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a: \(B=\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+3-1}{x+3}\)
\(=\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)}\cdot\dfrac{1}{x+2}\)
\(=\dfrac{3x+6}{\left(x-3\right)\left(x+2\right)}=\dfrac{3}{x-3}\)
b: |2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>2x=4 hoặc 2x=-6
=>x=-3(loại) hoặc x=2(nhận)
KHi x=2 thì \(B=\dfrac{3}{2-3}=-3\)
d: Để B<0thì x-3<0
hay x<3
a: \(B=\left(\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right):\dfrac{x+3-1}{x+3}\)
\(=\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+2}\)
\(=\dfrac{3x+6}{\left(x-3\right)}\cdot\dfrac{1}{x+2}=\dfrac{3}{x-3}\)
b: |2x+1|=5
=>2x+1=-5 hoặc 2x+1=5
=>2x=-6 hoặc 2x=4
=>x=2(nhận) hoặc x=-3(loại)
Khi x=2 thì \(B=\dfrac{3}{2-3}=\dfrac{3}{-1}=-3\)
d: Để B<0 thì x-3<0
hay x<3
a: \(B=\left(\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x+1}{x+3}\right):\left(1-\dfrac{1}{x+3}\right)\)
\(=\dfrac{21+x^2-x-12-x^2+2x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+3-1}{x+3}\)
\(=\dfrac{x+12}{x-3}\cdot\dfrac{1}{x+2}=\dfrac{\left(x+12\right)}{\left(x-3\right)\left(x+2\right)}\)
b: |2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>2x=-6 hoặc 2x=4
=>x=2
Khi x=2 thì \(B=\dfrac{2+12}{\left(2-3\right)\left(2+2\right)}=\dfrac{14}{-4}=-\dfrac{7}{2}\)
a) \(B=\left[\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x-4}{x-3}-\frac{\left(x-1\right)}{x+3}\right]:\left(\frac{x+3-1}{x+3}\right)\)
ĐK: \(\hept{\begin{cases}x\ne3\\x\ne-3\end{cases}}\)
\(=\left[\frac{21+x-4-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+2}{x+3}\right)\)
\(=\left[\frac{21+x-4-x^2+3x+x-3}{\left(x+3\right)\left(x-3\right)}\right]\times\left(\frac{x+3}{x+2}\right)\)
\(=\left(\frac{-x^2+5x+14}{x-3}\right)\left(\frac{1}{x+2}\right)\)
\(=\frac{-\left(x^2+2x-7x-14\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{-\left(x+2\right)\left(x-7\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{7-x}{x-3}\)
b) \(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Mà \(x\ne-3\)
\(\Rightarrow x=2\)
Thế \(x=2\)vào B ta được:
\(B=\frac{7-2}{2-3}=-5\)
c) \(B=\frac{7-x}{x-3}=\frac{-3}{5}\)
\(\Leftrightarrow5\left(7-x\right)=-3\left(x-3\right)\)
\(\Leftrightarrow35-5x+3x-9=0\)
\(\Leftrightarrow-2x=-26\)
\(\Leftrightarrow x=13\)
Vậy để \(B=\frac{-3}{5}\)thì \(x=13\)
d) B<0\(\Rightarrow\frac{7-x}{x-3}< 0\)
TH1: \(\hept{\begin{cases}7-x< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x>7\\x>3\end{cases}\Rightarrow}x>7}\)
TH2: \(\hept{\begin{cases}7-x>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 7\\x< 3\end{cases}\Rightarrow}x< 3}\)
Để B<0 thì x>7 hoặc x<3
a) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\) ĐKXĐ: x khác =-3; x khác -2
\(B=\frac{21+x^2-x-12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)
\(B=\frac{3x+6}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)
\(B=\frac{3\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+2}\)
\(B=\frac{3}{x-3}\)
b) bước đầu tiên ta phải tìm x:
\(\left|2x+1\right|=5\)
TH1: 2x+1=5 TH2: 2x+1=-5
2x=4 2x=-6
x=2 (nhận) x=-3 (loại)
thay x=2 vào biểu thức B, ta được:
\(B=\frac{3}{2-3}=\frac{3}{-1}=-3\)
vậy B=-3 tại x=2
c) Để \(B=-\frac{3}{5}\)thì \(\frac{3}{x-3}=-\frac{3}{5}\)
\(\Leftrightarrow-3\left(x-3\right)=15\)
\(\Leftrightarrow x-3=-5\)
\(\Leftrightarrow x=-2\)
vậy \(x=-2\)thì \(B=-\frac{3}{5}\)
d) để B<0 thì \(\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)
vậy để B<0 thì x phải < 3 và x khác -3
đề có sai 1 chút nha bn .
a) điều kiện : \(x\ne\pm3\)
ta có : \(B=\left(\dfrac{21}{x^2-9}-\dfrac{x-4}{3-x}-\dfrac{x-1}{3+x}\right):\left(1-\dfrac{1}{x+3}\right)\)
\(B=\left(\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right):\left(\dfrac{-1+x+3}{x+3}\right)\)
\(B=\left(\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{x+2}{x+3}\right)\)
\(B=\left(\dfrac{3\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\right).\left(\dfrac{x+3}{x+2}\right)=\dfrac{3}{x-3}\)
b) ta có : \(\left|2x+1\right|=2\Leftrightarrow\left[{}\begin{matrix}2x+1=2\\2x+1=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
với : \(x=\dfrac{1}{2}\Rightarrow B=\dfrac{3}{x-3}=\dfrac{3}{\dfrac{1}{2}-3}=\dfrac{-6}{5}\)
với : \(x=\dfrac{-3}{2}\Rightarrow B=\dfrac{3}{x-3}=\dfrac{3}{-\dfrac{3}{2}-3}=\dfrac{-2}{3}\)
c) ta có : \(B=\dfrac{-3}{5}\Leftrightarrow\dfrac{3}{x-3}=\dfrac{3}{-5}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)
d) ta có : \(B< 0\Leftrightarrow\dfrac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)