rút gọn phân thức : \(\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
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Những câu hỏi liên quan
30 tháng 12 2021
a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
\(C=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
12 tháng 11 2021
a: \(P=\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
3 tháng 2 2022
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.
KT
1
22 tháng 2 2021
1) Ta có: \(\dfrac{x\left|x-2\right|}{x^2-5x+6}\)
\(=\left[{}\begin{matrix}\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x< 2\right)\\\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x>2\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}\dfrac{-x}{x-3}\\\dfrac{x}{x-3}\end{matrix}\right.\)
2) Ta có: \(\dfrac{a^{2x}-b^{2x}}{a^x-b^x}\)
\(=\dfrac{\left(a^x\right)^2-\left(b^x\right)^2}{a^x-b^x}\)
\(=\dfrac{\left(a^x-b^x\right)\left(a^x+b^x\right)}{a^x-b^x}=a^x+b^x\)
25 tháng 2 2023
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
5 tháng 10 2021
\(A=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5x\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\left(x>0;x\ne4\right)\\ A=\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5x\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\\ A=\dfrac{10x\left(\sqrt{x}-2\right)}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}\)
5 tháng 10 2021
\(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}-4}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
6 tháng 10 2021
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
3 tháng 1 2019
a) P xác định \(\Leftrightarrow\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Leftrightarrow x\ne\left\{-5;0\right\}}\)
b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+5x^2-x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+5\right)-x\left(x+5\right)}{2x\left(x+5\right)}\)
\(P=\frac{\left(x+5\right)\left(x^2-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x\left(x-1\right)}{2x}\)
\(P=\frac{x-1}{2}\)
c) Để P = 0 thì \(x-1=0\Leftrightarrow x=1\)( thỏa mãn ĐKXĐ )
Để P = 1/4 thì \(\frac{x-1}{2}=\frac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)( thỏa mãn ĐKXĐ )
d) Để P > 0 thì \(\frac{x-1}{2}>0\)
Mà 2 > 0, do đó để P > 0 thì \(x-1>0\Leftrightarrow x>1\)
Để P < 0 thì \(\frac{x-1}{2}< 0\)
Mà 2 > 0, do đó để P < 0 thì \(x-1< 0\Leftrightarrow x< 1\)
\(\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+5x-x-5\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x\left[x\left(x+5\right)-\left(x+5\right)\right]}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x-1}{2}\)
= \(\dfrac{x^{3^{ }}+2x^2+2x^2-10x+10x-50+50-5x}{2x\left(x+5\right)}\)
= \(\dfrac{x^3+4x^2}{2x\left(x+5\right)}\) = \(\dfrac{x^2+4x}{2x+10}\)