GIẢI PHƯƠNG TRÌNH:
\(x^2-x-\dfrac{1}{x}+\dfrac{1}{x^2}-10=0\)
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Những câu hỏi liên quan
12 tháng 4 2022
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b.\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12+\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x^2-4\right)\)
\(\Leftrightarrow x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\left(ktm\right)\)
Vậy pt vô nghiệm
DT
12 tháng 4 2022
a)
<=> x (x-2 ) = 0
<=> x =0
x = 2
b)
đkxđ : x khác 2 , x khác -2
<=> \(\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{12}{x^2-4}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
<=> \(\dfrac{x^2+3x+2}{....}-\dfrac{5x-10}{....}-\dfrac{12}{...}+\dfrac{x^2-4}{....}=0\)
<=> \(x^2+3x+2-5x+10-12+x^2-4=0\)
<=> \(2x^2-2x-4=0\)
<=> x =2 (ktm)
Vậy..
5 tháng 4 2021
|x-9|=2x+5
Xét 3 TH
TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)
TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)
TH3: x=9 =>0=23(L)
Vậy x= 4/3
5 tháng 4 2021
Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)
\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)
\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)
KK
30 tháng 3 2022
a)\(\dfrac{7x-1}{2}+2x=\dfrac{16-x}{3}\)
\(\dfrac{\left(7x-1\right).3}{2.3}+\dfrac{2x.6}{6}=\dfrac{\left(16-x\right)2}{3.2}\)
khử mẫu
=> (7x-1).3+12x=(16-x).2
=>21x-3+12x=-2x+32
=>21x-3+12x+2x-32=0
=>35x-35=0
KK
30 tháng 3 2022
b)\(\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
ĐKXĐ: x khác +-2
\(\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)
khử mẫu
(x+1).(x+2)+(x-1)(x-2)=2x2+4
=>x2+x+2+x+2+x2-2x-x+2=2x2+4
=>x2+x+2+x+2+x2-2x-x+2-2x2-4=0
=>(x2+x2-2x2)+(x+x-2x-x)+(2+2+2-4)=0
=>-x+2=0
=>-x=-2
=>x=2(loại)
vậy pt vô nghiệm
HT
2
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2019
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+...+\dfrac{x-2016}{1}=2016\)
\(\Leftrightarrow\dfrac{x-1}{2016}-1+\dfrac{x-2}{2015}-1+\dfrac{x-3}{2014}-1+...+\dfrac{x-2016}{1}-1=0\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}+...+\dfrac{x-2017}{1}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+...+1\right)=0\)
\(\Leftrightarrow x-2017=0\) (do \(\dfrac{1}{2016}+\dfrac{1}{2015}+...+1\ne0\))
\(\Rightarrow x=2017\)
KV
18 tháng 4 2022
a) \(5x-3=7\)
\(\Leftrightarrow5x=7+3\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=\dfrac{10}{5}\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow x+3=0\) hoặc \(x-4=0\)
*) \(x+3=0\)
\(x=0-3\)
\(x=-3\)
*) \(x-4=0\)
\(x=0+4\)
\(x=4\)
Vậy \(S=\left\{-3;4\right\}\)
c) \(\left|x^2+2014\right|=1\)
\(\Leftrightarrow x^2+2014=1\) hoặc \(x^2+2014=-1\)
*) \(x^2+2014=1\)
\(\Leftrightarrow x^2=1-2014\)
\(\Leftrightarrow x^2=-2013\) (vô lý)
*) \(x^2+2014=-1\)
\(\Leftrightarrow x^2=-1-2014\)
\(\Leftrightarrow x^2=-2015\) (vô lý)
Vậy \(S=\varnothing\)
d) \(\dfrac{2}{x+1}-\dfrac{1}{x-3}=\dfrac{3x-11}{x^2-2x-3}\) (1)
ĐKXĐ: \(x\ne-1;x\ne3\)
\(\left(1\right)\Leftrightarrow2\left(x-3\right)-\left(x+1\right)=3x-11\)
\(\Leftrightarrow2x-6-x-1=3x-11\)
\(\Leftrightarrow-2x=-11+7\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\) (nhận)
Vậy \(S=\left\{2\right\}\)
22 tháng 5 2021
