\(\frac{x+3}{x^2+x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)+1\)
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!
1,Giải PT
a,\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)
b,\(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
c,\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
a,\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)
ĐKXĐ: x≠1/4, x≠-1/4
⇔\(-\frac{3}{4x-1}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)
⇔\(\frac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\frac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\frac{3+6x}{16x^2-1}\)
⇒-12x-3=8x-2-3-6x
⇔8x-6x+12x=-3+2+3
⇔14x=2
⇔x=1/7(tmđk)
Vậy phương trình có nghiệm là x=1/7
b, \(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\) (2)
ĐKXĐ: x≠0, x≠2
(2)⇔\(\frac{2\left(5-x\right)}{2.4x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4.\left(x-1\right)}{4.2x\left(x-2\right)}+\frac{x}{8.x\left(x-2\right)}\)
⇒10-2x+7x-14=4x-4+x
⇔-2x+7x-4x-x=-4-10+14
⇔0x=0
⇔ x∈R
Vậy phương trình có nghiệm là x∈R và x≠0, x≠2
c, \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) (3)
ĐKXĐ: x≠0
(3)⇒x(x+1)(x2-x+1)-x(x-1)(x2+x+1)=3
⇔x4+x-x4+x=3
⇔2x=3
⇔x=3/2(tmđk)
Vậy phương trình có nghiệm là x=3/2
Tìm điều kiện xác định rồi giải các phương trình sau:
a) \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
b) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
c) \(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Help me!
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
Giải phương trình
a) \(\frac{4}{20-6x-2x^2}\)+ \(\frac{x^2+4x}{x^2+5x}-\frac{x+3}{2-x}+3=0\)
b)\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2-10x}+10=\frac{x+25}{2x^2-50}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
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 các phương trình sau:
a)\(\frac{\left(9x-0.7\right)}{4}-\frac{\left(5x-1.5\right)}{7}=\frac{\left(7x-1.1\right)}{3}-\frac{5\left(0.4-2x\right)}{6}\)
b)\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
c)\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}\)
d)\(\frac{8x^2}{3\left(1-4x\right)^2}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
Giải phương trình:
a,\(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
b,\(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
c,\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+3}\)
Giải phương trình:
a. \(\frac{x+4}{x^2-3x+2}-\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
b. \(\frac{1}{x-1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
c. \(\frac{x+2}{3\:\:}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)d. \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
b, \(\frac{1}{x-1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\left(ĐKXĐ:x\ne\pm1;x\ne2\right)\)
\(\Leftrightarrow\)\(\frac{1}{x-1}+\frac{5}{2-x}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(2-x\right)+5\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(2-x\right)\left(x-1\right)}=\frac{15\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(2-x\right)}\)
Suy ra:
\(\Leftrightarrow\)(x+1)(2-x)+5(x-1)(x+1) = 15(x-1)
\(\Leftrightarrow\)2x-x2-x+2+5x2-5 = 15x-15
\(\Leftrightarrow\)2x-x2-x+5x2-15x = -15+5-2
\(\Leftrightarrow\)4x2-14x = -12
\(\Leftrightarrow4x^2-14x+12=0\)
\(\Leftrightarrow4x^2-8x-6x+12=0\)
\(\Leftrightarrow\)4x(x-2) - 6(x-2) = 0
\(\Leftrightarrow\left(x-2\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(kotm\right)\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất x = \(\frac{3}{2}\)
\(\text{Giải phương trình:}\)
\(a,\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
\(b,\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
\(c,\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+3}\)
a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\) ĐKXĐ : x #0, x#2, x#-2
<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)
<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
=> 10 - 2x + 7x - 14 = 4x - 4 + x
<=>-2x + 7x - 4x + x = -4 - 10 + 14
<=>x=-14