\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.

Tìm \(x\) câu a:

\(\frac13.x\) + \(\frac25.\left(x+1\right)\) = 0

\(\frac{5}{15}x\) + \(\frac{6}{15}x\) + \(\frac25\) = 0

\(\frac{11}{15}x\) = - \(\frac25\)

\(x=-\frac25:\frac{11}{15}\)

\(x\) = - \(\frac25\times\frac{15}{11}\)

\(x\) = - \(\frac{6}{11}\)

Vậy \(x=-\frac{6}{11}\)

Tìm \(x\) câu b:

\(x\) x 25% = 0,5

\(x\times0,25\) = 0,5

\(x=0,5:0,25\)

\(x=2\)

Vậy \(x=2\)

Câu 1a:

1/3x + 2/5(x + 1) = 0

1/3x + 2/5x + 2/5 = 0

1/3x + 2/5x = - 2/5

x(1/3 + 2/5) = -2/5

x.(5/15 + 6/15) = -2/5

x.11/15 = - 2/5

x = - 2/5 : 11/15

x = - 6/11

Vậy x = -6/11

Câu b:

x . 25%. x = 0,5

x.x = 0,5 : 25%

x^2 = 2

x = - \(\sqrt2\); x = \(\sqrt2\)

Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )

Câu 1a:

1/3x + 2/5(x + 1) = 0

1/3x + 2/5x + 2/5 = 0

1/3x + 2/5x = - 2/5

x(1/3 + 2/5) = -2/5

x.(5/15 + 6/15) = -2/5

x.11/15 = - 2/5

x = - 2/5 : 11/15

x = - 6/11

Vậy x = -6/11

Câu b:

x . 25%. x = 0,5

x.x = 0,5 : 25%

x^2 = 2

x = - \(\sqrt2\); x = \(\sqrt2\)

Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )

khó @gmail.com

nè, không làm thôi ằ nhagg. khó thì đừng gửi câu trả lời làm gì cho mệt nha bạn 

Bài 1:

a; \(\dfrac{x}{3}\) = \(\dfrac{4}{y}\)

     \(xy\) = 12

12 = 2².3; 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:

(\(x\)\(;y\)) =(-12; -1);(-6; -2);(-4; -3);(-2; -6);(-1; 12);(1; 12);(2;6);(3;4);(4;3);(6;2);(12;1)

b; \(\dfrac{x}{y}\) = \(\dfrac{2}{7}\)

    \(x\) = \(\dfrac{2}{7}\).y 

     \(x\) \(\in\)z ⇔ y ⋮ 7

   y = 7k;

   \(x\) = 2k 

Vậy \(\left\{{}\begin{matrix}x=2k\\y=7k;k\in z\end{matrix}\right.\) 

Câu 1a:

\(-\frac36\) = \(\frac{x}{-10}\) = \(\frac{-7}{y}\) = \(\frac{z}{18}\)

-3/6 = x/-10 ⇒ x = -3/6 x (-10) = 5

- 3/6 = - 7/y ⇒ y = -7 : (-3/6) = 14

-3/6 = z/18 ⇒ z = -3/6 x 18 = - 9

Vậy (x; y; z) = (5; 14; -9)

Câu 1b:

\(\frac{x+5}{x-2}=\frac23\)

3.(x+5) = 2.(x -2)

3x + 15 = 2x - 4

3x - 2x = - 15 - 4

x = - 19

Vậy x = - 19