Tìm x biết:
\(\dfrac{x+1}{58}+\dfrac{x+2}{57}=\dfrac{x+3}{56}+\dfrac{x+4}{55}\)
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Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}-\dfrac{x-60}{56}-\dfrac{x-60}{55}-\dfrac{x-60}{54}=0\)
\(\Leftrightarrow x-60=0\)
hay x=60
\(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}=\dfrac{x-4}{56}-1+\dfrac{x-5}{55}-1+\dfrac{x-6}{54}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{56}+\dfrac{x-60}{55}+\dfrac{x-60}{54}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}-\dfrac{1}{54}\right)=0\)
\(\Leftrightarrow x-60=0\)
\(\Rightarrow x=60\)
vậy \(S=\left\{60\right\}\)
\(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(< =>\dfrac{x+1}{59}+1+\dfrac{x+3}{57}+1+\dfrac{x+5}{55}+1=\dfrac{x+7}{53}+1+\dfrac{x+9}{51}+1+\dfrac{x+11}{49}+1\)
\(< =>\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}=\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(< =>\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)=0\\ < =>x+60=0\\ < =>x=-60\)
Ta có : \(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(\Leftrightarrow\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}+3\text{=}\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}+3\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}\text{=}\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}-\dfrac{x+60}{53}-\dfrac{x+60}{51}-\dfrac{x-60}{49}\text{=}0\)
\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)\text{=}0\)
\(Do\) \(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\ne0\)
\(\Leftrightarrow\left(x+60\right)\text{=}0\)
\(x\text{=}-60\)
\(Vậy...\)
a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)
b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)
\(\dfrac{x+1}{60}+\dfrac{x+2}{59}=\dfrac{x+3}{58}+\dfrac{x+4}{57}\)
\(\Leftrightarrow\dfrac{x+1}{60}+1+\dfrac{x+2}{59}+1=\dfrac{x+3}{58}+1+\dfrac{x+4}{57}+1\)
\(\Leftrightarrow\dfrac{x+1+60}{60}+\dfrac{x+2+59}{59}=\dfrac{x+3+58}{58}+\dfrac{x+4+57}{57}\)
\(\Leftrightarrow\dfrac{x+61}{60}+\dfrac{x+61}{59}-\dfrac{x+61}{58}-\dfrac{x+61}{57}=0\)
\(\Leftrightarrow\left(x+61\right)\left(\dfrac{1}{60}+\dfrac{1}{59}-\dfrac{1}{58}-\dfrac{1}{57}\right)=0\)
\(\Leftrightarrow x+61=0\)
\(\Leftrightarrow x=-61\)
a,Phương trình đã cho tương đương với:
( \(\dfrac{x+1}{58}\)+1) + ( \(\dfrac{x+2}{57}\)+1) =(\(\dfrac{x+3}{56}\)+1) +(\(\dfrac{x+4}{55}\)+1)
\(\Leftrightarrow\)(x +59) (\(\dfrac{1}{58}\)+\(\dfrac{1}{57}\)- \(\dfrac{1}{56}\)-\(\dfrac{1}{55}\))
\(\Leftrightarrow\)(x+59)=0 \(\Rightarrow\)x = -59
b,
\(\dfrac{x+a}{a-2}\)+\(\dfrac{x-a}{a+2}\)=\(\dfrac{x+2}{a+2}\)+\(\dfrac{2x-4}{a-2}\)
\(\Leftrightarrow\)(x+a)(a+2)+(x-a)(a-2) = (x+2)(a-2) +(2x-4)(a+2)
\(\Leftrightarrow\)(a+2)x = 6(a+2)
\(\Leftrightarrow\)x=6 (do a+2 \(\ne\)0)
Vậy phương trình đã cho có nghiệm duy nhất x=6 (với a\(\ne\) \(\pm\)2)
a, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Leftrightarrow\left(\dfrac{59-x}{49}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{45}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right).\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
Mà \(\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)\ne0\)
\(\Rightarrow100-x=0\)
\(\Rightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
b, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)
\(\Leftrightarrow6x^2-5x+3=-7x+6x^2\)
\(\Leftrightarrow6x^2-5x+3+7x-6x^2=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
Vậy \(S=\left\{\dfrac{-3}{2}\right\}\)
a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)
b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)
a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền
b, - 1 hoặc + 1 vs mỗi phân số nha
\(\dfrac{x+1}{58}+\dfrac{x+2}{57}=\dfrac{x+3}{56}+\dfrac{x+4}{55}\)
\(\Leftrightarrow\left(\dfrac{x+1}{58}+1\right)+\left(\dfrac{x+2}{57}+1\right)=\left(\dfrac{x+3}{56}+1\right)+\left(\dfrac{x+4}{55}+1\right)\)
\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}-\dfrac{x+59}{56}-\dfrac{x+59}{55}=0\)
\(\Leftrightarrow\left(x+59\right)\left(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\right)=0\)
\(\Leftrightarrow x+59=0\)
\(\Leftrightarrow x=-59\)
\(\dfrac{x+1}{58}+\dfrac{x+2}{59}=\dfrac{x+3}{56}+\dfrac{x+4}{55}\)
\(\Leftrightarrow\dfrac{x+1}{58}+1+\dfrac{x+2}{57}+1=\dfrac{x+3}{56}+1+\dfrac{x+4}{55}+1\)
\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}=\dfrac{x+59}{56}+\dfrac{x+59}{55}\)
\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}-\dfrac{x+59}{56}-\dfrac{x+59}{55}=0\)
\(\Leftrightarrow\left(x+59\right)\left(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\right)=0\)
Mà \(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\ne0\)
\(\Rightarrow x+59=0\)
\(\Leftrightarrow x=-59\)
Vậy: \(S=\left\{-59\right\}\)