giải phương trình sau:
\(\frac{80}{x-2}-\frac{80}{x}=2\)
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NC
0
4 tháng 5 2020
ĐKXĐ: \(x\ne0\)
Ta có: \(\frac{10}{x}+2\left(\frac{1}{80}-\frac{1}{x}\right)=\frac{2}{15}\)
\(\Leftrightarrow\frac{8}{x}+\frac{1}{40}-\frac{2}{15}=0\)
\(\Leftrightarrow\frac{960}{120x}+\frac{3x}{120x}-\frac{16x}{120x}=0\)
\(\Leftrightarrow960+3x-16x=0\)
\(\Leftrightarrow960-13x=0\)
\(\Leftrightarrow13x=960\)
hay \(x=\frac{960}{13}\)(tm)
Vậy: \(x=\frac{960}{13}\)
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2019
ĐKXĐ: \(x\ne\left\{-10;-8;-3;-1\right\}\)
\(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\)
\(\Leftrightarrow\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=52\)
\(\Leftrightarrow x^2+11x-42=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\)
25 tháng 2 2019
\(\frac{x+5}{95}+\frac{x+3}{97}+\frac{x+1}{99}=\frac{x+15}{85}+\frac{x+20}{80}+\frac{x+25}{75}.\)
\(\frac{x+5}{95}+1+\frac{x+3}{97}+1+\frac{x+1}{99}+1-\frac{x+15}{85}-1-\frac{x+20}{80}-1-\frac{x+25}{75}-1=0\)
\(\frac{x+100}{95}+\frac{x+100}{97}+\frac{x+100}{99}-\frac{x+100}{85}-\frac{x+100}{80}-\frac{x+100}{75}=0\)
\(\left(x+100\right).\left(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\right)=0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
\(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\ne0\)
4 tháng 4 2020
\(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\left(ĐkXĐ:x\ne\pm4\right)\\ \Leftrightarrow\frac{3.80\left(x-4\right)+3.80\left(x+4\right)-25.\left(x^2-16\right)}{\left(x^2-16\right).3}=0\\ \Leftrightarrow\frac{240x-960+240x+960-25x^2+400}{3.\left(x^2-16\right)}=0\\ \Leftrightarrow\frac{-25x^2+480x+400}{3.\left(x^2-16\right)}=0\\ \Leftrightarrow\frac{-25.\left(x+\frac{4}{5}\right)\left(x-20\right)}{3.\left(x^2-16\right)}=0\\ \Leftrightarrow\left[{}\begin{matrix}x+\frac{4}{5}=0\\x-20=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\frac{4}{5}\left(Nhận\right)\\x=20\left(Nhận\right)\end{matrix}\right.\\ \Rightarrow S=\left\{-\frac{4}{5};20\right\}\)
HP
16 tháng 4 2016
Cộng thêm 1 vào 2 vế ta đc:
\(\left(\frac{x+2}{89}+1\right)+\left(\frac{x+5}{86}+1\right)>\left(\frac{x+8}{83}+1\right)+\left(\frac{x+11}{80}+1\right)\)
\(\Leftrightarrow\frac{x+91}{89}+\frac{x+91}{86}>\frac{x+91}{83}+\frac{x+91}{80}\)
\(\Leftrightarrow\frac{x+91}{89}+\frac{x+91}{86}-\frac{x+91}{83}-\frac{x+91}{80}>0\)
\(\Leftrightarrow\left(x+91\right).\left(\frac{1}{89}+\frac{1}{86}-\frac{1}{83}-\frac{1}{80}\right)>0\)
Vì \(\frac{1}{89}+\frac{1}{86}-\frac{1}{83}-\frac{1}{80}<0\)
=>x+91<0
=>x<-91
Vậy.......................
