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a) \(P=\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(1+x+x^2\right)}.\frac{x^2+x+1}{x+1}\right).\frac{x^2+2x+1}{2x+1}\)
\(=\left(\frac{1}{x-1}-\frac{x}{\left(x-1\right)\left(x+1\right)}\right).\frac{x^2+2x+1}{2x+1}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x}{\left(x-1\right)\left(x+1\right)}\right).\frac{x^2+2x+1}{2x+1}\)
\(=\frac{1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{x+1}{\left(x-1\right)\left(2x+1\right)}\)
b) \(Q=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{5x-5x}{2x\left(x+5\right)}\)
\(=\frac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3-x^2+5x^2-5x}{2x\left(x+5\right)}\)
\(=\frac{x^2\left(x-1\right)+5x\left(x-1\right)}{2x\left(x+5\right)}\)
\(=\frac{\left(x-1\right)\left(x^2+5x\right)}{2x\left(x+5\right)}\)
\(=\frac{x\left(x-1\right)\left(x+5\right)}{2x\left(x+5\right)}\)
\(=\frac{x-1}{2}\)
a) \(P=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)
\(P=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)
\(P=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)
\(P=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x\left(x+1\right)-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(P=\frac{x-1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x+1\right)}\)
b) \(Q=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(Q=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x^2+4x-5}{2\left(x+5\right)}\)
\(A=\left(\frac{x}{25+5x}+\frac{5x+50}{x^2+5x}-\frac{10-2x}{x}\right)\div\frac{3x+15}{7}\)
ĐK : \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(=\left(\frac{x}{5\left(x+5\right)}+\frac{5\left(x+10\right)}{x\left(x+5\right)}-\frac{2\left(5-x\right)}{x}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{5\cdot5\cdot\left(x+10\right)}{5x\left(x+5\right)}-\frac{2\left(5-x\right)\cdot5\left(x+5\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{25x+250}{5x\left(x+5\right)}-\frac{10\left(25-x^2\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\left(\frac{x^2+25x+250-250+10x^2}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)
\(=\frac{11x^2+25x}{5x\left(x+5\right)}\times\frac{7}{3\left(x+5\right)}\)
\(=\frac{77x^2+175x}{15x\left(x+5\right)^2}\)
\(=\frac{77x^2+175x}{15x\left(x^2+10x+25\right)}=\frac{77x^2+175x}{15x^3+150x^2+375x}\)
\(=\frac{77x+175}{15x^2+150x+375}\)
a) ĐKXĐ: \(\begin{cases}x\ne0\\x+5\ne0\end{cases}\Leftrightarrow\begin{cases}x\ne0\\x\ne-5\end{cases}\)
b)\(A=\frac{x^2+2x}{2x+10}+\frac{x+5}{x}-\frac{50-5x}{2x\left(x+5\right)}=\frac{x^2+2x}{2.\left(x+5\right)}+\frac{x+5}{x}-\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^2+2x}{2x.\left(x+5\right)}+\frac{2\left(x+5\right)^2}{2x\left(x+5\right)}-\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^2+2x+2x^2+20x+50-50+5x}{2x\left(x+5\right)}=\frac{3x^2+27x}{2x\left(x+5\right)}=\frac{3x.\left(x+9\right)}{2x\left(x+5\right)}=\frac{3x+27}{2x+10}\)
c)Để A=1 thì: \(\frac{3x+27}{2x+10}=1\Rightarrow3x+27=2x+10\Leftrightarrow x=-17\)(nhận)
Vậy x=-17 thì A=1
\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)
\(\frac{x^3+2x^2}{2x^2+10x}\)+\(\frac{2x^2-10x+10x-50}{2x^2-10x}\)+\(\frac{50-5x}{2x^2+10x}\)=\(\frac{x^3+4x^2-5x}{2x^2-10x}\)=\(\frac{x\left(x^2+4x-5\right)}{2x\left(x-5\right)}\)=\(\frac{x\left(x-1\right)\left(x-5\right)}{2x\left(x-5\right)}\)=\(\frac{x-1}{2}\)
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
\(A=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(=\frac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+5\left(10-x\right)}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3-3x+2x^2}{2x\left(x+5\right)}=\frac{x\left(x^2+2x-3\right)}{2x\left(x+5\right)}\)
\(=\frac{\left(x-1\right)\left(x+3\right)}{2\left(x+5\right)}\)