\(\frac{x^2-4x+1}{x^2}\)
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4 tháng 2 2020
Bài 1:
d)ĐKXĐ: \(x\ne8\)
Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)
\(\Leftrightarrow\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3x-24}=0\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)
MTC=24(x-8)
\(\Leftrightarrow\frac{36}{24\left(x-8\right)}+\frac{72x-480}{24\left(x-8\right)}+\frac{3x-24}{24\left(x-8\right)}-\frac{104x-816}{24\left(x-8\right)}=0\)
\(\Leftrightarrow36+72x-480+3x-24-104x+816=0\)
\(\Leftrightarrow348-29x=0\)
\(\Leftrightarrow-29x+348=0\)
\(\Leftrightarrow x=\frac{-348}{-29}=12\)
Vậy: x=12
e) ĐKXĐ: \(x\ne\pm1\)
Ta có: \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4x+4}+\frac{12x-1}{4-4x}=0\)
\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}+\frac{12x-1}{4\left(1-x\right)}=0\)
MTC=4(x+1)(x-1)
\(\Leftrightarrow\frac{24}{4\left(x-1\right)\left(x+1\right)}+\frac{20x^2-20}{4\left(x-1\right)\left(x+1\right)}-\frac{8x^2-9x+1}{4\left(x-1\right)\left(x+1\right)}-\frac{12x^2-11x-1}{4\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2+11x+1=0\)
\(\Leftrightarrow20x+4=0\)
\(\Leftrightarrow20x=-4\)
\(\Leftrightarrow x=-\frac{4}{20}=-0,2\)(loại)
Vậy: x không có giá trị
g) Ta có: \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)
\(\Leftrightarrow\frac{\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}}{\frac{x-1}{x-1}+\frac{x+1}{x-1}}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}}{\frac{2x}{x-1}}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-1}{2x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{4x\cdot\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\cdot2x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{2}=0\)
MTC=2(x+1)
\(\Leftrightarrow\frac{2}{2\left(x+1\right)}-\frac{x+1}{2\left(x+1\right)}=0\)
\(\Leftrightarrow2-x+1=0\)
\(\Leftrightarrow1-x=0\)
\(\Leftrightarrow x=1\)(loại vì không thỏa mãn ĐKXĐ)
Vậy: x không có giá trị
LN
9 tháng 5 2020
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
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 = \(\frac{3}{2}\)
11 tháng 1 2022
\(A=\left(\dfrac{1}{x^2-4x}+\dfrac{2}{16-x^2}+\dfrac{4}{4x+16}\right):\dfrac{1}{4x}\left(x\ne4;x\ne-4;x\ne0\right).\)
\(A=\left(\dfrac{1}{x\left(x-4\right)}+\dfrac{-2}{\left(x+4\right)\left(x-4\right)}+\dfrac{1}{x+4}\right).4x\).
\(A=\dfrac{x+4-2x+x^2-4x}{x\left(x-4\right)\left(x+4\right)}.4x.\)
\(A=\dfrac{x^2-5x+4}{\left(x-4\right)\left(x+4\right)}.4.\)
\(A=\dfrac{\left(x-4\right)\left(x-1\right)}{\left(x-4\right)\left(x+4\right)}.4.\)
\(A=\dfrac{4\left(x-1\right)}{x+4}.\)
S
14 tháng 12 2018
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
14 tháng 12 2018
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
Đề là tìm min phải không bạn.Nếu thế thì:
\(\frac{x^2-4x+1}{x^2}=\frac{x^2-4x+4-3}{x^2}=\frac{\left(x-2\right)^2-3}{x^2}\ge-\frac{3}{x^2}\) (do \(\left(x-2\right)^2\ge0\))
Dấu "=" xảy ra khi \(x-2=0\Leftrightarrow x=2\)
Ta có: \(-\frac{3}{x^2}=-3:\frac{x^2}{1}=-3:\frac{2^2}{1}=-\frac{3}{4}\)
Vậy ...