TG
5
30 tháng 4 2020
\(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\left(1\right)\)
ĐKXĐ: \(x\ne\pm4\)
\(\left(1\right)\Leftrightarrow\frac{80\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{80\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\frac{25}{3}\)
\(\Leftrightarrow\frac{80x-320+80x+320}{\left(x-4\right)\left(x+4\right)}=\frac{25}{3}\)
\(\Leftrightarrow\frac{160x}{\left(x-4\right)\left(x+4\right)}=\frac{25}{3}\)
\(\Rightarrow480x=25\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow480x=25\left(x^2-16\right)\)
\(\Leftrightarrow480x=25x^2-400\)
\(\Leftrightarrow25x^2-480x-400=0\)
\(\Leftrightarrow25x^2-480x+2304-2704=0\)
\(\Leftrightarrow\left(5x-48\right)^2=2704\)
\(\Leftrightarrow\left|5x-48\right|=52\)
+Trường hợp 1: Nếu \(x\ge\frac{48}{5}\)thì \(5x-48\ge0\Rightarrow\left|5x-48\right|=5x-48\)
Ta có phương trình;
\(5x-48=52\)
\(\Leftrightarrow5x=100\)
\(\Leftrightarrow x=20\)(thỏa măn)
+Trường hợp 2: Nếu\(x< \frac{48}{5}\)thì \(5x-48< 0\Rightarrow\left|5x-48\right|=48-5x\)
Ta có phương trình:
\(48-5x=52\)
\(\Leftrightarrow-5x=4\)
\(\Leftrightarrow x=-\frac{4}{5}\)(thỏa mãn)
\(S=\left\{20;-\frac{4}{5}\right\}\)
30 tháng 4 2020
Ta có: \(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\)
\(\Leftrightarrow80.\left[\frac{x-4+x+4}{\left(x+4\right).\left(x-4\right)}\right]=\frac{25}{3}\)
\(\Leftrightarrow\frac{2x}{x^2-16}=\frac{25}{3}:80\)
\(\Leftrightarrow\frac{2x}{x^2-16}=\frac{5}{48}\)
\(\Rightarrow48.2x=5.\left(x^2-16\right)\)
\(\Leftrightarrow96x=5x^2-80\)
\(\Leftrightarrow5x^2-96x-80=0\)
\(\Leftrightarrow\left(5x^2-100x\right)+\left(4x-80\right)=0\)
\(\Leftrightarrow5x.\left(x-20\right)+4.\left(x-20\right)=0\)
\(\Leftrightarrow\left(5x+4\right).\left(x-20\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x+4=0\\x-20=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}5x=-4\\x=20\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{5}\left(TM\right)\\x=20\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{-\frac{4}{5};20\right\}\)
7 tháng 8 2020
Bài làm:
PT:
đkxđ: \(x\ne0;x\ne2\)
Ta có: \(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x=2+x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(vl\right)\\x+1=0\end{cases}}\Rightarrow x=-1\)
BPT:
Ta có: \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{x+1}{2}-x-\frac{1}{2}\le0\)
\(\Leftrightarrow\frac{x+1-2x-1}{2}\le0\)
\(\Leftrightarrow\frac{-x}{2}\le0\)
\(\Rightarrow-x\le0\)
\(\Rightarrow x\ge0\)
MN
7 tháng 8 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{1}{x}-\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\frac{2+x-2-x^2-2x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow-x^2-x=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}}\)
Vậy \(S=\left\{-1\right\}\)
b) \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow x+1-2x-1\le0\)
\(\Leftrightarrow-x\le0\)
\(\Leftrightarrow x\ge0\)
Vậy \(x\ge0\)
22 tháng 2 2019
Điều kiện: x khác 0
