\(\frac{x^4-4x^2+3}{x^2-8x+15}\ge0\)
K
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4 tháng 5 2019
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 = \(\frac{3}{2}\)
B
2
KT
14 tháng 2 2018
ĐKXĐ: \(x\ne-1;\) \(x\ne-3;\)\(x\ne-5;\)\(x\ne-7\)
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)
\(\Leftrightarrow\)\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)
\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}\right)=\frac{3}{16}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}-\frac{1}{x+7}=\frac{3}{8}\)
\(\Leftrightarrow\)\(\frac{6}{\left(x+1\right)\left(x+7\right)}=\frac{3}{8}\)
\(\Rightarrow\)\(3\left(x+1\right)\left(x+7\right)=48\)
\(\Leftrightarrow\)\(x^2+8x+7=16\)
\(\Leftrightarrow\)\(x^2+8x-9=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x-9\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=9\left(TMĐKXĐ\right)\end{cases}}\)
Vậy...
14 tháng 2 2018
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)
\(\Leftrightarrow\frac{1}{x^2+x+3x+3}+\frac{1}{x^2+3x+5x+15}+\frac{1}{x^2+5x+7x+35}=\frac{3}{16}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}+\frac{\left(x+1\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}\)
\(=\frac{3\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}\)
Mẫu của mỗi phân thức bằng nhau nên => tử của mỗi phân thức cũng phải bằng nhau
=> Đến đây thì dễ rồi, bạn giải ra tìm x
9 tháng 10 2020
\(\frac{x+3}{x^2+x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)+1\)
ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\right)+1\)
\(=\frac{x+3}{x^2+x+6}\div\frac{6}{\left(x-2\right)\left(x+2\right)}+1\)
\(=\frac{\left(x+3\right)\left(x-2\right)\left(x+2\right)}{6\left(x^2+x+6\right)}+1\)
\(=\frac{x^3+3x^2-4x-12}{6x^2+6x+36}+\frac{6x^2+6x+36}{6x^2+6x+36}\)
\(=\frac{x^3+3x^2-4x-12+6x^2+6x+36}{6x^2+3x+36}\)
\(=\frac{x^3+9x^2+2x+24}{6x^2+3x+36}\)
9 tháng 10 2020
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