4. Giải phương trình
a) \(\frac{10}{x-2}\)= \(\frac{x^2-16}{\left(x-2\right)\left(x-1\right)}\)- \(\frac{5}{x+1}\)
b) x3 - 6x2 + 11x - 6 = 0
Giải các phương trình,bất phương trình:
c,\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
d,\(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\)
e,\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}=0\)
g,\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
h,\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
i,\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
k,\(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)
l,\(\left(x^2-2x+1\right)-4=0\)
m,\(4x^2+4x++1=x^2\)
Xin đáy ai giúp mình đi
Giải phương trình:
1.\(\frac{x-5}{x-5}+\frac{x-6}{x-5}+\frac{x-7}{x-5}+...+\frac{1}{x-5}=4\left(x\in N\right)\)
2.\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
3.\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{31}{16}\left(x\in N\right)\)
4.\(8\left(x^2+\frac{1}{x^2}\right)-34\left(x+\frac{1}{x}\right)+51=0\)
5.\(6x^4-5x^3-38x^2-5x+6=0\)
Bài 2 : Giải các bất phương trình sau :
11 , \(\left(2x-7\right)\left(4-5x\right)\ge0\)
12 , \(x^2-x-20>2\left(x-11\right)\)
13 , \(3x\left(2x+7\right)\left(9-3x\right)\ge0\)
14 , \(x^3+8x^2+17x+10< 0\)
15 , \(x^3+6x^2+11x+6>0\)
16 , \(\frac{\left(2x-5\right)\left(x+2\right)}{-4x+3}>0\)
17 , \(\frac{x-3}{x+1}>\frac{x+5}{x-2}\)
18 , \(\frac{x-3}{x+5}< \frac{1-2x}{x-3}\)
19 , \(\frac{3x-4}{x-2}>1\)
20 , \(\frac{2x-5}{2-x}\ge-1\)
Giaỉ các phương trình sau:
a) \(\left(x^2+11x+12\right)\left(x^2+9x+20\right)\left(x^2+13x+42\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)
b) \(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)^2+48\cdot\frac{x^2-4}{x^2-1}=0\)
Giải các phương trình:
\(a.\left(x^2+1\right)\left(x^2-4x+4\right)=0\)
\(b.\left(3x-2\right)\left(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right)=0\)
\(c.\left(3,3-11x\right)\left(\frac{7x+2}{5}+\frac{2\left(1-3x\right)}{3}\right)=0\)
a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)
b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)
c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)
giải phương trình
a)\(\frac{7x+10}{x+1}\left(x^2-x-2\right)=\frac{7x+10}{x+1}\left(2x^2-3x-5\right)\)
b)\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
c)\(x^2+\frac{1}{x^2}+\frac{9x}{2}-\frac{9}{2x}+7=0\)
1.Giải phương trình: \(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
2.Giải phương trình: \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{8}\right|+......+\left|x+\frac{1}{110}\right|=11x\)
GIẢI PHƯƠNG TRÌNH
Ta có vế trái của pt luôn \(\ge0\)
Do đó : \(11x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=x+\frac{1}{2}\\...\\\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\end{cases}}\)
Khi đó pt trở thành :
\(x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}=11x\)
\(\Leftrightarrow10x+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=11x\)
\(\Leftrightarrow x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Leftrightarrow x=1-\frac{1}{11}=\frac{10}{11}\) ( thỏa mãn )
Vậy : pt đã cho có nghiệm \(S=\left\{\frac{10}{11}\right\}\)
Dễ thấy \(VT>0\forall x\)
\(\Rightarrow11x>0\Rightarrow x>0\)
Phương trình trở thành \(10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)
\(\Rightarrow x=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow x=1-\frac{1}{11}=\frac{10}{11}\)
Vậy \(x=\frac{10}{11}\)
Giải phương trình
a) \(\sin \left( {2x + \frac{\pi }{4}} \right) = \sin x\)
b) \(\sin 2x = \cos 3x\)
c) \({\cos ^2}2x = {\cos ^2}\left( {x + \frac{\pi }{6}} \right)\)
a)
\(\sin \left( {2x + \frac{\pi }{4}} \right) = \sin x \Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{4} = x + k2\pi \\2x + \frac{\pi }{4} = \pi - x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\3x = \pi - \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{4} + \frac{{k2\pi }}{3}\end{array} \right.;k \in Z\)
b)
\(\begin{array}{l}\sin 2x = \cos 3x\\ \Leftrightarrow \cos 3x = \cos \left( {\frac{\pi }{2} - 2x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{2} - 2x + k2\pi \\3x = - \left( {\frac{\pi }{2} - 2x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} + k2\pi \\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\end{array}\)
c)
\(\begin{array}{l}{\cos ^2}2x = {\cos ^2}\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = - \cos \left( {x + \frac{\pi }{6}} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\pi - \left( {x + \frac{\pi }{6}} \right)} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right)\end{array} \right.\end{array}\)
Với \(\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right) \Leftrightarrow \left[ \begin{array}{l}2x = - \left( {x + \frac{\pi }{6}} \right) + k2\pi \\2x = x + \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{18}} + \frac{{k2\pi }}{3}\\x = \frac{\pi }{6} + k2\pi \end{array} \right.\)
Với \(\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right) \Leftrightarrow \left[ \begin{array}{l}2x = \frac{{5\pi }}{6} - x + k2\pi \\2x = - \left( {\frac{{5\pi }}{6} - x} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = \frac{{5\pi }}{6} + k2\pi \\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}\\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right.\)