Giải các phương trình sau:
a)\(\frac{4}{-25x^2+30x-3}=\frac{3}{5x-1}-\frac{2}{5x-1}\)
b)\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+4}-\frac{2}{x^2-4x+3}\)
Giải phương trình sau:
a) \(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\)
b) \(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}\)
c)\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
d)\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
giải phương trình sau:
a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\\\)
b) \(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(x-3\right)}\)
c)\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8+6x}{16x^2-1}\)
d) \(5+\frac{76}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
Bài làm
a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)
\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)
\(\Leftrightarrow6x+4=0\)
\(\Leftrightarrow x=-\frac{4}{6}\)
\(\Leftrightarrow x=-\frac{2}{3}\)
Vậy x = -2/3 là nghiệm.
@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x2-4
Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)
b) Bạn kiểm tra lại đề bài
c) \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8}{16x^2-1}\left(x\ne\pm\frac{1}{4}\right)\)
\(\Leftrightarrow\frac{3}{1-4x}-\frac{2}{4x+1}+\frac{8}{16x^2-1}=0\)
\(\Leftrightarrow\frac{-3}{4x+1}-\frac{2}{4x+1}+\frac{8}{\left(4x+1\right)\left(4x-1\right)}=0\)
\(\Leftrightarrow\frac{-3\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}-\frac{2\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)
\(\Leftrightarrow\frac{-12x+3}{\left(4x-1\right)\left(4x+1\right)}-\frac{8x-2}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)
\(\Leftrightarrow\frac{-12x+3-8x+2+8}{\left(4x-1\right)\left(4x+1\right)}=0\)
=> -20x+13=0
<=> -20x=-13
<=> \(x=\frac{13}{20}\left(tmđk\right)\)
Giải các phương trình sau:
a) \(\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\)
b) \(\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\)
c) \(\sin 3x - \cos 5x = 0\)
d) \({\cos ^2}x = \frac{1}{4}\)
e) \(\sin x - \sqrt 3 \cos x = 0\)
f) \(\sin x + \cos x = 0\)
a)
\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
b) \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
c)
\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x = - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x = - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)
d)
\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x = - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x = - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)
e)
\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)
f)
\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x = - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)
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
Bài 1. Giải các phương trình sau :
a) 7x - 35 = 0 b) 4x - x - 18 = 0
c) x - 6 = 8 - x d) 48 - 5x = 39 - 2x
Bài 2. Giải các phương trình sau :
a) 5x - 8 = 4x - 5 b) 4 - (x - 5) = 5(x - 3x)
c) 32 - 4(0,5y - 5) = 3y + 2 d) 2,5(y - 1) = 2,5y
Bài 3. Giải các phương trình sau :
a) \(\frac{3x-7}{5}=\frac{2x-1}{3}\)
b) \(\frac{4x-7}{12}- x=\frac{3x}{8}\)
Bài 4. Giải các phương trình sau :
a) \(\frac{5x-8}{3}=\frac{1-3x}{2}\)
b) \(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
Bài 5. Giải các phương trình sau :
a) 6(x - 7) = 5(x + 2) + x b) 5x - 8 = 2(x - 4) + 3
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
có bị viết nhầm thì thông cảm nha!
la`thu'hai nga`y 19 nhe
bài 1 giải phương trình
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\frac{3}{5x-1}+\frac{3}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8+6x}{16x^2-1}\)
\(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
Giải:
a) ⇔⇔ 9x2 + 12x + 4 - 18x + 12 = 9x2 ⇔ 9x2 + 12x + 4 - 18x + 12 - 9x2 = 0
⇔ 16 + 6x = 0 ⇔ 2(8 + 3x) = 0 ⇔ 8 + 3x = 0 ⇔ x = \(\frac{-8}{3}\)
Vậy nghiệm của phương trình là x = \(\frac{-8}{3}\) .
b) \(\frac{3}{5x-1}+\frac{3}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\text{⇔ }\frac{-3}{1-5x}+\frac{-3}{5x-3}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)
⇔ \(\frac{9-15x}{\left(1-5x\right)\left(5x-3\right)}+\frac{15x-3}{\left(1-5x\right)\left(5x-3\right)}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\) ⇔ 9 - 15x + 15x - 3 = 4
⇔ 8 = 4 ( vô lí)
Vậy phương trình trên vô nghiệm.
