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⇔ \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
⇔ \(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
⇔ \(\dfrac{4}{x^2+8x+12}=\dfrac{1}{8}\)
⇔ \(x^2+8x+12=32\)
⇔ \(x^2+8x-20=0\)
⇔ \(\left(x-2\right)\left(x+10\right)=0\)
⇔ \(\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
ĐKXĐ : \(x\ne\left\{2;3;4;5;6\right\}\)
\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right).\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right).\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x-20=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(\text{loại}\right)\\x=-10\end{matrix}\right.\Leftrightarrow x=-10\)
Vậy tập nghiệm phương trình S = {10}
a) x - 5 = 7 - x
<=> 2x = 12
<=> x = 6
Vậy tập nghiệm phương trình S = {6}
b) 3x - 15 = 2x(x - 5)
<=> 3(x - 5) = 2x(x - 5)
<=> (2x - 3)(x - 5) = 0
<=> \(\left[{}\begin{matrix}2x-3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=5\end{matrix}\right.\)
Tập nghiệm phương trình \(S=\left\{\dfrac{3}{2};5\right\}\)
a: \(x-3\left(2x-6\right)=21-\left(5x+3\right)\)
=>\(x-6x+18=21-5x-3\)
=>18=18(luôn đúng)
=>\(x\in R\)
b: \(\left(x-2\right)\left(x+2\right)-\left(x-1\right)^2=2\left(x+1\right)\)
=>\(x^2-4-x^2+2x-1=2x+2\)
=>2x-5=2x+2
=>-7=0(vô lý)
=>\(x\in\varnothing\)
c: \(\dfrac{9x+4}{6}=1-\dfrac{3x-5}{9}\)
=>\(\dfrac{3\left(9x+4\right)}{18}=\dfrac{18}{18}-\dfrac{2\left(3x-5\right)}{18}\)
=>3(9x+4)=18-2(3x-5)
=>27x+12=18-6x+10
=>27x+12=-6x+28
=>33x=16
=>\(x=\dfrac{16}{33}\left(nhận\right)\)
d: ĐKXĐ: \(x\notin\left\{2;5\right\}\)
\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(\dfrac{6x+1}{\left(x-2\right)\left(x-5\right)}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(6x+1+5\left(x-5\right)=3\left(x-2\right)\)
=>6x+1+5x-25=3x-6
=>11x-24=3x-6
=>8x=18
=>\(x=\dfrac{9}{4}\left(nhận\right)\)
a: x−3(2x−6)=21−(5x+3)
=>x−6x+18=21−5x−3
=>18=18(luôn đúng)
=>x∈R
b: (x−2)(x+2)−(x−1)2=2(x+1)
=>x2−4−x2+2x−1=2x+2
=>2x-5=2x+2
=>-7=0(vô lý)
=>x∈∅
c: 9x+46=1−3x−59
=>3(9x+4)18=1818−2(3x−5)18
=>3(9x+4)=18-2(3x-5)
=>27x+12=18-6x+10
=>27x+12=-6x+28
=>33x=16
=>x=1633(nhận)
d: ĐKXĐ: x∉{2;5}
6x+1x2−7x+10+5x−2=3x−5
=>6x+1(x−2)(x−5)+5x−2=3x−5
=>6x+1+5(x−5)=3(x−2)6
=>6x+1+5x-25=3x-6
=>11x-24=3x-6
=>8x=18
=>x=94(nhận)
a/ \(\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}\)
\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+4}\)
Vậy..
b/ \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+5}\)
Vậy..
c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
PT \(\Leftrightarrow\dfrac{m\left(1-mx\right)+1+mx}{\left(1+mx\right)\left(1-mx\right)}=\dfrac{1}{\left(1-mx\right)\left(1+mx\right)}\)
\(\Rightarrow m-m^2x+1+mx=1\)
\(\Leftrightarrow x\left(m-m^2\right)+m=0\)
Để phương trình vô nghiệm \(\Leftrightarrow\left\{{}\begin{matrix}m-m^2=0\\m\ne0\end{matrix}\right.\)
\(\Leftrightarrow m=1\)
Vậy \(m=1\)
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
\(đkxđ:x\ne1;2;3;4;5\\ \Leftrightarrow\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}=\dfrac{1}{15}\\ \Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}=\dfrac{1}{15}\\ \Leftrightarrow\dfrac{1}{x-5}-\dfrac{1}{x-1}=\dfrac{1}{15}\\ \Leftrightarrow60=x^2-6x+5\\ \)
\(\Leftrightarrow60=x^2-6x+5\\ \Leftrightarrow\left[{}\begin{matrix}x-11=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\\ \Rightarrow D\)