ĐKXĐ: \(\hept{\begin{cases}x\ne10\\x\ne-70\end{cases}}\)

\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)

\(\Rightarrow\left(x-20\right)\left(x+70\right)=\left(x+40\right)\left(x-10\right)\)

\(\Rightarrow x^2+70x-20x-1400=x^2-10x+40x-400\)

\(\Rightarrow20x=1000\Rightarrow x=50\)

                                                 Vậy x = 50

Vì 56⋮x,70⋮x ➞ x∈ƯC(56;70)

Ta có:

56=2³.7

70=2.5.7

ƯCLN(56;70)=2.7=14

ƯC(56;70)=Ư(14)={1;2;7;14}

Vì 56⋮x,70⋮x và 10<x<20

⇔x=14

Tham khảo:

Ta có:

56 = 2³ . 7

70 = 2 . 5 . 7

⇒ ƯCLN ( 56 ; 70 ) = 2. 7 =14

⇒ ƯC ( 56 ; 70 ) = { 1 ; 2 ; 7 ; 14 }

mà 10<x<20

⇒ x = 14

Ư(56) = {1; 2; 4; 7; 8; 14; 28; 56}

Ư(70) = {1; 2; 5; 7; 10; 14; 35}

=>x ∈ ƯC (56; 70) = {1; 2; 7; 14}

Mà 10 < x < 20; x ∈ N nên x ∈ 14

\(\frac{x-20}{x-10}=\frac{x+40}{x+70}=\frac{-20-40}{-10-70}=\frac{6}{8}=\frac{3}{4}.\)

\(\Rightarrow4\cdot\left(x-20\right)=3\cdot\left(x-10\right)\Leftrightarrow4x-80=3x-30\Leftrightarrow x=50.\)

bn dinh thuy linh gioi that, toàn dung ngôn ngữ toán hoc nên bai lam rat ấn tuong

\(\left(x-20\right):\left(x-10\right)=\left(x+140\right):\left(x+70\right)\text{ hay }\frac{x-20}{x-10}=\frac{x+140}{x+70}\)

\(\text{Áp dụng tính chất của dãy tỉ số bằng nhau ta có:}\)

\(\frac{x-20}{x-10}=\frac{x+140}{x+70}=\frac{x-20-\left(x+140\right)}{x-10-\left(x+70\right)}=\frac{x-20-x-140}{x-10-x-70}=\frac{-160}{-80}=2\)

\(\text{Suy ra:}\frac{x-20}{x-10}=2\Rightarrow x-20=2.\left(x-10\right)\)

\(x-20=2x-20\)

\(x-2x=-20+20\)

\(-x=0\)

\(x=0\)

x=0              

\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)

\(\Rightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)

\(\Rightarrow x^2+70x-20x-1400=x^2+40x-10x-400\)

\(\Rightarrow x^2+70x-20x-x^2-40x+10x=1400-400\)

\(\Rightarrow20x=1000\Rightarrow x=50\)

a) (x + 15) : x = 4 : 3

=> x : x + 15 : x = \(\frac{4}{3}\)

=> 1 + 15 : x = \(\frac{4}{3}\)

=> 15 : x = \(\frac{4}{3}\)- 1 = \(\frac{1}{3}\)

=> x = 15 : \(\frac{1}{3}\)

=> x = 45

phượng quyên là otaku sa( hình đại diện so cute)

 =>x=45

\(a,\frac{x+15}{x}=\frac{4}{3}\Rightarrow4x=3x+45\Leftrightarrow x=45\)

\(b,\frac{7,5-x}{3,5+x}=\frac{5}{6}\Rightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Rightarrow x=2,5\)

\(c,\frac{x-20}{x-10}=\frac{x+40}{x+70}\Rightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)

\(\Leftrightarrow x^2+50x-1400=x^2+30x-400\)

\(\Leftrightarrow20x=1000\)

\(\Rightarrow x=50\)