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11 tháng 3 2018
đkxđ với mọi x
đặt a=x2+x+1
\(\dfrac{a}{a+1}+\dfrac{a+1}{a+2}=\dfrac{7}{6}\)
<=> \(\dfrac{6a\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}+\dfrac{6\left(a+1\right)^2}{6\left(a+1\right)\left(a+2\right)}=\dfrac{7\left(a+1\right)\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}\)
=> 6a(a+2) +6(a+1)2 =7(a+1)(a+2)
<=> 6a2+12a +6a2 +12a+6 =a2 +21a+14
<=> 12a2 -a2+24a-21a+6-14=0
<=> 11a2+3a-8=0
<=> 11a2 +11a-8a-8=0
<=> (11a2 +11a)-(8a+8)=0
<=> 11a(a+1)-8(a+1)=0
<=> (a+1)(11a-8)=0
=> a=-1 và a=\(\dfrac{8}{11}\)
thay a=x2+x+1 ta đc
x2+x+1=-1
<=> x2+x+2 =0 (vô nghiệm)
và x2+x+\(\dfrac{3}{11}\) =0(vô nghiệm )
vậy pt trên vô nghiệm
12 tháng 3 2018
c) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\left(2\right)\)ĐKXĐ : x # 0
( 2) <=> \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)
\(< =>8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right).\left(-2\right)=\left(x+4\right)^2\)
\(< =>8.\left[\left(x+\dfrac{1}{x}\right)^2-x^2-\dfrac{1}{x^2}\right]=\left(x+4\right)^2\)
\(< =>16=\left(x+4\right)^2\)
<=> x2 + 8x = 0
<=> x( x + 8) = 0
<=> x = 0 ( KTM ) hoặc x = - 8 ( TM )
Vậy,....
8 tháng 3 2018
\(\Leftrightarrow\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{x\left(x+1\right)}=-1\left(đkxđ:x\ne\pm1;0;2;3\right)\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x}-\dfrac{1}{x+1}=-1\)
\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x+1}=-1\)
\(\Leftrightarrow\dfrac{4}{x^2-2x-3}=-1\)
\(\Leftrightarrow x^2-2x-3=-4\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\left(loai\right)\)
Vậy không có giá trị x thỏa mãn
TL
3
TA
11 tháng 6 2021
ĐK: `x \ne \pm 3`
`(2(9+2x))/(x^2-9)=2/(x-3)-1/(x+3)`
`<=>2(2x+9)=2(x+3)-(x-3)`
`<=>4x+18=2x+6-x+3`
`<=>4x+18=x+9`
`<=>3x=-9`
`x=-3 (L)`
Vậy `S=∅`.
KH
11 tháng 6 2021
\(\dfrac{2\left(9+2x\right)}{x^2-9}=\dfrac{2}{x-3}-\dfrac{1}{x+3}\)
\(\Leftrightarrow\dfrac{18+4x}{\left(x-3\right)\left(x+3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Rightarrow18+4x-2x-6+x-3=0\)
\(\Leftrightarrow3x+9=0\)
\(\Leftrightarrow3\left(x+3\right)=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
\(S=\left\{-3\right\}\)
HN
10 tháng 3 2017
\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{130}\)
ĐK: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\\x\ne-3\\x\ne-4\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{130\left(x+3\right)\left(x+4\right)+130\left(x+1\right)\left(x+4\right)+130\left(x+1\right)\left(x+2\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{3\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}\)
\(\Leftrightarrow3x^2+15x-378=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-14\end{matrix}\right.\)
10 tháng 3 2017
@ngonhuminh @Nguyễn Huy Thắng @Đức Minh@Hoang Hung Quan@Nguyễn Huy Tú@Hoàng Thị Ngọc Anh.... và mb khác giúp mik đi mà, cần gấp lắm T_T
13 tháng 12 2021
\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
10 tháng 12 2017
Bài 1:
\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)
10 tháng 12 2017
đc bn , nhg mà đề bài câu a b2 sao tự nhiên lại có " n "
bn xem lại đề đi
ĐKXĐ: x≠0, x≠2.
\(\dfrac{x+2}{x-2}\) - \(\dfrac{1}{x}\) = \(\dfrac{2}{x\left(x-2\right)}\)
<=> \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}\) - \(\dfrac{x-2}{x\left(x-2\right)}\)= \(\dfrac{2}{x\left(x-2\right)}\)
=> x (x+2) - x + 2 = 2
<=> x^2+2x-x+2=2
<=> x^2+2x-x=2-2
<=> x^2+x=0
<=> x(x+1)=0
<=>x=0 hoặc x+1=0
<=> x=0( ko thỏa mãnĐKXĐ) x=-1( thỏa mãn ĐKXĐ)
Vậy phương trình có tập nghiệm S={-1}