Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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.
a) Phân thức B xác định \(\Leftrightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\left\{\pm1\right\}\\x\ne-1\end{cases}\Leftrightarrow}x\ne\left\{\pm1\right\}}\)
b) \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\cdot\frac{4x^2-4}{5}\)
\(B=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{3\cdot2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(2x\right)^2-2^2}{5}\)
\(B=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(2x-2\right)\left(2x+2\right)}{5}\)
\(B=\frac{10\cdot2\left(x-1\right)\cdot2\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)\cdot5}\)
\(B=\frac{40\left(x-1\right)\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}\)
\(B=4\)
Vậy với mọi giá trị của x thì B luôn bằng 4
Vậy giá trị của B không phụ thuộc vào biến ( đpcm )
\(Giải:\)
\(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right]=\left[\frac{x+1}{2x-2}+\frac{12}{4x^2-4}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{x+1}{2x-2}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{x+3}{2x+2}\right]\)
\(=\left[\frac{\left(x+1\right)\left(2x+2\right)}{\left(2x+2\right)\left(2x-2\right)}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]\)
\(=\frac{2x^2+4x+14-2x^2+2x-6x+6}{\left(2x-2\right)\left(2x+2\right)}\)
\(=\frac{6}{\left(2x-2\right)\left(2x+2\right)}\)
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 \(x-1< 0\Leftrightarrow x< 1\)
a: ĐKXĐ: x<>-1
b: \(P=\left(1-\dfrac{x+1}{x^2-x+1}\right)\cdot\dfrac{x^2-x+1}{x+1}\)
\(=\dfrac{x^2-x+1-x-1}{x^2-x+1}\cdot\dfrac{x^2-x+1}{x+1}=\dfrac{x^2-2x}{x+1}\)
c: P=2
=>x^2-2x=2x+2
=>x^2-4x-2=0
=>\(x=2\pm\sqrt{6}\)
a.
P được xác định khi \(\left[{}\begin{matrix}2x+3=0\\2x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
vậy ĐKXĐ là: \(x\ne\pm\dfrac{3}{2}\)
b.
\(P=\dfrac{2}{2x+3}+\dfrac{3}{2x-3}-\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\\ P=\dfrac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\dfrac{3\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}-\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(P=\dfrac{2\left(2x-3\right)+3\left(2x+3\right)-6x-5}{\left(2x+3\right)\left(2x-3\right)}\\ P=\dfrac{4x-6+6x+9-6x-5}{\left(2x+3\right)\left(2x-3\right)}=\dfrac{4x-2}{\left(2x+3\right)\left(2x-3\right)}\)
c.
theo đề bài, ta có:
\(\dfrac{4x-2}{\left(2x+3\right)\left(2x-3\right)}=4\\ \Leftrightarrow4x-2=4\left(2x+3\right)\left(2x-3\right)\)
\(\Leftrightarrow4x-2=4\left(4x^2-6x+6x-9\right)\\ \Leftrightarrow2x-1=8x^2-18\)
\(\Leftrightarrow8x^2-2x-17=0\\ \Leftrightarrow x^2-\dfrac{1}{4}x=\dfrac{17}{8}\)
\(\Leftrightarrow x^2-2.\dfrac{1}{8}+\dfrac{1}{64}=\dfrac{17}{8}+\dfrac{1}{64}\\ \Leftrightarrow\left(x-\dfrac{1}{8}\right)^2=\dfrac{137}{64}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{8}=\dfrac{\sqrt{137}}{8}\\x-\dfrac{1}{8}=-\dfrac{\sqrt{137}}{8}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{137}+1}{8}\\x=\dfrac{1-\sqrt{137}}{8}\end{matrix}\right.\)
vậy P=4 khi \(x=\dfrac{\sqrt{137}+1}{8}\) và \(x=\dfrac{1-\sqrt{137}}{8}\)
\(P=\dfrac{2}{2x+3}+\dfrac{3}{2x-3}-\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)
ĐKXĐ \(2x+3\ne0\) và \(2x-3\ne0\)
Suy ra \(x\ne\dfrac{-3}{2}\) và \(x\ne\dfrac{3}{2}\)
MC: (2x+3)(2x-3)
\(\dfrac{2.\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\dfrac{3.\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}-\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(=\dfrac{4x-6}{\left(2x+3\right)\left(2x-3\right)}+\dfrac{6x+9}{\left(2x+3\right)\left(2x-3\right)}\dfrac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(=\dfrac{4x-6}{\left(2x+3\right)\left(2x-3\right)}\)
a) P xác định khi và chỉ khi \(\hept{\begin{cases}2x+3\ne0\\2x+1\ne0\end{cases}}\Rightarrow x\ne\frac{-3}{2};x\ne\frac{-1}{2}\)
b) \(P=\frac{2}{2x+3}+\frac{3}{2x+1}-\frac{6x+5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\frac{2\left(2x+1\right)+3\left(2x+3\right)-\left(6x+5\right)}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\frac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\frac{4x+6}{\left(2x+3\right)\left(2x+1\right)}=\frac{2\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}\)
\(=\frac{2}{2x+1}\)
Vậy \(P=\frac{2}{2x+1}\)
c) \(P=1\Leftrightarrow\frac{2}{2x+1}=1\Leftrightarrow2x+1=2\Leftrightarrow x=\frac{1}{2}\left(tmdkxđ\right)\)
\(P=-3\Leftrightarrow\frac{2}{2x+1}=-3\Leftrightarrow2x+1=\frac{-2}{3}\Leftrightarrow x=\frac{-5}{6}\left(tmđkđ\right)\)
Vậy \(x=\frac{1}{2}\)thì P = 1; \(x=\frac{-5}{6}\)thì P = -3
d) \(P>0\Leftrightarrow\frac{2}{2x+1}>0\Leftrightarrow2x+1>0\Leftrightarrow x>\frac{-1}{2}\)
Vậy \(x>\frac{-1}{2}\)thì P > 0
Sửa đề: \(P=\dfrac{2}{2x+3}+\dfrac{3}{2x+1}-\dfrac{6x+5}{\left(2x+3\right)\left(2x+1\right)}\)
a: ĐKXĐ: \(x\notin\left\{-\dfrac{3}{2};-\dfrac{1}{2}\right\}\)
b: \(A=\dfrac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}=\dfrac{4x+6}{\left(2x+3\right)\left(2x+1\right)}=\dfrac{2}{2x+1}\)
c: Để P=-1 thì 2x+1=-2
=>2x=-3
hay x=-3/2(loại)