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đkxd: \(x\ne\left\{\pm3\right\}\)
a) B= \(\frac{21+\left(x-4\right)\left(x+3\right)-\left(x+1\right)\left(x-3\right)}{x^2-9}:\left(\frac{x+3-1}{x+3}\right)\)
=\(\frac{21+x^2-x-12-x^2+2x+3}{x^2-9}.\frac{x+3}{x+2}\)
=\(\frac{x+12}{x-3}\)
b)|2x+1|=5
<=> \(\left[\begin{array}{nghiempt}2x+1=-5\\2x+1=5\end{array}\right.\)<=> x=-3 hoặc x=2
với x=-3 thì B=\(\frac{-3}{2}\)
với x=2 thì B=-14
\(a,ĐKXĐ:\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow x\ne\pm1}\)
\(b,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}+\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
\(=\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}:\frac{x-1-x\left(x+1\right)+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{x-1-x^2-x+2}\)
\(=\frac{4x}{1-x^2}\)
\(c,A\ge0\Leftrightarrow\frac{4x}{1-x^2}\ge0\)
\(\Leftrightarrow\hept{\begin{cases}4x\ge0\\1-x^2\ge0\end{cases}\left(h\right)\hept{\begin{cases}4x\le0\\1-x^2\le0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2\le1\end{cases}\left(h\right)\hept{\begin{cases}x\le0\\x^2\ge1\end{cases}}}\)
\(\Leftrightarrow0\le x\le1\left(h\right)x\le-1\)
Vậy ///////
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\1-\frac{1}{x+3}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm3\\x\ne-2\end{cases}}}\)
a ) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\frac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+3-1}{x+3}\)
\(=\frac{21+x^2-x-12-\left(x^2-4x+3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+2}{x+3}\)
\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)
\(=\frac{3.\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}\)
\(=\frac{3}{x-3}\)
b ) \(B=-\frac{3}{5}\Leftrightarrow\frac{3}{x-3}=-\frac{3}{5}\)
\(\Leftrightarrow x-3=-5\Leftrightarrow x=-2\) ( do \(x\ne\pm3;x\ne-2\) )
c ) \(B< 0\Leftrightarrow\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow\) \(\hept{\begin{cases}x< 3\\x\ne-2\\x\ne-3\end{cases}}\)
a) ĐKXĐ: x∉{3;-3}
Ta có: \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)
\(=\frac{21+x^2-x-12-\left(x^2-4x+3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+2}{x+3}\)
\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x+2}\)
\(=\frac{3\left(x+2\right)}{x-3}\cdot\frac{1}{x+2}=\frac{3}{x-3}\)
b) Ta có: |2x+1|=5
⇔\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Do x=-3 không thỏa mãn ĐKXĐ nên ta chỉ tính giá trị của B tại x=2
Thay x=2 vào biểu thức \(B=\frac{3}{x-3}\), ta được:
\(\frac{3}{2-3}=\frac{3}{-1}=-3\)
Vậy: -3 là giá trị của biểu thức \(B=\frac{3}{x-3}\) tại x=2
c) Ta có: \(B=\frac{-3}{5}\)
⇔\(\frac{3}{x-3}=\frac{-3}{5}\)
\(\Leftrightarrow x-3=\frac{5\cdot3}{-3}=\frac{15}{-3}=-5\)
hay x=-2(tm)
Vậy: Khi \(B=\frac{-3}{5}\) thì x=-2
d) Để B<0 thì \(\frac{3}{x-3}< 0\)
mà 3>0
nên x-3<0
hay x<3
Vậy: Khi x<3 và x≠-3 thì B<0
Ta có:
a) M = \(\left(\frac{6x}{x^2-9}-\frac{1}{x+3}+\frac{5}{3-x}\right):\frac{4}{x^2-3x}\)
M = \(\left(\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{\left(x+3\right)\left(x-3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x^2-3x}{4}\)
M = \(\left(\frac{6x-x+3-5x-15}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x\left(x-3\right)}{4}\)
M = \(\frac{-12.x\left(x-3\right)}{\left(x-3\right)\left(x+3\right).4}\)
M = \(-\frac{3x}{x+3}\)
b) Với x = 2 => M = \(-\frac{3.2}{3+2}=-\frac{6}{5}\)
Đề sai sửa luôn !
\(a,M=\left(\frac{21}{x^2-9}+\frac{4-x}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21-\left(4-x\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+3-1}{x+3}\right)\)
\(=\frac{21-4x-12+x^2+3x-x^2+3x+x-3}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)
\(=\frac{3x+6}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{3}{x-3}\)
\(b,x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Kết hợp ĐKXĐ => x = 2
Thay vào \(M=\frac{3}{2-3}=\frac{3}{-1}=-3\)
Vậy ...........................
biết đề ghê vậy :D ?!