Lời giải:

Gọi số tự nhiên cần tìm là $a$

$a\vdots 5,7,9$

$\Rightarrow a=BC(5,7,9)$

$\Rightarrow a\vdots BCNN(5,7,9)$

$\Rightarrow a\vdots 315$

Vậy $a=315k$ với $k$ là stn.

Vì $a$ có 4 chữ số nên $315k< 10000$

$\Rightarrow k< 31,7$

Để $a$ lớn nhất thì $k$ lớn nhất

$\Rightarrow k=31$

$a=315.31=9756$

a ⋮ 5; 7; 9 ⇒ a \(\in\) BC(5; 7;9) 

5= 5; 7 = 7; 9 =3² ⇒ BCNN(5; 7; 9) = 3².5.7 = 315

a   \(\in\) BC(315) = {0; 315; 630;1890;...;9765;10080}

Vì a là số tự nhiên lớn nhất có 4 chữ số nên a = 9765

1) -x - 3 = 7

-x = 7 + 3

-x = 10

x = -10

2) x + 5 = -10

x = -10 - 5

x = -15

3) 2x - 7 = 713

2x = 713 + 7

2x = 720

x = 720 : 2

x = 360

4) -129 - (35 - x) = 55

35 - x = -129 - 55

35 - x = -184

x = 35 - (-184)

x = 219

5) 103 - x = 16 - (13 - 8)

103 - x = 16 - 5

103 - x = 11

x = 103 - 11

x = 192

\(\left(12+x\right)-\left(22+x\right)\)

\(=12+x-22-x\)

\(=-10\)

Vậy giá trị của biểu thức đã cho là \(-10\) với mọi \(x\)

\(ƯCLN\left(1;2;3;...;100\right)+ƯCLN\left(1^2;2^2;3^2;...;100^2\right)+ƯCLN\left(1^3;2^3;3^3;...;100^2\right)+...+ƯCLN\left(1^{100};2^{100};3^{100};...;100^{100}\right)\)

\(=1+1+1+...+1\) (100 chữ số 1)

\(=100\)

Bài 10

a) \(5-2x=13\)

\(2x=5-13\)

\(2x=-8\)

\(x=-\dfrac{8}{2}\)

\(x=-4\)

b) \(-45:\left(3x-17\right)=\left(-3\right)^2\)

\(-45:\left(3x-17\right)=9\)

\(3x-17=-45:9\)

\(3x-17=-5\)

\(3x=-5+17\)

\(3x=12\)

\(x=\dfrac{12}{3}\)

\(x=4\)

c) \(-12x-23=\left(-3\right)^3+\left(-7\right).8\)

\(-12x-23=-27-56\)

\(-12x-23=-83\)

\(-12x=-83+23\)

\(-12x=-60\)

\(x=\dfrac{-60}{-12}\)

\(x=5\)

d) \(x^2-16=3^8:3^6\)

\(x^2-16=9\)

\(x^2=9+16\)

\(x^2=25\)

\(x=5;x=-5\)

e) \(x^3+15=-12\)

\(x^3=-12-15\)

\(x^3=-27\)

\(x^3=\left(-3\right)^3\)

\(x=-3\)

f) \(3x^2-7=5\)

\(3x^2=5+7\)

\(3x^2=12\)

\(x^2=\dfrac{12}{3}\)

\(x^2=4\)

\(x=2;x=-2\)

g) \(\left(x-5\right)\left(x^2-9\right)=0\)

\(x-5=0;x^2-9=0\)

*) \(x-5=0\)

\(x=5\)

*) \(x^2-9=0\)

\(x^2=9\)

\(x=-3;x=3\)

Vậy \(x=-3;x=3;x=5\)

Bài 9

a) \(\left(x-2\right)\left(y+3\right)=5\)

\(\Rightarrow\left(x-2;y+3\right)\in\left\{\left(1;5\right);\left(-1;-5\right);\left(5;1\right);\left(-5;-1\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(3;2\right);\left(1;-8\right);\left(7;-2\right);\left(-3;-4\right)\right\}\)

b) \(\left(x+4\right)\left(y-5\right)=3\)

\(\Rightarrow\left(x+4;y-5\right)\in\left\{\left(1;3\right)\left(-1;-3\right);\left(3;1\right);\left(-3;-1\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-3;8\right);\left(-5;2\right);\left(-1;6\right);\left(-7;4\right)\right\}\)

c) \(\left(2x+1\right)\left(y-3\right)=8\)