Mình chỉ làm 2 câu a, b thôi nhé! Các bài tập này cách làm giống nhau, bạn tự hoàn thành những bài còn lại nhé!
\(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\)
\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}=0\)
\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
giải phương trình:\(\frac{2x}{6x^2-x+3}+\frac{5x}{4x^2+5x+2}+\frac{x}{2x^2+3x+1}=\frac{1}{3}\)
b, \(\frac{1}{x+1}+\frac{2}{x+2}+\frac{1}{x+3}=\frac{1}{x+4}+\frac{2}{x+5}+\frac{1}{x+6}\)
c, \(x^2+\frac{9x^2}{\left(x+3\right)^2}=7\)
d,\(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}=0\)
e,\(\frac{9x}{x^2-2x+3}=\frac{5x^2+9x+15}{x^2+3x+3}\)
a,ĐKXĐ \(x\ne-1;-\frac{1}{2}\)
Ta thấy x=0 không là nghiệm của PT
Xét \(x\ne0\)
Khi đó PT
<=> \(\frac{2}{6x-1+\frac{3}{x}}+\frac{5}{4x+5+\frac{2}{x}}+\frac{1}{2x+3+\frac{1}{x}}=\frac{1}{3}\)
Đặt \(2x+\frac{1}{x}=a\)
=> \(\frac{2}{3a-1}+\frac{5}{2a+5}+\frac{1}{a+3}=\frac{1}{3}\)
<=> \(3\left(25a^2+75a+10\right)=6a^3+31a^2+34a-15\)
<=> \(6a^3-44a^2-191a-45=0\)
Xin lỗi đến đây tớ ra nghiệm không đẹp
c, \(x^2+\frac{9x^2}{\left(x+3\right)^2}=7\) ĐKXĐ \(x\ne-3\)
<=> \(\left(x-\frac{3x}{x+3}\right)^2+2.\frac{3x^2}{x+3}=7\)
<=> \(\left(\frac{x^2}{x+3}\right)^2+6.\frac{x^2}{x+3}-7=0\)
<=> \(\left(\frac{x^2}{x+3}+7\right)\left(\frac{x^2}{x+3}-1\right)=0\)
<=> \(\orbr{\begin{cases}x^2+7x+21=0\\x^2-x-3=0\end{cases}}\)
\(S=\left\{\frac{1\pm\sqrt{13}}{2}\right\}\)thỏa mãn ĐKXĐ
b,\(\frac{1}{x+1}+\frac{2}{x+2}+\frac{1}{x+3}=\frac{1}{x+4}+\frac{2}{x+5}+\frac{1}{x+6}\)ĐKXĐ \(x\ne-1;-2;-3;-4;-5;-6\)
<=>\(\left(\frac{1}{x+1}-\frac{1}{x+6}\right)+2\left(\frac{1}{x+2}-\frac{1}{x+5}\right)+\left(\frac{1}{x+3}-\frac{1}{x+4}\right)=0\)
<=>\(\frac{5}{x^2+7x+6}+\frac{6}{x^2+7x+10}+\frac{1}{x^2+7x+12}=0\)
Đặt \(x^2+7x+6=a\)
=> \(\frac{5}{a}+\frac{6}{a+4}+\frac{1}{a+6}=0\)
<=> \(12a^2+90a+120=0\)
<=> \(a=\frac{-15\pm\sqrt{65}}{4}\)
Thay vào tính x nhưng bài này tớ ra nghiệm không đẹp