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ĐKXĐ:\(x\ne\pm\dfrac{1}{2}\)
\(\dfrac{1+8x}{4+8x}-\dfrac{4x}{12x-6}+\dfrac{32x^2}{3\left(4-16x^2\right)}=0\)
\(\Leftrightarrow\dfrac{1+8x}{4\left(2x+1\right)}-\dfrac{4x}{6\left(2x-1\right)}+\dfrac{32x^2}{-6\cdot\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow\dfrac{6\cdot\left(1+8x\right)\left(2x-1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{4\cdot4x\left(2x+1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{32x^2\cdot4}{24\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow96x^2-36x-6-36x^2-16x-144x^2=0\)
\(\Leftrightarrow-84x^2-52x-6=0\)
\(\Leftrightarrow\Delta=688\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{52-\sqrt{688}}{-168}=\dfrac{-13+\sqrt{43}}{42}\\x_2=\dfrac{52+\sqrt{688}}{-168}=\dfrac{-13-\sqrt{43}}{43}\end{matrix}\right.\)
Vậy pt có 2 nghiệm phân biệt............
1) \(\frac{8xy\left(3x-1\right)^3}{12x^3\left(1-3x\right)}=-\frac{8xy\left(3x-1\right)^3}{12x^3\left(3x-1\right)}=-\frac{2y\left(3x-1\right)^2}{3x^2}\)
2) \(\frac{5x^3+5x}{x^4-1}=\frac{5x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}=\frac{5x}{x^2-1}\)
3) \(\frac{9-\left(x+5\right)^2}{x^2+4x+4}=\frac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}=\frac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=-\frac{x+8}{x+2}\)
3) \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{2x\left(16-4x+x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)
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!
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
a/
\(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2+2\left(x^2+2x\right)+1=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Rightarrow x=1\)
b/
\(y^2+2y+1+\left(2^x\right)^2-2.2^x+1=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\2^x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-1\\x=0\end{matrix}\right.\)
c/
ĐKXĐ: \(x\ne\left\{-2;-4;-6;-8\right\}\)
\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}=\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\frac{2}{x+2}+x+8+\frac{8}{x+8}=x+4+\frac{4}{x+4}+x+6+\frac{6}{x+6}\)
\(\Leftrightarrow\frac{1}{x+2}+\frac{4}{x+8}=\frac{2}{x+4}+\frac{3}{x+6}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{2}{x+4}+\frac{4}{x+8}-\frac{3}{x+6}=0\)
\(\Leftrightarrow\frac{-x}{\left(x+2\right)\left(x+4\right)}+\frac{x}{\left(x+8\right)\left(x+6\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{\left(x+2\right)\left(x+4\right)}=\frac{1}{\left(x+6\right)\left(x+8\right)}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x+2\right)\left(x+4\right)=\left(x+6\right)\left(x+8\right)\)
\(\Leftrightarrow8x=-40\Rightarrow x=-5\)
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 = \(\frac{3}{2}\)
\(\frac{x+3}{x^2+x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)+1\)
ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\frac{6}{\left(x-2\right)\left(x+2\right)}+1\)
\(=\frac{\left(x+3\right)\left(x-2\right)\left(x+2\right)}{6\left(x^2+x+6\right)}+1\)
\(=\frac{x^3+3x^2-4x-12}{6x^2+6x+36}+\frac{6x^2+6x+36}{6x^2+6x+36}\)
\(=\frac{x^3+3x^2-4x-12+6x^2+6x+36}{6x^2+3x+36}\)
\(=\frac{x^3+9x^2+2x+24}{6x^2+3x+36}\)
ơ lag hai dòng cuối :v
bạn sửa mẫu thức ở hai dòng cuối thành 6x2 + 6x + 36 dùm mình nhé -.- dạo này lú quá