\(\Rightarrow\left(2x+1;y-3\right)\in\left\{\left(-8;-1\right);\left(-4;-2\right);\left(-2;-4\right);\left(-1;-8\right);\left(1;8\right);\left(2;4\right)\left(4;2\right);\left(8;1\right)\right\}\)

\(\Rightarrow\left(2x;y\right)\in\left\{\left(-9;2\right);\left(-5;1\right);\left(-3;-1\right);\left(-2;-5\right);\left(0;11\right);\left(1;7\right);\left(3;5\right);\left(7;4\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-\dfrac{9}{2};2\right);\left(-\dfrac{5}{2};1\right);\left(-\dfrac{3}{2};-1\right);\left(-1;-5\right);\left(0;11\right);\left(\dfrac{1}{2};7\right);\left(\dfrac{3}{2};5\right);\left(\dfrac{7}{2};4\right)\right\}\)

Mà \(x;y\in Z\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-1;-5\right);\left(0;11\right)\right\}\)

d) \(\left(3x-2\right)\left(2y-3\right)=12\)

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

\(\Rightarrow\left(3x;2y\right)\in\left\{\left(-10;2\right);\left(-4;1\right);\left(-2;0\right);\left(-1;-1\right);\left(0;-3\right);\left(1;-9\right);\left(3;15\right);\left(4;9\right);\left(5;7\right);\left(6;6\right);\left(8;5\right);\left(14;4\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-\dfrac{10}{3};1\right);\left(-\dfrac{4}{3};\dfrac{1}{2}\right);\left(-\dfrac{2}{3};0\right);\left(-\dfrac{1}{3};-\dfrac{1}{2}\right);\left(0;-\dfrac{3}{2}\right);\left(\dfrac{1}{3};-\dfrac{9}{2}\right);\left(1;\dfrac{15}{2}\right);\left(\dfrac{4}{3};\dfrac{9}{2}\right);\left(\dfrac{5}{3};\dfrac{7}{2}\right);\left(2;3\right):\left(\dfrac{8}{3};\dfrac{5}{2}\right);\left(\dfrac{14}{3};2\right)\right\}\)

Mà \(x,y\in Z\)

\(\Rightarrow\left(x;y\right)=\left(2;3\right)\)

e) \(\left(x+7\right)\left(2x-y\right)=3\)

\(\Rightarrow\left(x+7;2x-y\right)\in\left\{\left(-3;-1\right);\left(-1;-3\right);\left(1;3\right);\left(3;1\right)\right\}\)

\(\Rightarrow\left(x;2x-y\right)\in\left\{\left(-10;-1\right);\left(-8;-3\right);\left(-6;3\right);\left(-4;1\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-10;-19\right);\left(-8;-13\right);\left(-6;-15\right);\left(-4;-9\right)\right\}\)

f) \(\left(2x-1\right)\left(y^2-13\right)=4\)

\(\Rightarrow\left(2x-1;y^2-13\right)\in\left\{\left(-4;-1\right);\left(-2;-2\right);\left(-1;-4\right);\left(1;4\right);\left(2;2\right)\left(4;1\right)\right\}\)

\(\Rightarrow\left(2x;y^2\right)\in\left\{\left(-3;12\right);\left(-1;11\right);\left(0;7\right);\left(2;17\right);\left(3;15\right);\left(5;14\right)\right\}\)

Do \(y\in Z\Rightarrow y^2\) là số chính phương

Mà \(12;11;7;17;15;14\) không phải số chính phương

\(\Rightarrow\) Không tìm được cặp giá trị \(\left(x;y\right)\in Z\) thỏa mãn yêu cầu đề bài

Lời giải:
$B=1+(5+5^2+5^3)+(5^4+5^5+5^6)+....+(5^{88}+5^{89}+5^{90})$

$=1+5(1+5+5^2)+5^4(1+5+5^2)+....+5^{88}(1+5+5^2)$

$=1+(1+5+5^2)(5+5^4+....+5^{88})$

$=1+31(5+5^4+...+5^{88})\not\vdots 31$
Ta có đpcm.

134(-1+51-48)=134*2=268

(-134)+51.134+(-134).48

=134.(-1)+51.134+134.(-48)

=134.[(-1)+51+(-48)]

=134.2

 =268