\(\frac{5+x}{1-x}\)
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Những câu hỏi liên quan
3) frac{1-x}{x+1}-frac{3+2x}{x+1}0
13) frac{x+2}{x}-frac{x^2+5x+4}{xleft(x+2right)}frac{x}{x+2}
14) frac{1}{x+1}-frac{5}{x-2}frac{20}{left(x+1right)left(2-xright)}
16) frac{x+5}{x-5}-frac{x-5}{x+5}frac{20}{x^2-25}
17) frac{3x+2}{3x-2}-frac{6}{2+3x}frac{9x^2}{9x^2-4}
18) frac{x-1}{x}+frac{1}{x+1}frac{2x-1}{2x^2+2}
19) frac{2}{x+1}-frac{3x+1}{left(x+1right)}frac{1}{left(x+1right)left(x-2right)}
20) frac{x+5}{3x-6}-frac{1}{2}frac{2x-3}{2x-4}
Đọc tiếp
3) \(\frac{1-x}{x+1}-\frac{3+2x}{x+1}=0\)
13) \(\frac{x+2}{x}-\frac{x^2+5x+4}{x\left(x+2\right)}=\frac{x}{x+2}\)
14) \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{20}{\left(x+1\right)\left(2-x\right)}\)
16) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
17) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
18) \(\frac{x-1}{x}+\frac{1}{x+1}=\frac{2x-1}{2x^2+2}\)
19) \(\frac{2}{x+1}-\frac{3x+1}{\left(x+1\right)}=\frac{1}{\left(x+1\right)\left(x-2\right)}\)
20) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\)
Giải phương trình:1.frac{x-5}{x-5}+frac{x-6}{x-5}+frac{x-7}{x-5}+...+frac{1}{x-5}4left(xin Nright)2.frac{1}{x^2+3x+2}+frac{1}{x^2+5x+6}+frac{1}{x^2+7x+12}+...+frac{1}{x^2+15x+56}frac{1}{14}3.left(1+frac{1}{1.3}right)left(1+frac{1}{2.4}right)left(1+frac{1}{3.5}right)...left(1+frac{1}{xleft(x+2right)}right)frac{31}{16}left(xin Nright)4.8left(x^2+frac{1}{x^2}right)-34left(x+frac{1}{x}right)+5105.6x^4-5x^3-38x^2-5x+60
Đọc tiếp
Giải phương trình:
1.\(\frac{x-5}{x-5}+\frac{x-6}{x-5}+\frac{x-7}{x-5}+...+\frac{1}{x-5}=4\left(x\in N\right)\)
2.\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
3.\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{31}{16}\left(x\in N\right)\)
4.\(8\left(x^2+\frac{1}{x^2}\right)-34\left(x+\frac{1}{x}\right)+51=0\)
5.\(6x^4-5x^3-38x^2-5x+6=0\)
Cho \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
CMR: \(\frac{1}{x^5}+\frac{1}{y^5}+\frac{1}{z^5}=\frac{1}{x^5+y^5+z^5}\)
Áp dụng BĐT Cauchy dạng Engel , ta được
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^3}{x+y+z}=\frac{1}{x+y+z}\)
Dấu "=" xảy ra khi \(\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\) => \(x=y=z\).(*)
Áp dụng BĐT Cauchy dạng Engel , ta được : \(\frac{1}{x^5}+\frac{1}{y^5}+\frac{1}{z^5}\ge\frac{\left(1+1+1\right)^3}{x^5+y^5+z^5}\) \(=\frac{1}{x^5+y^5+z^5}\)
Dấu "=" xảy ra khi x=y=z ( đã có ở (*) )
Vậy \(\frac{1}{x^5}+\frac{1}{y^5}+\frac{1}{z^5}=\frac{1}{x^5+y^5+z^5}\) ( đpcm) với x=y=z
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Bài này gần giống câu hỏi số 965642 bn xem đi nhé
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mai mik kiểm tra rùi giúp mik vs pls
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
`=> x^2 +x -x-1 -x-2x+1=0`
`<=> x^2 -3x =0`
`<=> x(x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)
__
`(x+2)(5-3x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
__
\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)
\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)
`<=> 2x- 40x + 6x = 9x - 45 -24`
`<=> 2x- 40x + 6x-9x + 45 +24=0`
`<=>-41x+69=0`
`<=>-41x=-69`
`<=> x=69/41`
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a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
