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T
1
MA
3 tháng 2 2018
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)-2=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)-2\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\times\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\right)\)
\(\Leftrightarrow x=-100\)
Vậy......
7 tháng 8 2019
\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Leftrightarrow6\left(x^2+8\right)\left(x^3+3\right)+12\left(x^2+2\right)\left(x^2+3\right)=3\left(x^2+2\right)\left(x^2+8\right)\left(x^2+3\right)-7\left(x^2+2\right)\left(x^2+8\right)\)
\(\Leftrightarrow18x^4+126x^2+216=3x^6+32x^4+68x^2+32\)
\(\Leftrightarrow18x^4+126x^2+216-3x^6-32x^4-68x^2-32=0\)
\(\Leftrightarrow-14x^4+58x^2+184-3x^6=0\)
\(\Leftrightarrow x=\pm2\)
Vậy: nghiệm phương trình là: \(\left\{\pm2\right\}\)
7 tháng 8 2019
Giải như bạn trên cũng được, nhưng mình nghĩ làm cách này đỡ tốn sức hơn :
\(2,\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Rightarrow\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1+\frac{7}{x^2+3}-1=0\)
\(\Rightarrow\frac{6-x^2-2}{x^2+2}+\frac{12-x^2-8}{x^2+8}+\frac{7-x^2-3}{x^2+3}=0\)
\(\Rightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}+\frac{-x^2+4}{x^2+3}=0\)
\(\Rightarrow-\left(x^2-4\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)
Vì \(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\left(>0\forall x\right)\)
\(\Rightarrow x^2-4=0\Rightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x=\pm2\)
TH
23 tháng 3 2019
a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
TN
24 tháng 4 2017
A . 3x + 2(x + 1) = 6x - 7
<=> 3x + 2x + 2 = 6x -7
<=> 5x - 6x = -7 - 2
<=> -x = -9
<=> x =9
B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)
=> 3(x +3) < 5(5 -x)
<=> 3x+9 < 25 - 5x
<=> 3x + 5x < 25 - 9
<=> 8x < 16
<=> x < 2
C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)
<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)
<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{\left(x+1\right)\left(x-4\right)}\)= \(\frac{2}{x-4}\)
<=> 5(x - 4) + 2x = 2(x +1)
<=> 5x - 20 + 2x = 2x + 2
<=>7x - 2x = 2 + 20
<=> 5x = 22
<=> x =\(\frac{22}{5}\)
NN
1
WR
11 tháng 6 2017
1)
a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)
(đk:x khác \(\frac{1}{2}\))
\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)
Vậy x=\(\frac{25}{7}\)
b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)
(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))
\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)
Vậy x=4
2)
\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)
\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)
\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)
Ta có: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
<=> \(\frac{6\left(x+4\right)-30x+120}{30}=\frac{10x-15x+30}{30}\)
<=> 6x + 24 - 30x + 120 = -5x + 30
<=> -24x + 5x = 30 - 144
<=> -19x = -114
<=> x = 6
Vậy S = {6}