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\(\frac{x-20}{x-10}=\frac{x+40}{x+70}=\frac{-20-40}{-10-70}=\frac{6}{8}=\frac{3}{4}.\)
\(\Rightarrow4\cdot\left(x-20\right)=3\cdot\left(x-10\right)\Leftrightarrow4x-80=3x-30\Leftrightarrow x=50.\)
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-3 = -1
=> x = 2
hoặc x-3 = 0
x = 3
hoặc x-3 = 1
x = 4
Vậy x\(\in\){2; 3; 4} thì (x-3)20 = (x-3)10
\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)
\(\Rightarrow x+5=20\)
\(\Rightarrow x=20-5\)
\(\Rightarrow x=15\)
\(B=\left|x-10\right|+\left|20-x\right|\ge\left|x-10+20-x\right|=\left|10\right|=10\)
dấu = xảy ra khi \(\left(x-10\right).\left(20-x\right)\ge0\)
\(\Rightarrow10\le x\le20\)
Vậy min B =10 khi \(10\le x\le20\)