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\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{x^2+5x+6}-\frac{2}{x-3}=0\)
\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{x^4-8x^2+16-25}-\frac{3\left(x+2\right)}{x^2+2x+3x+6}-\frac{2}{x-3}=0\)
\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x^4-8x^2+16\right)-5^2}-\frac{3\left(x+2\right)}{x\left(x+2\right)+3\left(x+2\right)}-\frac{2}{x-3}=0\)
\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x^2-4\right)^2-5^2}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\)
\(\Leftrightarrow\frac{13-x}{x+3}-\frac{3}{x+3}+\frac{6\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-9\right)}-\frac{2}{x-3}=0\)
\(\Leftrightarrow\frac{10-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x-3}=0\)
\(\Leftrightarrow\frac{\left(10-x\right)\left(x-3\right)+6-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow10x-30-x^2+3x+6-2x-6=0\)
\(\Leftrightarrow-x^2+11x-30=0\)
\(\Leftrightarrow-x^2+5x+6x-30=0\)
\(\Leftrightarrow-x\left(x-5\right)+6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(-x+6\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-5=0\\-x+6=0\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=5\\x=6\end{matrix}\right.\)
Vậy x=5 ;x=6
Bài 1:
a) Để giá trị của phân thức A được xác định <=> \(7x^2+7x\ne0\) <=> \(7x.\left(x+1\right)\ne0\)<=> \(x\ne0\)và \(x\ne-1\)
=> Để giá trị của phân thức A được xác định thì x phải khác -1 và 0.
b) Để phân thức A = 0 => x - 3 = 0 => x = 3 (thỏa mãn đkxd)
=> Để giá trị phân thức A = 0 thì x = 3
Bạn viết z chắc mỏi tay lắm. Mik sẽ giải cho bạn b3 nhé
a) \(2x^3-12x^2+18x=2x.\left(x^2-6x+9\right)=2x.\left(x-3\right)^2\)
b) \(16y^2-4x^2-12x-9=16y^2-\left(4x^2+12x+9\right)=16y^2-\left(2x+3\right)^2\)
\(=\left(4y+2x+3\right).\left(4y-2x-3\right)\)
A=\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\)\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x-3\right)\left(x+3\right)\left(x^2+1\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\) ( với \(x^4-8x^2-9=x^4-9x^2+x^2-9=x^2\left(x^2-9\right)+\left(x^2-9\right)=\left(x^2-9\right)\left(x^2+1\right)=\left(x-3\right)\left(x+3\right)\left(x^2+1\right)\)
A= \(\frac{13-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}-\frac{2}{x-3}=0\) \(\Leftrightarrow\frac{10-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x-3}=0\) \(\Leftrightarrow\left(10x-30\right)\left(x-3\right)+6-2\left(x+3\right)=0\Leftrightarrow-x^2+11x-30=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=5\end{array}\right.\)
ĐK: \(x\ne-3,3,-2\)
Ta có: \(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{x^2+5x+6}-\frac{2}{x-3}=0\)
=>\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-9x^2+x^2-9}-\frac{3x+6}{x^2+3x+2x+6}-\frac{2}{x-3}=0\)
=>\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^2.\left(x^2-9\right)+\left(x^2-9\right)}-\frac{3x+6}{x.\left(x+3\right)+2.\left(x+3\right)}-\frac{2}{x-3}=0\)
=>\(\frac{13-x}{x+3}+\frac{6.\left(x^2+1\right)}{\left(x^2+1\right).\left(x^2-9\right)}-\frac{3.\left(x+2\right)}{\left(x+2\right).\left(x+3\right)}-\frac{2}{x-3}=0\)
=>\(\frac{13-x}{x+3}+\frac{6}{x^2-9}-\frac{3}{x+3}-\frac{2}{x-3}=0\)
=>\(\left(\frac{13-x}{x+3}-\frac{3}{x+3}\right)+\left(\frac{6}{x^2-9}-\frac{2}{x-3}\right)=0\)
=>\(\frac{13-x-3}{x+3}+\left[\frac{6}{x^2-9}-\frac{2.\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\right]=0\)
=>\(\frac{10-x}{x+3}+\left[\frac{6}{x^2-9}-\frac{2x+6}{x^2-9}\right]=0\)
=>\(\frac{10-x}{x+3}+\frac{6-2x-6}{x^2-9}=0\)
=>\(\frac{\left(10-x\right).\left(x-3\right)}{\left(x+3\right).\left(x-3\right)}+\frac{-2x}{x^2-9}=0\)
=>\(\frac{13x-x^2-30}{x^2-9}-\frac{2x}{x^2-9}=0\)
=>\(\frac{13x-x^2-30-2x}{x^2-9}=0\)
=>\(\frac{11x-x^2-30}{x^2-9}=0\)
Vì \(x\ne-3,3=>x^2\ne0\)
=>11x-x2-30=0
=>6x-30-x2+5x=0
=>6.(x-5)-x.(x-5)=0
=>(6-x).(x-5)=0
=>6-x=0=>x=6
hoặc x-5=0=>x=5
Vậy tập nghiệm của phương trình S=6; 5
bài 1 ( tự luận )
a, Để \(\frac{3x+3}{x^2-1}\)Xác định
\(\Rightarrow\orbr{\begin{cases}x+1\ne0\\x-1\ne0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ne-1\\x\ne1\end{cases}}\)
\(\frac{3x+3}{x^2-1}=\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{3}{x-1}\)
Thay \(\frac{3}{x-1}=2\)......
\(c,\)Để \(\frac{3}{x-1}\)nguyên
\(\Rightarrow3⋮x-1\Rightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(x-1=1\Rightarrow x=2\)
\(x-1=-1\Rightarrow x=0\)
\(x-1=3\Rightarrow x=4\)
\(x-1=-3\Rightarrow x=-2\)
\(KL:x\in\left\{0;4;\pm2\right\}\)
1/. PT <=> \(\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x^4+x^2\right)-\left(9x^2+9\right)}-\frac{3\left(x+2\right)}{\left(x^2+2x\right)+\left(3x+6\right)}-\frac{2}{x-3}=0\)
<=> \(\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{x^2\left(x^2+1\right)-9\left(x^2+1\right)}-\frac{3\left(x+2\right)}{x\left(x+2\right)+3\left(x+2\right)}-\frac{2}{x-3}=0\)
<=> \(\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-9\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\)
<=>\(\frac{\left(13-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\) (1)
ĐKXĐ: \(x\ne3vàx\ne-3\)
(1) => \(13x-39-x^2+3x+6-3x+9-2x-6=0\)
<=> \(x^2-11x+30=0\)
<=> (x2-5x) -(6x - 30) = 0
<=> x(x - 5) -6 (x - 5) = 0
<=> (x-5) (x - 6) = 0
<=> x = 5 hay x = 6 (nhận )
Vậy pt đã cho có tập nghiệm S = {5;6}