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\(\dfrac{1.2}{1.1}.\dfrac{2.3}{2.2}.\dfrac{3.4}{3.3}.\dfrac{4.5}{4.4}...\dfrac{10.11}{10.10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow\dfrac{2.3.4...11}{1.2.3...10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11x-22=-20x+40\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
\(=>\dfrac{2\cdot1}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot......\cdot\dfrac{10\cdot11}{10\cdot10}\cdot\left(x-2\right)=-20\left(x+1\right)+60\)=>11*(x-2)=-20*(x+1)+60
=>11x-22=-20x-20+60
=>31x=62
=>x=2
\(\dfrac{1.2}{1^2}.\dfrac{2.3}{2^2}.\dfrac{3.4}{3^2}...\dfrac{9.10}{9^2}.\dfrac{10.11}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)
\(\Leftrightarrow\dfrac{1.2^2.3^2.4^2...10^2.11}{1^2.2^2.3^2....10^2}\left(x-2\right)+20\left(x+1\right)=60\)
\(\Leftrightarrow11\left(x-2\right)+20\left(x+1\right)=60\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
a) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x=5\)
b) \(\dfrac{x+4}{2000}+\dfrac{x+8}{1996}=\dfrac{x+12}{1992}+\dfrac{x+16}{1988}\)
\(\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+8}{1996}+1=\dfrac{x+12}{1992}+1+\dfrac{x+16}{1988}+1\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{1996}-\dfrac{x+2004}{1992}-\dfrac{x+2004}{1988}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\right)=0\)
\(\Leftrightarrow x+2004=0\)(vì \(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\ne0\))
\(\Leftrightarrow x=-2004\)
b: \(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3\left(x^2+x-6\right)\)
\(\Leftrightarrow3x^2-10x+3=3x^2+3x-18\)
=>-13x=-21
hay x=21/13
c: \(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\)
=>x-100=0
hay x=100
a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)
=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)
=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)
=>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\)
=>x=0
b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\)
=>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\)
=>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\)
Đến đây thì bạn giải giống câu a
\(\dfrac{x+1}{60}+\dfrac{x+2}{59}=\dfrac{x+3}{58}+\dfrac{x+4}{57}\)
\(\Leftrightarrow\dfrac{x+1}{60}+1+\dfrac{x+2}{59}+1=\dfrac{x+3}{58}+1+\dfrac{x+4}{57}+1\)
\(\Leftrightarrow\dfrac{x+1+60}{60}+\dfrac{x+2+59}{59}=\dfrac{x+3+58}{58}+\dfrac{x+4+57}{57}\)
\(\Leftrightarrow\dfrac{x+61}{60}+\dfrac{x+61}{59}-\dfrac{x+61}{58}-\dfrac{x+61}{57}=0\)
\(\Leftrightarrow\left(x+61\right)\left(\dfrac{1}{60}+\dfrac{1}{59}-\dfrac{1}{58}-\dfrac{1}{57}\right)=0\)
\(\Leftrightarrow x+61=0\)
\(\Leftrightarrow x=-61\)
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
a) 0,25x+1,5=0
=> x = (0 - 1,5) : 0,25 = -1,5 : 0,25 = -6
Vậy x = -6.
b) 6,36−5,3x=0
=> x = (0 + 6,36) : 5,3 = 6,36 : 5,3 =\(\dfrac{6}{5}=1,2\)
Vậy x = 1,2.
c) 43x−56=12
=> x = \(\left(\dfrac{1}{2}+\dfrac{5}{6}\right)\): \(\dfrac{4}{3}\) = \(\dfrac{4}{3}:\dfrac{4}{3}=1\)
Vậy x = 1.
d) −59x+1=23x−10
=> \(\dfrac{-5}{9}x-\dfrac{2}{3}x=\dfrac{-11}{9}x=-10-1=-11\)
=> \(x=-11:\dfrac{-11}{9}=9\)
Vậy x = 9.
Đáp số x=-40 em nhé, lúc nãy anh nhầm dạng =)
Nhg đây là giải pt ạ. Nếu đã là giải phương trình ko đc nhân chéo đâu ạ