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tim xa)2frac{3}{4}x-1frac{5}{8}x1b)2frac{3}{4}x-1frac{5}{8}x1c)frac{3}{2}x-frac{2}{5}frac{1}{3}x-frac{1}{4}d)2x-frac{1}{4}frac{5}{6}-frac{1}{2}xe)frac{-5}{6}+3xfrac{2}{3}-frac{1}{2}xf)frac{5}{2}x-frac{3}{2}x+frac{29}{10}g)frac{2}{3}+frac{7}{3}xfrac{5}{4}x+frac{1}{6}h)frac{1}{3}.x+frac{2}{5}left(x-1right)0giup mk nha moi nguoi
Đọc tiếp
tim x
a)\(2\frac{3}{4}x-1\frac{5}{8}x=1\)
b)\(2\frac{3}{4}x-1\frac{5}{8}x=1\)
c)\(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\)
d)\(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\)
e)\(\frac{-5}{6}+3x=\frac{2}{3}-\frac{1}{2}x\)
f)\(\frac{5}{2}x-\frac{3}{2}=x+\frac{29}{10}\)
g)\(\frac{2}{3}+\frac{7}{3}x=\frac{5}{4}x+\frac{1}{6}\)
h)\(\frac{1}{3}.x+\frac{2}{5}\left(x-1\right)=0\)
giup mk nha moi nguoi
giải phương trình chứa ẩna) frac{x^2+5}{25-x^2}frac{3}{x+5}+frac{x}{x-5}b)frac{3}{x+1}-frac{2}{x+2}frac{4x+5}{x^2+3x+2}c)frac{2cdotleft(x^2+x+6right)}{x^3-8}+frac{2}{2-x}frac{3}{x^2+2x+4}d)frac{6}{x^3+1}-frac{1-x}{^{x^2}-x+1}frac{5}{x+1}e)frac{1}{x^23x+2}-frac{3}{x^2-x-2}frac{-1}{x^2-4}
Đọc tiếp
giải phương trình chứa ẩn
a) \(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)
b)\(\frac{3}{x+1}-\frac{2}{x+2}=\frac{4x+5}{x^2+3x+2}\)
c)\(\frac{2\cdot\left(x^2+x+6\right)}{x^3-8}+\frac{2}{2-x}=\frac{3}{x^2+2x+4}\)
d)\(\frac{6}{x^3+1}-\frac{1-x}{^{x^2}-x+1}=\frac{5}{x+1}\)
e)\(\frac{1}{x^23x+2}-\frac{3}{x^2-x-2}=\frac{-1}{x^2-4}\)
\(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)
\(\Leftrightarrow\frac{x^2+5}{\left(5-x\right)\left(5+x\right)}=\frac{3}{5+x}-\frac{x}{5-x}\)
\(\Leftrightarrow\frac{x^2+5}{\left(5-x\right)\left(5+x\right)}=\frac{3\left(5-x\right)-x\left(5+x\right)}{\left(5-x\right)\left(5+x\right)}\)
\(\Rightarrow x^2+5=3\left(5-x\right)-x\left(5+x\right)\)
\(\Leftrightarrow x^2+5=15-3x-5x-x^2\)
\(\Leftrightarrow15-3x-5x-x^2-x^2-5=0\)
\(\Leftrightarrow10-8x-2x^2=0\)
\(\Leftrightarrow2x^2+8x-10=0\)
\(\Leftrightarrow2\left(x^2+4x-5\right)=0\)
\(\Leftrightarrow2\left(x^2+5x-x-5\right)=0\)
\(\Leftrightarrow x^2-x+5x-5=0\)
\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)
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Giải phương trình sau:
a) frac{10}{left(x+5right)left(x-1right)}+frac{3}{1-x}frac{5}{x+5}
b) frac{x-1}{x+2}+frac{x+3}{x-4}frac{2}{left(x-2right)left(4-xright)}
c) frac{7x-3}{x-x^3}frac{1}{x-1}-frac{5}{xleft(x-1right)}
d) frac{1}{left(x+2right)}+frac{1}{left(x+3right)}frac{1}{left(x+2right)left(x+3right)}
Đọc tiếp
Giải phương trình sau:
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
b) \(\frac{x-1}{x+2}+\frac{x+3}{x-4}=\frac{2}{\left(x-2\right)\left(4-x\right)}\)
c) \(\frac{7x-3}{x-x^3}=\frac{1}{x-1}-\frac{5}{x\left(x-1\right)}\)
d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
Giai phương trình sau :
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
ĐKXĐ : \(x\ne1;x\ne-5\)
Với điều kiện trên ta có :
\(\Leftrightarrow\)\(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{-3}{x-1}=\frac{5}{x+5}\)
\(\Leftrightarrow10-3\left(x+5\right)=5\left(x-1\right)\)
\(\Leftrightarrow10-3x-15=5x-5\)
\(\Leftrightarrow-8x=0\)
\(\Leftrightarrow x=0\) (nhận)
Vậy : \(S=\left\{0\right\}\)
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d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
