\(2\dfrac{3}{x}=\dfrac{13}{x}\)(đk \(x\ne0\))

\(\dfrac{2x+3}{x}=\dfrac{13}{x}\)

\(\Rightarrow\left(2x+3\right).x=13.x\)

\(\Rightarrow2x+3=13\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\left(tmđk\right)\)

Vậy x=5

mn giúp em vs ạ chiều em pk nộp rồi ạ!!!

gợi ý nè:

thử cộng chúng lại xem

\(\dfrac{x}{y+z+1}\) = \(\dfrac{y}{x+z+2}\) = \(\dfrac{z}{x+y-3}\) = \(x+y+z\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{y+z+1}\)=\(\dfrac{y}{x+z+2}\)=\(\dfrac{z}{x+y-3}\)=\(\dfrac{x+y+z}{y+z+1+x+z+2+x+y-3}\)

\(x+y+z\) = \(\dfrac{x+y+z}{2.\left(x+y+z\right)}\) = \(\dfrac{1}{2}\) (1)

\(\dfrac{x}{y+z+1}\) = \(\dfrac{1}{2}\) ⇒ 2\(x\) = y+z+1 

⇒ 2\(x\) + \(x\) = \(x+y+z+1\) (2)

 Thay (1) vào (2) ta có: 2\(x\) + \(x\) = \(\dfrac{1}{2}\) + 1

                                      3\(x\)      = \(\dfrac{3}{2}\) ⇒ \(x=\dfrac{1}{2}\)

\(\dfrac{y}{x+z+2}\) = \(\dfrac{1}{2}\) ⇒ 2y = \(x+z+2\) ⇒ 2y+y = \(x+y+z+2\) (3)

Thay (1) vào (3) ta có: 2y + y = \(\dfrac{1}{2}\) + 2 

                                   3y = \(\dfrac{5}{2}\) ⇒ y = \(\dfrac{5}{6}\)

Thay \(x=\dfrac{1}{2};y=\dfrac{5}{6}\) vào (1) ta có: \(\dfrac{1}{2}+\dfrac{5}{6}+z\) = \(\dfrac{1}{2}\)

                                                              \(\dfrac{5}{6}\) + z = 0 ⇒ z = - \(\dfrac{5}{6}\)

Kết luận: (\(x;y;z\)) = (\(\dfrac{1}{2}\); \(\dfrac{5}{6}\); - \(\dfrac{5}{6}\))

\(2\dfrac{3}{x}\) là hỗn số hay là \(2\times\dfrac{3}{x}\) vậy bạn .

5

\(\dfrac{5}{2}m=2,5m\)

Chọn Câu A,B,F

B = \(\dfrac{2x-3}{x-1}\) ( đkxđ \(x\) \(\ne\) 1)

B \(\in\) Z ⇔ 2\(x\) - 3 ⋮ \(x\) - 1

         2\(x\) - 2 - 1 ⋮ \(x\) - 1

     2(\(x-1\)) - 1 ⋮ \(x\) - 1

                    1 ⋮ \(x-1\)

     \(x-1\)  \(\in\) Ư(1) = { -1; 1}

      \(x\) \(\in\) { 0; 2}

Để biểu thức B thỏa mãn điều kiện gì vậy em?

B = (2\(x\) - 5)² - 1

(2\(x\) - 5)² ≥ 0 ∀ \(x\) ⇒ (2\(x\) - 5)² - 1 ≥ -1  ∀ \(x\)

B_min = -1 ⇔ \(x\) = \(\dfrac{5}{2}\)

A = (\(x\) + 5)² + 4

(\(x\) + 5)² ≥ 0 ∀ \(x\) ⇒ (\(x\) + 5)² + 4 ≥ 4 ∀ \(x\)

A_min = 4 ⇔ \(x=-5\)

\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{2}\)

\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)

Vậy \(B=3\)

\(\dfrac{x-1}{3}=\dfrac{2-x}{-2}\\ =>-2\left(x-1\right)=3\left(2-x\right)\\ =>-2x+2=6-3x\\ =>3x-2x=6-2\\ =>x=4\)

x = 4