline{73\cdot}⋮2;9\)

\(\Rightarrow\cdot\in\left\{0;2;4;8\right\}\)mà \(\overline{73\cdot}⋮9\)

nên *=8

2)\(\overline{589\cdot}\)\(⋮\)2;3

\(\Rightarrow\cdot\in\left\{0;2;4;6\right\}\)và *\(\in\left\{2;5;8\right\}\)

\(\Rightarrow\)*=2

3)8*9* \(⋮2;3;5;9\)

Để *\(⋮2;5\)thì * sau =0

Ta có:8*90

Để: *\(⋮\)3 thì *\(\in\left\{1;4;7\right\}\)

Để *\(⋮9\)thì *\(=1\)

\(\Rightarrow\)*=1

Ta có:8190

1, *=8

2,*=2

3,*=1;*=0

1 . Số 738 

2 . Số 5892 ; 5898

3 . Số 8190