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\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}+\dfrac{1}{x-20}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
Đến đây cạn rồi?! ==''
\(\dfrac{x+2}{x-1}=\dfrac{x-3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=\left(x-1\right)\left(x-3\right)\)
\(\Rightarrow x^2+x+2x+2=x^2-3x-x+3\)
\(\Rightarrow x^2+3x-x^2+4x=3-2\)
\(\Rightarrow7x=1\Rightarrow x=\dfrac{1}{7}\)
Chúc bạn học tốt!!!
a) \(\dfrac{-2}{3}:x+\dfrac{5}{8}=\dfrac{-7}{12}\) b)\(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
=> \(\dfrac{-2}{3}:x=\dfrac{-7}{12}-\dfrac{5}{8}=\dfrac{-29}{24}\) => \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\dfrac{3}{2}\right)^2\)
=> \(x=\dfrac{-2}{3}:\dfrac{-29}{24}\) => \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
=> \(x=\dfrac{-2}{3}.\dfrac{-24}{29}=\dfrac{16}{29}\) => \(\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
=> \(\dfrac{3}{2}x=\dfrac{-13}{10}\)
=> \(x=\dfrac{-13}{10}:\dfrac{3}{2}\)
=> \(x=\dfrac{-13}{10}.\dfrac{2}{3}=\dfrac{-13}{15}\)
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x+2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=19\)
Chúc bạn học tốt!!!
a, \(\dfrac{x+1}{5}+\dfrac{x+3}{4}=\dfrac{x+5}{3}+\dfrac{x+7}{2}\)
\(\Rightarrow\dfrac{x+1}{5}+2+\dfrac{x+3}{4}+2=\dfrac{x+5}{3}+2+\dfrac{x+7}{2}+2\)
\(\Rightarrow\dfrac{x+11}{5}+\dfrac{x+11}{4}-\dfrac{x+11}{3}-\dfrac{x+11}{2}=0\)
\(\Rightarrow\left(x+11\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Vậy x = -11
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
\(B=\dfrac{1}{x^2+2}\le\dfrac{1}{2}\)
\("="\Leftrightarrow x=0\)
\(C=\dfrac{x^2+15}{x^2+3}=\dfrac{x^2+3+12}{x^2+3}=1+\dfrac{12}{x^2+3}\le1+\dfrac{12}{3}=5\)
\("="\Leftrightarrow x=0\)
\(D=\dfrac{x^2+y^2+5}{x^2+y^2+3}=\dfrac{x^2+y^2+3+2}{x^2+y^2+3}=1+\dfrac{2}{x^2+y^2+3}\le1+\dfrac{2}{3}=\dfrac{5}{3}\)
\("="\Leftrightarrow x=y=0\)
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
a: =>\(-\dfrac{6+x}{2}-\dfrac{3}{2}=2\)
=>-x-6-3=4
=>-x-9=4
=>-x=5
hay x=-5
b: =>(x+1)2=16
=>x+1=4 hoặc x+1=-4
=>x=3 hoặc x=-5
c: \(\Leftrightarrow\left(\dfrac{x-2}{27}-1\right)+\left(\dfrac{x-3}{26}-1\right)+\left(\dfrac{x-4}{25}-1\right)+\left(\dfrac{x-5}{24}-1\right)+\left(\dfrac{x-44}{5}+3\right)=0\)
=>x-29=0
hay x=29
Áp dụng t/c dtsbn:
\(\dfrac{x-1}{8}=\dfrac{x+1}{12}=\dfrac{x+1-x+1}{12-8}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(\Rightarrow x-1=\dfrac{1}{2}.8=4\Rightarrow x=4+1=5\)
12(x-1)=8(x+1)
12x - 12 =8x + 8
12x - 8x = 8 +12
4x. = 20
x. = 20 :4
x. = 5