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\(B=\left(\frac{21}{x^2-9}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+2}{x+3}\)
\(B=\frac{2x^2-5x+12}{x^2-9}\cdot\frac{x+3}{x+2}\)
\(B=\frac{2x^2-5x-12}{\left(x-3\right)\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{x^2-x-6}\)
Thik thì tách tiếp nha
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
- \(B=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+3-1}{x+3}\)\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)
- Điều kiện \(x\ne3\) \(\Rightarrow\frac{-3}{5}=\frac{3}{x-3}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)
- \(B=\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)
a) B=(\(\frac{21}{x^2-9}\)-\(\frac{x-4}{3-x}\)-\(\frac{x-1}{3+x}\)) : (1-\(\frac{1}{x+3}\)) (ĐK: x khác +-3)
=(\(\frac{21}{\left(x-3\right).\left(x+3\right)}\)+\(\frac{x-4}{x-3}\)-\(\frac{x-1}{x+3}\)) : (1-\(\frac{1}{x+3}\))
=(\(\frac{21+\left(x+4\right).\left(x+3\right)-\left(x-1\right).\left(x-3\right)}{\left(x-3\right).\left(x+3\right)}\):(\(\frac{x+3-1}{x+3}\))
=(\(\frac{3x+6}{\left(x-3\right).\left(x+3\right)}\)) . (\(\frac{x+3}{x+2}\))
=(\(\frac{3.\left(x+2\right)}{\left(x-3\right).\left(x+3\right)}\). \(\frac{x+3}{x+2}\)
=\(\frac{3}{x-3}\)
b) B=\(\frac{3}{x-3}\)=\(\frac{-3}{5}\)
(=) \(\frac{3.5}{x-3}\)=-3
(=) -3.(x-3) = 15
(=) -3x=6
(=) x=-2
vậy x=2 thì B=\(\frac{-3}{5}\)
c) B=\(\frac{3}{x-3}\)<0
(=) 3 < x - 3
(=) -x < - 3 - 3
(=) x > 6
Vậy với x > 6 thì B < 0
a) \(\left(x-1\right)^2-\left(x-2\right)\left(x+2\right)=x^2-2x+1-x^2+4=5-2x\)
mình nghĩ là câu b bạn ghi đề sai vì như thế không có hằng đẳng thức nhé
b)\(\left(x^2+\frac{1}{3}x+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3=x^3-\frac{1}{27}-x^3+\frac{1}{27}+x^2-\frac{1}{3}x=x^2-\frac{1}{3}x\)
b,\(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)
\(=\)\(\left(x-\frac{1}{3}\right)\left[\left(x^2+\frac{1}{x}+\frac{1}{9}\right)-\left(x-\frac{1}{3}\right)^2\right]\)
\(=\)\(\left(x-\frac{1}{3}\right)\left(x^2+\frac{1}{x}+\frac{1}{9}-x^2+\frac{2}{3}x-\frac{1}{9}\right)\)
\(=\left(x-\frac{1}{3}\right)\left(\frac{1}{x}+\frac{2}{3}x\right)\) \(=1+\frac{2}{3}x^2-\frac{1}{3x}-\frac{2}{9}x\)
\(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}+\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)
\(B=\left(\frac{21}{x^2-9}+\frac{x-4}{x-3}+\frac{x-1}{x+3}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)
\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right)\div\frac{x+2}{x+3}\)
\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-x-12}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)
\(B=\left(\frac{21+x^2-x-12+x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)
\(B=\frac{2x^2-5x+12}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{\left(x-3\right)\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{x^2-x-6}\)
Đến đây là chịu ạ :(