Đặt \(\frac{x^2+1}{x}=t\Rightarrow\frac{x}{x^2+1}=\frac{1}{t}\)
Khi đó: \(\frac{x^2+1}{x}+\frac{x}{x^2+1}=\frac{5}{2}\)
\(\Leftrightarrow t+\frac{1}{t}=\frac{5}{2}\)
\(\Leftrightarrow\frac{t^2+1}{t}=\frac{5}{2}\Rightarrow2t^2+2=5t\)
\(\Leftrightarrow2t^2-5t+2=0\Leftrightarrow\left(2t-1\right)\left(t-2\right)=0\Leftrightarrow\orbr{\begin{cases}t=\frac{1}{2}\\t=2\end{cases}}\)
Nếu \(t=\frac{1}{2}\Rightarrow\frac{x^2+1}{x}=\frac{1}{2}\Rightarrow2x^2+2=x\)
\(\Leftrightarrow2x^2-x+2=0\)
Mà \(2x^2-x+2=2\left(x-\frac{1}{4}\right)^2+\frac{15}{8}>0\forall x\)
Nên \(x\in\varnothing\)
Nếu \(t=2\Rightarrow\frac{x^2+1}{x}=2\Rightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)(thỏa mãn ĐKXĐ)
Tập nghiệm của pt: \(S=\left\{1\right\}\)
\(\)
T
23 tháng 2 2019
Theo BĐT AM-GM,ta có: \(x^2+1\ge2\left|x\right|\ge2x\Rightarrow\frac{x^2+1}{x}\ge2\)
Đặt \(\frac{x^2+t}{x}=t\left(t\ge2\right)\).Bài toán trở thành:
\(t+\frac{1}{t}=\frac{5}{2}\Leftrightarrow\left(\frac{1}{t}+\frac{t}{4}\right)+\frac{3t}{4}=\frac{5}{2}\)
Áp dụng BĐT AM-GM: \(VT\ge1+\frac{3t}{4}\ge1+\frac{6}{4}=\frac{5}{2}\)
Mà \(VT=\frac{5}{2}\) .Dấu "=" xảy ra khi \(\frac{1}{t}=\frac{t}{4}\Leftrightarrow t=2\Leftrightarrow\frac{x^2+1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x=1\)
Vậy tập hợp nghiệm của phương trình: \(S=\left\{1\right\}\)
P
22 tháng 4 2017
\(a,\Leftrightarrow5\left(x-2\right)-15x\le9+10\left(x+1\right)\)
\(\Leftrightarrow5x-10-15x\le9+10x+10\)
\(\Leftrightarrow-20x\le29\)
\(\Leftrightarrow x\ge-1,45\)
Vậy ...........
\(b,\Rightarrow\left(x+2\right)-3\left(x-3\right)=5\left(x-2\right)\)
\(\Leftrightarrow x+2-3x+9-5x+10=0\)
\(\Leftrightarrow-7x+21=0\)
\(\Leftrightarrow x=3\)
Vậy ..............
23 tháng 4 2017
\(\frac{x-2}{6}-\frac{x}{2}\le\frac{3}{10}+\frac{x+1}{3}\Leftrightarrow\frac{5\left(x-2\right)}{30}-\frac{15x}{30}\le\frac{9}{30}+\frac{10\left(x+1\right)}{30}\)
\(\Leftrightarrow5x-10-15x-9-10x-10\le0\)
\(\Leftrightarrow-20x-29\le0\Leftrightarrow\left(-20x\right)\cdot\frac{-1}{20}\ge29\cdot-\frac{1}{20}\)
\(\Leftrightarrow x\ge-\frac{29}{20}\)
ĐKXĐ : \(x\ne0;x\ne2\)
Khi đó \(\frac{80}{x-2}-\frac{80}{x}=2\)
=> \(\frac{80x-80\left(x-2\right)}{x\left(x-2\right)}=\frac{2x\left(x-2\right)}{x\left(x-2\right)}\)
<=> \(\frac{160}{x\left(x-2\right)}=\frac{2x^2-4x}{x\left(x-2\right)}\)
=> 2x2 - 4x = 160
<=> x2 - 2x = 80
<=> x2 - 2x - 80 = 0
<=> x2 - 2x + 1 - 81 = 0
<=> (x - 1)2 - 92 = 0
<=> (x - 1 + 9)(x - 1 - 9) = 0
<=> (x + 8)(x - 10) = 0
<=> \(\orbr{\begin{cases}x+8=0\\x-10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-8\\x=10\end{cases}}\)
Vậy \(x\in\left\{-8;10\right\}\)là nghiệm phương trình
\(\frac{80}{x-2}-\frac{80}{x}=2\)ĐK : \(x\ne0;2\)
\(\Leftrightarrow\frac{80x-80\left(x-2\right)}{x\left(x-2\right)}=\frac{2x\left(x-2\right)}{x\left(x-2\right)}\)
\(\Rightarrow160=2x^2-4x\Leftrightarrow x^2-2x-80=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+8\right)=0\Leftrightarrow x=10;x=-8\)( tm )
Vậy tập nghiệm của phương trình là S = { -8 ; 10 }