ĐKXĐ : \(x\ne-2;x\ne-3\)
\(\Leftrightarrow x+3+x+2=1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\) (không nhận)
Vậy : \(S=\varnothing\)
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24 tháng 12 2016 lúc 18:31
Tìm x :
\(x\times\frac{1}{5}+\frac{1}{5}\times x-x\times\frac{1}{5}=\frac{1}{5}\)
\(x\times\frac{1}{5}+\frac{1}{5}\times x-x\times\frac{1}{5}=\frac{1}{5}\)
\(x\times\left(\frac{1}{5}+\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{5}\)
\(x\times\frac{1}{5}=\frac{1}{5}\)
\(x=\frac{1}{5}:\frac{1}{5}\)
\(x=1\)
Vậy x = 1
ủng hộ mik nha các bn
chúc các bn giáng sinh vui vẻ
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x × 1/5 + 1/5 × x − x × 1/5 = 1/5
x × (1/5 + 1/5 − 1/5) = 1/5
x × 1/5 = 1/5
x = 1/5 : 1/5
x = 1
Mery christmas
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Xx1/5+1/5xX-Xx1/5=1/5
Xx(1/5+1/5-1/5)=1/5
Xx1/5=1/5
X=1/5:1/5
X=1
đúng!100%
(^-^)
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giải pt
1,frac{1}{x-1}+frac{2x^2-5}{x^3-1}frac{4}{x^2+x+1}
2,frac{x+4}{2x^2-5x+2}+frac{x+1}{2x^2-7x+3}frac{2x+5}{2x^2-7x+3}
3,frac{x+1}{x-1}-frac{x-1}{x+1}3xleft(1-frac{x-1}{x+1}right)
4,frac{2x}{x-1}+frac{4}{x^2+2x-3}frac{2x-5}{x+3}
5,frac{1}{x-1}-frac{7}{x+2}frac{3}{x^2+x-2}
6,frac{x+3}{x-4}+frac{x-1}{x-2}frac{2}{6x-8-x^2}
7,frac{1}{x-1}-frac{7}{x+2}frac{3}{1-x^2}
Đọc tiếp
giải pt
1,\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
2,\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
3,\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)
4,\(\frac{2x}{x-1}+\frac{4}{x^2+2x-3=}=\frac{2x-5}{x+3}\)
5,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)
6,\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
7,\(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)
Bài 1:
ĐKXĐ: x≠1
Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4\left(x-1\right)}{\left(x^2+x-1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow x^2+x+1+2x^2-5-4\left(x-1\right)=0\)
\(\Leftrightarrow x^2+x+1+2x^2-5-4x+4=0\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
Vì 3≠0
nên \(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)
Vậy: x=0
Bài 2:
ĐKXĐ: x≠2; x≠3; \(x\ne\frac{1}{2}\)
Ta có: \(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-\left(2x+5\right)}{\left(x-3\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-2x-5}{\left(x-3\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(-x-4\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-12-x^2-2x+8=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(tm)
Vậy: x=-4
Bài 3:
ĐKXĐ: x≠1; x≠-1
Ta có: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x-\frac{3x\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-3x+\frac{3x\left(x-1\right)}{x+1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)-3x\left(x^2-1\right)+3x\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-3x^3+3x+3x^3-6x^2+3x=0\)
\(\Leftrightarrow-6x^2+10x=0\)
\(\Leftrightarrow2x\left(-3x+5\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{5}{3}\right\}\)
Bài 4:
ĐKXĐ: x≠1; x≠-3
Ta có: \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=0\)
\(\Leftrightarrow2x^2+6x+4-\left(2x^2-7x+5\right)=0\)
\(\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0\)
\(\Leftrightarrow13x-1=0\)
\(\Leftrightarrow13x=1\)
hay \(x=\frac{1}{13}\)(tm)
