\(\frac{1}{3}x^4-\frac{1}{2}x^2+\frac{1}{6}=0\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
\(1.\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}
\)
2.\(\frac{2x^4}{\left(x+1\right)^2}-\frac{5x^2}{x+1}+2=0\)
3.\(\left(x+\frac{1}{x}\right)^2-6\left(x+\frac{1}{x}\right)+8=0\)
4.\(\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+6=0\)
5.\(\frac{2x}{3x^2-x+2}-\frac{7x}{3x^2+5x+2}=1\)
\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)
\(\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}\)
\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\) \(Đkxđ:.......\)
Đặt: \(t=x^2-3x+2\left(t\ne0\right)\)
\(\Rightarrow2t=2x^2-6x+4\)
\(\Rightarrow2x^2-6x+1=2t-3\)
\(Pt:\Leftrightarrow\frac{4}{7}-\frac{3}{2t-3}+1=0\)
\(\Leftrightarrow4\left(2t-3\right)-3t+t\left(2t-3\right)=0\)
\(\Leftrightarrow8t-12-3t+2t^2-3t=0\)
\(\Leftrightarrow2t^2+2t-12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\left(tm:\left[{}\begin{matrix}t\ne0\\t\ne\frac{3}{2}\end{matrix}\right.\right)\)
+ Với \(t=2\) thì: \(x^2-3x+2=2\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\left(tmđk\right)\)
+ Với \(t=-3\) thì \(x^2-3x+2=-3\)
\(\Leftrightarrow x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{11}{4}=0\left(vô-lí\right)\)
Vậy pt có nghiệm: \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Bài 2:
ĐKXĐ: $x\neq 1;2;3;6$
PT $\Leftrightarrow \frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}-\frac{1}{x-1}$
$\Leftrightarrow \frac{5x-12}{x^2-5x+6}=\frac{5x}{x^2-7x+6}$
Đặt $x^2+6=t$ thì $\frac{5x-12}{t-5x}=\frac{5x}{t-7x}$
$\Rightarrow (5x-12)(t-7x)=5x(t-5x)$
$\Leftrightarrow 10x^2+12t+84x=0$
$\Leftrightarrow 10x^2+12(x^2+6)+84x=0$
$\Leftrightarrow 22x^2+84x+72=0$
$\Leftrightarrow 11x^2+42x+36=0$
$\Rightarrow x=\frac{-21\pm 3\sqrt{5}}{11}$
k, x3 - x2 - 17x - 15 = 0
l, x3 +4x2+x- 6=0
m, x4+2x3-13x2 -14x+ 24 =0
n, \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
i, (x-4) (x-5) (x-6) (x-7) = 1680
p, \(\frac{1}{x^2-5x-6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
trong quá trình bạn xem bài mk thấy chỗ nào sai dấu thì sửa giùm mk nha trong quá trình làm mk cx có thể sai sót nhầm lẫn nha
\( m){x^4} + 2{x^3} - 13{x^2} - 14x + 24 = 0\\ \Leftrightarrow {x^4} - {x^3} + 3{x^3} - 3{x^2} - 10{x^2} + 10x - 24x + 24 = 0\\ \Leftrightarrow {x^3}\left( {x - 1} \right) + 3{x^2}\left( {x - 1} \right) - 10x\left( {x - 1} \right) - 24\left( {x - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^3} + 3{x^2} - 10x - 24} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^3} + 2{x^2} + {x^2} + 2x - 12x - 24} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left[ {{x^2}\left( {x + 2} \right) + x\left( {x + 2} \right) - 12\left( {x + 2} \right)} \right] = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {{x^2} + x - 12} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {{x^2} + 4x - 3x - 12} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left[ {x\left( {x + 4} \right) - 3\left( {x + 4} \right)} \right] = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)\left( {x + 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 1 = 0\\ x + 2 = 0\\ x - 3 = 0\\ x + 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - 2\\ x = 3\\ x = - 4 \end{array} \right. \)
\(n)\dfrac{x+1}{99}+\dfrac{x+2}{98}=\dfrac{x+3}{97}+\dfrac{x+4}{96}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right) = \left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}=\dfrac{x+100}{97}+\dfrac{x+100}{96}\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}-\dfrac{x+100}{97}-\dfrac{x+100}{96}=0\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{97}-\dfrac{1}{96}\right)=0 \)
Mà \(\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{97}-\dfrac{1}{96}\ne0 \\ \)
\(\Rightarrow x+100=0\\ \Rightarrow x=-100\\ \)
Vậy \(x=-100\)
\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)
\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)
\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(-\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)0
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{5}x+1=0\)
a) \(\left(\frac{1}{7}x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\)TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\) TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\) TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\)
\(\frac{1}{7}x=\frac{2}{7}\) \(-\frac{1}{5}x=\frac{3}{5}\) \(\frac{1}{3}x=\frac{4}{3}\)
\(x=\frac{2}{7}\cdot7\) \(x=\frac{3}{5}\cdot-5\) \(x=\frac{4}{3}\cdot3\)
\(x=2\) \(x=-3\) \(x=4\)
Vậy x = 2 hoặc x = -3 hoặc x = 4
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(x\cdot\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{5}\right)=1\)
\(x\cdot\frac{5+3-24}{30}=1\)
\(x\cdot\frac{-8}{15}=1\)
\(x=1\cdot\frac{-15}{8}=\frac{-15}{8}\)
Vậy x = \(\frac{-15}{8}\)
\(\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\)
\(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{\left(x-4\right)}{x\left(x+2\right)}=0\)
\(\frac{1}{x-2}\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)
Tìm x biết
a) (8-5x).(x+2)+4.(x-2).(x+1)+2.(x-2).(x+2)=0
b)\(\left(-\frac{2}{5}+x\right):\frac{7}{9}+\left(-\frac{3}{5}+\frac{5}{6}\right):\frac{7}{9}=0\)
c)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2004}\)