Ta có :

\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{b^2}{c}\\d=\dfrac{c^2}{b}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{d}=\dfrac{b^2}{c}:\dfrac{c^2}{b}\)

\(\Rightarrow\dfrac{a}{d}=\dfrac{b^2}{c}.\dfrac{b}{c^2}\)

\(\Rightarrow\dfrac{a}{d}=\dfrac{b^3}{c^3}=\dfrac{8b^3}{8c^3}=\dfrac{a^3}{b^3}=\dfrac{125c^3}{125d^3}\)

\(\Rightarrow\dfrac{a}{d}=\dfrac{a^3+8b^3+125c^3}{b^3+8c^3+125d^3}\left(dpcm\right)\)

Vì x;y là 2 đại lượng tỉ lệ nghịch nên

\(x_1.y_1=x_2.y_2\)

\(\Rightarrow y_2=\dfrac{x_1.y_1}{x_2}=\dfrac{45}{9}=5\)

`#040911`

`-12(11 - 18) - (x + 1)^2 = 59`

`\Rightarrow -12. (-7) - (x + 1)^2 = 59`

`\Rightarrow 84 - (x + 1)^2 = 59`

`\Rightarrow (x + 1)^2 = 84 - 59`

`\Rightarrow (x + 1)^2 =25`

`\Rightarrow (x + 1)^2 =`\(\left(\pm5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5-1\\x=-5-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-6;4\right\}.\)

(-11)² - 3.(\(x\) - 4) = 76

121 - 3\(x\) + 12 = 76

  121 + 12 - 76 = 3\(x\)

3\(x\)           = 57

    \(x\)         = 57 : 3

     \(x\)         = 19

\(121-3\left(x-4\right)=76\)

\(3\left(x-4\right)=126-76\)

\(3\left(x-4\right)=50\)

\(x-4=\dfrac{50}{3}\)

\(x=\dfrac{62}{3}\)

Gọi \(a\) là cạnh hình vuông lớn (cm)

\(\Rightarrow a-7\) là cạnh hình vuông nhỏ (cm)

Theo đề bài ta có :

\(a^2-\left(a-7\right)^2=175\)

\(\Rightarrow a^2-\left(a^2-14a+49\right)=175\)

\(\Rightarrow a^2-a^2+14a-49=175\)

\(\Rightarrow14a=224\)

\(\Rightarrow a=16\)

Chu vi mảnh đất là :

\(4a=4.16=64\left(m\right)\)

Sau khi bán buổi sáng thi số lê còn lại là \(\dfrac{2}{5}\)

Phân số chỉ số lê bán được so với số lê ban đầu:

\(\dfrac{2}{5}\cdot\dfrac{3}{4}=\dfrac{3}{10}\)

Phân số chỉ 9 quả lê bán được so với số lê ban đầu:

\(1-\dfrac{3}{5}-\dfrac{3}{10}=\dfrac{1}{10}\)

Số lê ban đầu:

\(9:\dfrac{1}{10}=90\left(quả\right)\)

1)

xy + x - 4y = 12

x + y(x - 4) = 12

y(x - 4) = 12 - x

\(y=\dfrac{-x+12}{x-4}\)

Vì \(x,y\inℕ\) nên

\(\left(-x+12\right)⋮\left(x-4\right)\)

\(\left(-x+12\right)-\left(x-4\right)⋮\left(x-4\right)\)

\(16⋮\left(x-4\right)\)

\(\left(x-4\right)\inƯ\left(16\right)\)

\(\left(x-4\right)\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)

\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12\right\}\)

\(y\in\left\{\dfrac{-5+12}{5-4};\dfrac{-3+12}{3-4};\dfrac{-6+12}{6-4};\dfrac{-2+12}{2-4};\dfrac{-8+12}{8-4};\dfrac{-0+12}{0-4};\dfrac{-12+12}{12-4};\dfrac{4+12}{-4-4};\dfrac{-20+12}{20-4};\dfrac{12+12}{-12-4}\right\}\)

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

\(\left(x;y\right)\in\left\{\left(5;7\right);\left(3;-9\right);\left(6;3\right);\left(2;-5\right);\left(8;1\right);\left(0;-3\right);\left(12;0\right);\left(-4;-2\right);\left(20;-\dfrac{1}{2}\right);\left(-12;-\dfrac{7}{5}\right)\right\}\)

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)

2)

(2x + 3)(y - 2) = 15

\(\left(2x+3\right)\inƯ\left(15\right)\)

\(\left(2x+3\right)\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

Ta lập bảng

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)

các thầy cô ơi giúp em vs ạ mai em phải nộp r ạ!!!

\(M=x^{10}-25x^9+25x^8-25x^7+...-25x^3+25x^2-25x+25\)

Ta thấy : \(x=24\Rightarrow x+1=25\)

\(\Rightarrow M=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)

\(M=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(\Rightarrow M=1\)

Vậy \(M=1\left(tạix=24\right)\)

M=x 

10

 −25x 

9

 +25x 

8

 −25x 

7

 +...−25x 

3

 +25x 

2

 −25x+25

Ta thấy : 

x

=

24

⇒

x

+

1

=

25

x=24⇒x+1=25

⇒

M

=

x

10

−

(

x

+

1

)

x

9

+

(

x

+

1

)

x

8

−

(

x

+

1

)

x

7

+

.

.

.

−

(

x

+

1

)

x

3

+

(

x

+

1

)

x

2

−

(

x

+

1

)

x

+

(

x

+

1

)

⇒M=x 

10

 −(x+1)x 

9

 +(x+1)x 

8

 −(x+1)x 

7

 +...−(x+1)x 

3

 +(x+1)x 

2

 −(x+1)x+(x+1)

M

=

x

10

−

x

10

−

x

9

+

x

9

+

x

8

−

x

8

−

x

7

+

.

.

.

−

x

4

−

x

3

+

x

3

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+

x

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1

M=x 

10

 −x 

10

 −x 

9

 +x 

9

 +x 

8

 −x 

8

 −x 

7

 +...−x 

4

 −x 

3

 +x 

3

 +x 

2

 −x 

2

 −x+x+1

⇒

M

=

1

⇒M=1

Vậy 

M

=

1

(

t

ạ

i

x

=

24

)

M=1(tạix=24)

\(M=x^{2023}-2023.\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)

Ta có : \(x=2022\Rightarrow x+1=2023\)

\(\Rightarrow M=x^{2023}-\left(x+1\right).\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)

\(\Rightarrow M=x^{2023}-\left(x+1\right)x^{2022}+\left(x+1\right)x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}+...-\left(x+1\right)x^2+\left(x+1\right)x\)

\(\Rightarrow M=x^{2023}-x^{2023}-x^{2022}+x^{2022}+x^{2021}-x^{2021}-x^{2020}+x^{2020}+x^{2019}-x^{2019}-...-x^3-x^2+x^2+x\)

\(\Rightarrow M=x\)

\(\Rightarrow M=2022\)

Vậy \(M=2022\left(tạix=2022\right)\)