Vậy: \(x=\frac{1}{13}\)
Bài 5:
ĐKXĐ: x≠1; x≠-2
Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)
\(\Leftrightarrow\frac{x+2}{\left(x-1\right)\left(x+2\right)}-\frac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{3}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Leftrightarrow x+2-7\left(x-1\right)-3=0\)
\(\Leftrightarrow x+2-7x+7-3=0\)
\(\Leftrightarrow-6x+6=0\)
\(\Leftrightarrow-6\left(x-1\right)=0\)
Vì -6≠0
nên x-1=0
hay x=1(ktm)
Vậy: x∈∅
Bài 6:
ĐKXĐ: x≠4; x≠2
Ta có: \(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{6x-8-x^2}=0\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{-\left(x^2-6x+8\right)}=0\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)
Vậy: x=0
Bài 7:
ĐKXĐ: x≠1; x≠-2; x≠-1
Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{7}{x+2}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}-\frac{7\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)\left(x+1\right)}+\frac{3\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x^2+3x+2-7\left(x^2-1\right)+3x+6=0\)
\(\Leftrightarrow x^2+3x+2-7x^2+7x+3x+6=0\)
\(\Leftrightarrow-6x^2+13x+8=0\)
\(\Leftrightarrow-6x^2+16x-3x+8=0\)
\(\Leftrightarrow2x\left(-3x+8\right)+\left(-3x+8\right)=0\)
\(\Leftrightarrow\left(-3x+8\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+8=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{3};\frac{-1}{2}\right\}\)
\( 1)\dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{4}{{{x^2} + x + 1}}\\ DK:x \ne 1\\ \Leftrightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{{4\left( {x - 1} \right)}}{{{x^3} - 1}}\\ \Leftrightarrow {x^2} + x + 1 + 2{x^2} - 5 = 4x - 4\\ \Leftrightarrow 3{x^2} - 3x = 0\\ \Leftrightarrow 3x\left( {x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\left( {tm} \right)\\ x = 1\left( {ktm} \right) \end{array} \right.\\ 2)\dfrac{{x + 4}}{{2{x^2} - 5x + 2}} + \dfrac{{x + 1}}{{2{x^2} - 7x + 3}} = \dfrac{{2x + 5}}{{2{x^2} - 7x + 3}}\\ + DK:x \ne \dfrac{1}{2};x \ne 2;x \ne 3\\ \Leftrightarrow \dfrac{{x + 4}}{{\left( {2x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}} = \dfrac{{2x + 5}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}}\\ \Leftrightarrow \left( {x + 4} \right)\left( {x - 3} \right) + \left( {x + 1} \right)\left( {x - 2} \right) = \left( {2x + 5} \right)\left( {x - 2} \right)\\ \Leftrightarrow {x^2} + x - 12 + {x^2} - x - 2 = 2{x^2} + x - 10\\ \Leftrightarrow x = - 4\left( {tm} \right)\\ 3)\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = 3x\left( {1 - \dfrac{{x - 1}}{{x + 1}}} \right)\\ DK:x \ne \pm 1\\ \Leftrightarrow {\left( {x + 1} \right)^2} - {\left( {x - 1} \right)^2} = 3x\left( {x - 1} \right)\left( {x + 1 - x + 1} \right)\\ \Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x - 1 = 6x\left( {x - 1} \right)\\ \Leftrightarrow 4x = 6{x^2} - 6x\\ \Leftrightarrow 2x\left( {3x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{5}{3} \end{array} \right.\left( {tm} \right) \)
Còn lại tương tự mà làm nhé!
Giải phương trình sau:
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
b) \(\frac{x-1}{x+2}+\frac{x+3}{x-4}=\frac{2}{\left(x-2\right)\left(4-x\right)}\)
c) \(\frac{7x-3}{x-x^3}=\frac{1}{x-1}-\frac{5}{\left(x-1\right)}\)
d) \(\frac{1}{x+2}+\frac{1}{x+3}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)