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a) Phân thức xác định \(\Leftrightarrow2x^2+2x\ne0\)
\(\Leftrightarrow2x\left(x+1\right)\ne0\)
\(\Rightarrow\left[{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
b) \(M=\dfrac{5x+5}{2x^2+2x}=\dfrac{5\left(x+1\right)}{2x\left(x+1\right)}=\dfrac{5}{2x}\)
Vì ĐKXĐ x khác 0 nên ta chỉ xét trường hợp x = 5
\(M=\dfrac{5}{2x}=\dfrac{5}{2\cdot5}=\dfrac{1}{2}\)
Vậy........
a, PT xác định khi 2x2+2x ≠0 ⇔2x(x+2) ≠0 ⇔\([\)\(\dfrac{x\ne0}{x\ne-2}\)
b, x=5 PT trở thành \(\dfrac{5.5+5}{2.5^2+2.5}\) =\(\dfrac{30}{60}\) =0,5
do x ≠0 nên x=0 không được
a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)
b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)
\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)
\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)
c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)
Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.
Bài 1:
a: ĐKXĐ: x<>1/3; x<>-1/3
b: \(M=\left(\dfrac{-3x}{3x-1}+\dfrac{2x}{3x+1}\right)\cdot\dfrac{\left(3x-1\right)^2}{2\left(3x^2+5\right)}\)
\(=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x-1\right)\left(3x+1\right)}\cdot\dfrac{\left(3x-1\right)^2}{2\left(3x^2+5\right)}\)
\(=\dfrac{-3x^2-5x}{\left(3x+1\right)}\cdot\dfrac{1}{2\left(3x^2+5\right)}=\dfrac{-3x^2-5x}{2\left(3x+1\right)\left(3x^2+5\right)}\)
Bài 2:
\(P=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right)\cdot\dfrac{x\left(x+5\right)}{2x-5}+\dfrac{x}{5-x}\)
\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}+\dfrac{x}{5-x}\)
\(=\dfrac{5}{x-5}+\dfrac{x}{5-x}=-1\)
a: ĐKXĐ: x<>1; x<>-1
\(A=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x-1}\)
\(=\dfrac{x+1}{x-1}-\dfrac{1}{x-1}=\dfrac{x}{x-1}\)
b: x^2+3x+2=0
=>x=-1(loại) hoặc x=-2(nhận)
Khi x=-2 thì A=-2/(-3)=2/3
a) \(x\ne0\) , \(x\ne-1\) , \(x\ne1\)
b)
\(A=\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right).\dfrac{3x}{1-2x+x^2}\)
\(=\left(\dfrac{1}{x\left(x+1\right)}-\dfrac{2-x}{x+1}\right).\dfrac{3x}{\left(x-1\right)^2}\)
\(=\left(\dfrac{1-\left(2-x\right).x}{x\left(x+1\right)}\right).\dfrac{3x}{\left(x-1\right)^2}\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}.\dfrac{3x}{\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^2.3x}{x\left(x+1\right)\left(x-1\right)^2}\)
\(=\dfrac{3x}{x\left(x+1\right)}=\dfrac{3}{x+1}\)
Với x =5 , ta có :
\(A=\dfrac{3}{5+1}=\dfrac{3}{6}=\dfrac{1}{2}\)
Với x =0, ta có ;
\(A=\dfrac{3}{0+1}=3\)
Vậy x = 5 \(\Leftrightarrow\) \(A=\dfrac{1}{2}\)
\(x=0\Leftrightarrow A=3\)
ý a bn giải ra hộ mik đi ạ