\(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}\)
⇔ \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=\dfrac{4x+2}{7}\)
⇔ \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=\dfrac{4x+2}{7}\)
⇔ \(\dfrac{140x-84}{168}-\dfrac{294x-42}{168}=\dfrac{96x+48}{168}\)
⇔ 140x-84-294x+42=96x+48
⇔ -154x-42=96x+48
⇔ -250x=90
⇔ x=\(\dfrac{-9}{26}\)
Vậy phương trình đã cho có tập nghiệm S={\(\dfrac{-9}{26}\)}
MN
1
13 tháng 2 2022
\(\left(dk:x\ne-\dfrac{2}{3};x\ne-1\right)pt\Leftrightarrow\dfrac{2x}{3x^2-x+2}-\dfrac{7x-3x^2-5x-2}{3x^2+5x+2}=0\Leftrightarrow\dfrac{2x}{3x^2-x+2}-\dfrac{3x^2+12x+2}{3x^2+5x+2}=0\left(1\right)\)
\(x=0\) \(không\) \(là\) \(nghiệm\left(1\right)\)
\(x\ne0\Rightarrow\left(1\right)\Leftrightarrow\dfrac{2}{3x-1+\dfrac{2}{x}}-\dfrac{3x+12+\dfrac{2}{x}}{3x+5+\dfrac{2}{x}}=0\)
\(đặt:3x+\dfrac{2}{x}=t\) \(do:x\ne-\dfrac{2}{3};x\ne-1;\Rightarrow t\ne-5\)
\(x>0\Rightarrow t\ge2\sqrt{3.2}=2\sqrt{6}\)
\(x< 0\Rightarrow-t\ge2\sqrt{6}\Rightarrow t\le-2\sqrt{6}\Rightarrow\left[{}\begin{matrix}t\ne-5;t\le-2\sqrt{6}\\t\ge2\sqrt{6}\end{matrix}\right.\)
\(\Rightarrow\dfrac{2}{t-1}-\dfrac{t+12}{t+5}=0\Rightarrow2\left(t+5\right)-\left(t+12\right)\left(t-1\right)=0\Leftrightarrow\left[{}\begin{matrix}t=-11\left(tm\right)\\t=2\left(ktm\right)\end{matrix}\right.\)
\(t=-11=3x+\dfrac{2}{x}\Leftrightarrow3x^2+2=-11x\Leftrightarrow3x^2+11x+2=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{97}}{6}\left(tm\right)\\x=\dfrac{-11-\sqrt{97}}{6}\left(tm\right)\end{matrix}\right.\)
DT
13 tháng 2 2022
bài nó dàiiiiiiii , khôg hiểu chỗ nèo hỏi lại mình hen
\(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)
\(\Leftrightarrow\left(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{\left(3x+2\right)\left(x+1\right)}\right)=1\)
\(\Leftrightarrow\dfrac{2x\left(3x+2\right)\left(x+1\right)-\left(7x.\left(3x^2-x+2\right)\right)}{\left(3x^2-x+2\right).\left(3x+2\right)\left(x+1\right)}=\dfrac{-15x^3+17x^2-10x}{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{-15x^3+17^2-10x }{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}-1=0\)
rồi quy đồng tùm lum từa lưa nữa được như này:
\(\Leftrightarrow\dfrac{-9x^4-27x^3+10x^2-18x-4}{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}=0\)
\(\Leftrightarrow-9x^4-27x^3+10x^2-18x-4=0\)
\(\Leftrightarrow x^2+\dfrac{5}{3}.x+\dfrac{25}{26}=0\)
\(\Leftrightarrow x+\left(\dfrac{5}{6}\right)^2=\dfrac{1}{36}\)
Sử dụng công thức bậc 2 hen:
\(\Leftrightarrow x=\dfrac{-5\pm\sqrt{1}}{6}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{1}}{6}\\x_2=\dfrac{-5-\sqrt{1}}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-\dfrac{2}{3}\\x_2=-1\end{matrix}\right.\)
\(x^2-x-\dfrac{1}{x}+\dfrac{1}{x^2}-10=0\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}-\left(x+\dfrac{1}{x}\right)-10=0\)
Đặt \(t=x+\dfrac{1}{x}\)
\(\Leftrightarrow t^2-2=x^2+\dfrac{1}{x^2}\)
Thế vào ta dược : \(t^2-t-12=0\)
Tới đây dễ r .
\(t^2-t-12=0\)
\(\Rightarrow t^2-t-\dfrac{1}{4}-\dfrac{47}{4}=0\)
\(\Rightarrow\left(t-\dfrac{1}{2}\right)^2=\dfrac{47}{4}\)
\(\Rightarrow\left(t-\dfrac{1}{2}-\sqrt{\dfrac{47}{4}}\right)\left(t-\dfrac{1}{2}+\sqrt{\dfrac{47}{4}}\right)=0\)