Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
- \(B=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+3-1}{x+3}\)\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)
- 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,\(M=\left(\frac{4}{x-4}-\frac{4}{x+4}\right).\frac{x^2+8x+16}{32}\)
\(M=\left(\frac{4\left(x+4\right)-4\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\right).\frac{\left(x+4\right)^2}{32}\)
\(M=\frac{4x+16-4x+16}{\left(x+4\right)\left(x-4\right)}.\frac{\left(x+4\right)^2}{32}\)
\(M=\frac{32\left(x+4\right)^2}{32\left(x+4\right)\left(x-4\right)}=\frac{x+4}{x-4}\)
b,
Để M = \(\frac{1}{3}\)
\(\Rightarrow x-4=3x+12\)
\(\Rightarrow2x=16\Leftrightarrow x=8\)
\(c,\)\(\frac{x+4}{x-4}=\frac{x-4+8}{x-4}\)
\(\Rightarrow x-4\inƯ\left(8\right)=\left(1;-1;2;-2;4;-4;8;-8\right)\)
\(\Rightarrow x-4\in\left(5;3;6;2;8;0;12;-4\right)\)
Vậy để M thuộc Z thì x phải thỏa mãn các điều kiện trên .
a.
ĐKXĐ: \(x\ne\pm4\)
\(C=\left(\dfrac{4\left(x+4\right)-4\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\right)\cdot\dfrac{\left(x+4\right)^2}{32}\) có lẽ là nhân
\(\dfrac{4x+16-4x+16}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{\left(x+4\right)^2}{32}\)
\(=\dfrac{32}{\left(x+4\right)\left(x-4\right)}\cdot\dfrac{\left(x+4\right)^2}{32}=\dfrac{x+4}{x-4}\)
b.
\(C=1\Leftrightarrow x+4=x-4\Leftrightarrow0=-8\left(vo-li\right)\)
c.
\(C=\dfrac{1}{3}\Leftrightarrow3\left(x+4\right)=x-4\Leftrightarrow2x=-16\Leftrightarrow x=-8\)
d.
\(C>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4>0\\x-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+4< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>4\\x< -4\end{matrix}\right.\)
Luân Đàotran nguyen bao quanDƯƠNG PHAN KHÁNH DƯƠNG
KHUÊ VŨNguyễn Huy TúAkai HarumaAce LegonaNguyễn Thanh HằngMashiro Shiina giúp mk vs
a) \(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right)\div\left(1-\frac{x}{x+2}\right)\)
\(A=\left(\frac{x}{\left(x-2\right)\cdot\left(x+2\right)}+\frac{1}{x+2}-\frac{2}{x-2}\right)\div\left(1-\frac{x}{x+2}\right)\)
\(A=\frac{x+x-2-2\cdot\left(x+2\right)}{\left(x-2\right)\cdot\left(x+2\right)}\div\frac{x+2-x}{x+2}\)
\(A=\frac{2x-2-2x-4}{\left(x-2\right)\cdot\left(x+2\right)}\div\frac{2}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\cdot\left(x+2\right)}\cdot\frac{x+2}{2}\)
\(\Rightarrow A=\frac{-3}{x-2}\)
b) Với x = -4 . Ta có :
\(A=\frac{-3}{x-2}=\frac{-3}{-4-2}=\frac{-3}{-6}=\frac{1}{2}\)
cho tam giác ABC có 3 góc nhọn , 2 đường cao BE và CF cắt nhau tại H
a/ Chứng minh tam giác AEB ~ tam giác AFC
b/ chứng minh tam giác DEF ~ tam giác ABC
c/ Tia AH cắt BC tại D. Chứng minh FC là tia phân giác góc DFE ?
I don't now
...............
.................
oc cho