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ĐKXĐ : \(\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Rightarrow x\ne0;x\ne-2\left(1\right)}\)
Ta có P = \(\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50+5x}{2x\left(x+5\right)}\)
\(=\frac{x^2+2x}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50+5x}{2x\left(x+5\right)}\)
\(=\frac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\frac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}+\frac{50+5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x^2+2x^2-50+50+5x}{2x\left(x+5\right)}=\frac{x^3+4x^2+5x}{2x\left(x+5\right)}=\frac{x\left(x^2+4x+5\right)}{2x\left(x+5\right)}\)
\(=\frac{x^2+4x+5}{2\left(x+5\right)}\)
c) P = 1
<=> \(\frac{x^2+4x+5}{2\left(x+5\right)}=1\Rightarrow x^2+4x+5=2\left(x+5\right)\)
=> x2 + 4x + 5 - 2x - 10 = 0
=> x2 + 2x - 5 = 0
=> x2 + 2x + 1 - 6 = 0
=> (x + 1)2 = 6
=> \(\orbr{\begin{cases}x+1=\sqrt{6}\\x+1=-\sqrt{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{6}-1\\x=-\sqrt{6}-1\end{cases}}\)(tm (1))
d) P = -1/2
<=> \(\frac{x^2+4x+5}{2\left(x+5\right)}=-\frac{1}{2}\)
=> 2(x2 + 4x + 5) = -2(x + 5)
=> 2x2 + 8x + 10 = -2x - 10
=> 2x2 + 8x + 10 + 2x + 10 = 0
=> 2x2 + 10x + 20 = 0
=> 2(x2 + 5x + 10) = 0
=> x2 + 5x + 10 = 0
=> \(x^2+2.\frac{5}{2}x+\frac{25}{4}+\frac{15}{4}=0\)
=> \(\left(x+\frac{5}{2}\right)^2+\frac{15}{4}=0\)
=> \(x\in\varnothing\left(\text{Vì }\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\forall x\right)\)
Vậy không tồn tại x để P = -1/2
\(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50+5x}{2x\left(x+5\right)}\)
a) ĐK : x ≠ 0 ; x ≠ -5
b) \(P=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50+5x}{2x\left(x+5\right)}\)
\(=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{50+5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x^2}{2x\left(x+5\right)}+\frac{2\left(x^2-25\right)}{2x\left(x+5\right)}+\frac{50+5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x^2+2x^2-50+50+5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+4x^2+5x}{2x\left(x+5\right)}=\frac{x\left(x^2+4x+5\right)}{2x\left(x+5\right)}\)
\(=\frac{x^2+4x+5}{2x+10}\)
c) Để P = 1
thì \(\frac{x^2+4x+5}{2x+10}=1\)
=> x2 + 4x + 5 = 2x + 10
=> x2 + 4x + 5 - 2x - 10 = 0
=> x2 - 2x - 5 = 0
=> ( x2 - 2x + 1 ) - 6 = 0
=> ( x - 1 )2 - ( √6 )2 = 0
=> ( x - 1 - √6 )( x - 1 + √6 ) = 0
=> x = 1 + √6 hoặc x = 1 - √6
Cả hai giá trị đều thỏa x ≠ 0 ; x ≠ -5
Vậy x = 1 + √6 hoặc x = 1 - √6
d) Để P = -1/2
thì \(\frac{x^2+4x+5}{2x+10}=\frac{-1}{2}\)
=> 2( x2 + 4x + 5 ) = -2x - 10
=> 2x2 + 8x + 10 + 2x + 10 = 0
=> 2x2 + 10x + 20 = 0
=> 2( x2 + 5x + 10 ) = 0
=> x2 + 5x + 10 = 0 (*)
Ta có : x2 + 5x + 10 = ( x2 + 5x + 25/4 ) + 15/4 = ( x + 5/2 )2 + 15/4 ≥ 15/4 > 0 ∀ x
tức (*) không xảy ra
Vậy không có giá trị của x để P = -1/2
a)
\(\left\{{}\begin{matrix}x-1\ne0\\x+2\ne0\end{matrix}\right.\)
b)
x khác 1
c)
x khác 0; x khác 5
d) x khác 5 ; x khác -5
\(a,Đkxđ:x\ne\pm2\)
\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+1\right)^2}{x^2-4}\)
b, Ta có: \(\left(x-2\right)\left(x+2\right)< 0;\forall-2< 2< 2;x\ne-1\)
Mà: \(\left(x+1\right)^2>0\left(\forall x\ne-1\right)\)
\(\Rightarrow\frac{\left(x+1\right)^2}{\left(x+2\right)\left(x-2\right)}< 0;\forall-2< x< 2;x\ne-1\)
Vậy ............
\(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(=\left(x^4-x^2+1\right)\left(x^4+2x^2-x^2+1\right)\)
\(=\left(x^4-x^2+1\right)[\left(x^2+1\right)^2-x^2]\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
a.2x#+_2 . quy đồng khử mẫu tchung : (x+2)(x+1)+(x-1)(x-2)--->2x^2 + 4=2(x^2+2). --> s={x thuộc R/ X#+_2}
a) ĐKXĐ \(\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x\left(x^2+2\right)=0\)
\(\Leftrightarrow x^2+3x+2+x^2-3x+2-2x^2-4=0\)
\(\Leftrightarrow0x=0\)(vô số nghiệm)
nghiệm x thỏa mãn phương trình S \(\in\)R với \(\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
b) ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\Rightarrow\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}=\frac{1}{2x\left(x-2\right)}-\frac{7}{8x}\)
\(\Rightarrow2\left(5-x\right)-x-4\left(x-1\right)+7\left(x-2\right)=0\)
\(\Leftrightarrow10-2x-x-4x+4+7x-14=0\)
\(\Leftrightarrow0x=0\)(phương trìh vô số nghiệm)
nghiệm x thỏa mãn phương trình S \(\in\)R với \(\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
Đáp án: A
Với x ≠ -1, x ≠ 1 ta có x 2 – 1 ≠ 0 .
Vậy phân thức
luôn có nghĩa với mọi x ≠ -1, x ≠ 1