Nhanh giúp mình với mn

Tìm x hả?

a, 2\(xy\) - 2\(x\) + 3\(y\) = -9

(2\(xy\) - 2\(x\)) + 3\(y\) - 3 = -12

2\(x\)(\(y-1\)) + 3(\(y-1\)) = -12

(\(y-1\))(2\(x\) + 3) = -12

Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12}

Lập bảng ta có:

Theo bảng trên ta có: Các cặp \(x\);\(y\) nguyên thỏa mãn đề bài là:

(\(x;y\)) = (-1; -11); (0; -3); (-3; 5); ( -2; 13)

b, (\(x+1\))²(\(y\) - 3) = -4 

    Ư(4) = {-4; -2; -1; 1; 2; 4}

Lập bảng ta có: 

Theo bảng trên ta có: các cặp \(x;y\) nguyên thỏa mãn đề bài là: 

(\(x;y\)) = (0; -1); (-3; 2); (1; 2)

Tính P = 

P=1+1/3+1/6+1/10+…..+1/2021×2022÷2

P/2=1/2+1/6+1/12+1/20+…..+1/2021×2022

P/2=1/1×2+1/2×3+1/3×4+…….+1/2021×2022

P/2=1-1/2+1/2-1/3+1/3-1/4+….+1/2021-1/2022=1-1/2022=2021/2022

P=2021/1011

Chúc bn học tốt

a,gt <=> (x-1)^3-x(x^2-4)=5

<=>x^3-3x^2+3x-1-x^3+4x-5=0

<=>-3x^2+7x-6=0

<=>x^2-7/3x+2=0

<=>x^2-7/3x+49/36+23/36=0

<=>(x-7/6)^2=-23/36(vl)

=>không tìm được x

b,gt <=> (x-1)^3-(x+3)^3+3(x^2-4)=2

<=>x^3-3x^2+3x-1-x^3-9x^2-27x-27+3x^2-12-2=0

<=>-9x^2-24x-42=0

<=>9x^2+24x+16=-26

<=>(3x+4)^2=-26(vl)

=> điều tương tự câu a

1. \(\frac{x-3}{x+2}=\frac{3}{4}\)

=> \(4(x-3)=3(x+2)\)

=> 4x - 12 = 3x + 6

=> 4x - 12 - 3x = 6

=> 4x - 3x - 12 = 6

=> x = 18

2. \(\frac{4-x}{x-1}=\frac{-5}{6}\)

=> 6[4 - x] = -5[x - 1]

=> 24 - 6x = 5 - 5x

=> 24 - 6x - 5 = 5x

=> 24 - 5 = 6x - 5x

=> 19 = x

=> x = 19

3. \(\frac{3-x}{-x+2}=\frac{-1}{3}\)

=> 3[3 - x] = -1[-x + 2]

=> 9 - 3x = x - 2

=> 9 - 3x + x = -2

=> 9 - 4x = -2

=> 4x = 9 - [ -2]

=> 4x = 9 + 2 = 11

=> x = 11/4

4. \(\frac{x-1}{12}=\frac{-3}{1-x}\)

=> [x-1][1-x] = -36

=> -[x-1]² = -36

=> [x-1]² = 36

=> [x-1]² = 6²

=> x -1 = 6 hoặc x - 1 = -6

=> x = 7 hoặc x = -5

5 . |x - 3 | - 8 = -4

=> |x - 3| = -4 + 8

=> |x - 3| = 4

=> x - 3 = 4 hoặc x - 3 = -4

=> x = 7 hoặc x = -1

Mấy bài sau tương tự

a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)

\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{-2}{3}\)

b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{-11}{20}\)

c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)

\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{-4}{5}\)

d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)

\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{9}\)

e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)

\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)

\(\Rightarrow x=\dfrac{-7}{2}\)

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2021}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2020}{2021}\)

\(A=\frac{1}{2021}\)