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

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 \(B=3\)

Gọi số chia là: \(x\) ( đk \(x\in\) N*; \(x\) > 5}

Thương là: \(\dfrac{59-5}{x}\) = \(\dfrac{54}{x}\) 

\(x\) \(\in\) Ư(54) = { 1; 2; 3; 6; 9; 18; 27; 54}

Vì \(x\) > 5 \(\Rightarrow\) \(x\) \(\in\) { 6; 9; 18; 27; 54}

Vậy số chia lần lượt là: 6; 9; 18; 27; 54

Thương tương ứng là: 9; 6; 3; 2; 1

Số bị chia bớt đi 5 là 59-5=54 chia hết cho số chia

=> số chia là ước của 54

=> SC = {1; 2; 3; 6; 9; 18; 27; 54}

=> thương = {54; 27; 12; 9; 6; 3; 2; 1}

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

x = 4

567 - 765 + 875543: 46 = \(\dfrac{26082}{46}\) - \(\dfrac{35190}{46}\) +  \(\dfrac{875543}{46}\) = \(\dfrac{866453}{46}\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{x-3\sqrt{x}+8}{x-7\sqrt{x}+10}-\dfrac{\sqrt{x}-1}{\sqrt{x}-5}\left(x\ge0,x\ne\left\{4;25\right\}\right)\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{x-3\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}-5}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)+x-3\sqrt{x}+8-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{x-5\sqrt{x}+x-3\sqrt{x}+8-\left(x-3\sqrt{x}+2\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{x-5\sqrt{x}+6}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}-5}\)

Để A nguyên : \(\dfrac{\sqrt{x}-3}{\sqrt{x}-5}=1+\dfrac{2}{\sqrt{x}-5}\in Z\)

\(=>\dfrac{2}{\sqrt{x}-5}\in Z=>\sqrt{x}-5\in\left\{1;-1;2;-2\right\}\)

\(=>\sqrt{x}\in\left\{6;4;7;3\right\}\\ =>x\in\left\{36;16;49;9\right\}\) (TMDK)

x ϵ {36;16;49;9}