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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,\(M=\left(\frac{4}{x-4}-\frac{4}{x+4}\right).\frac{x^2+8x+16}{32}\)
\(M=\left(\frac{4\left(x+4\right)-4\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\right).\frac{\left(x+4\right)^2}{32}\)
\(M=\frac{4x+16-4x+16}{\left(x+4\right)\left(x-4\right)}.\frac{\left(x+4\right)^2}{32}\)
\(M=\frac{32\left(x+4\right)^2}{32\left(x+4\right)\left(x-4\right)}=\frac{x+4}{x-4}\)
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 .
`B17:`
`a)` Với `x \ne +-3` có:
`A=[x+15]/[x^2-9]+2/[x+3]`
`A=[x+15+2(x-3)]/[(x-3)(x+3)]`
`A=[x+15+2x-6]/[(x-3)(x+3)]`
`A=[3x+9]/[(x-3)(x+3)]=3/[x-3]`
`b)A=[-1]/2<=>3/[x-3]=-1/2<=>-x+3=6<=>x=-3` (ko t/m)
`=>` Ko có gtr nào của `x` t/m
`c)A in ZZ<=>3/[x-3] in ZZ`
`=>x-3 in Ư_3`
Mà `Ư_3={+-1;+-3}`
`@x-3=1=>x=4`
`@x-3=-1=>x=2`
`@x-3=3=>x=6`
`@x-3=-3=>x=0`
________________________________
`B18:`
`a)M=1/3` `ĐK: x \ne +-4`
`<=>(4/[x-4]-4/[x+4]).[x^2+8x+16]/32=1/3`
`<=>[4(x+4)-4(x-4)]/[(x-4)(x+4)].[(x+4)^2]/32=1/3`
`<=>32/[x-4].[x+4]/32=1/3`
`<=>3x+12=x-4`
`<=>x=-8` (t/m)
Ta có : Để M=\(\left(\frac{4}{x-4}-\frac{4}{x+4}\right)\left(\frac{x^2+8x+16}{32}\right)=0\)
<=> M=\(\left(\frac{4\left(x+4\right)-4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)=0\)
<=>M=\(\left(\frac{4x+16-4x+16}{\left(x+4\right)\left(x-4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)\)
<=>M=\(\left(\frac{32}{\left(x-4\right)\left(x+4\right)}\right)\left(\frac{\left(x+4\right)^2}{32}\right)\)
<=>M=\(\frac{x+4}{x-4}\)
b) Thay x=\(\frac{-3}{8}\) vào M:
M=\(\frac{x+4}{x-4}=\frac{\frac{-3}{8}+4}{\frac{-3}{8}-4}=\frac{-29}{35}\)
c)Hình như sai!
d)
a: \(A=\dfrac{4x+16-4x+16}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{\left(x+4\right)^2}{32}\)
\(=\dfrac{x+4}{x-4}\)
b: Để A=2 thì 2x-8=x+4
=>x=12