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a)Có A=\(\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)(ĐKXĐ \(x\ne2,-2,-1\))
=\(\left(\frac{2-x}{\left(2-x\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}-\frac{x}{\left(2-x\right)\left(2+x\right)}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
=\(\frac{2-x+2x+4-x}{\left(2-x\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
=\(\frac{6\left(2-x\right)\left(x+1\right)}{6\left(2-x\right)\left(x+2\right)^2}\)
=\(\frac{x+1}{\left(x+2\right)^2}\)
b)Có A=\(\frac{x+1}{\left(x+2\right)^2}\)
Để A>0 <=> x+1>0 <=>x>-1
c) Có x2+3x+2=0
<=> x2+2x+x+2=0
<=> x(x+2)+(x+2)=0
<=>(x+1)(x+2)=0
<=> x=-1 hoặc x=-2
\(ĐKXĐ:x\ne0;x\ne\pm2\)
a) \(M=\left[\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right]:\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(\Leftrightarrow M=\left[\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(\Leftrightarrow M=\frac{3x^2-6x\left(x+2\right)+3x\left(x-2\right)}{3x\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)
\(\Leftrightarrow M=\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(\Leftrightarrow M=\frac{-18x\left(x+2\right)}{18x\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow M=-\frac{1}{x-2}\)
\(\Leftrightarrow M=\frac{1}{2-x}\)
b) Để M đạt giá trị lớn nhất
\(\Leftrightarrow2-x\)đạt giá trị nhỏ nhất
\(\Leftrightarrow x\)đạt giá trị lớn nhất
Vậy để M đạt giá trị lớn nhất thì x phải đạt giá trị lớn nhất \(\left(x\inℤ\right)\)
玉明, bạn làm sai rồi. Dấu ngoặc vuông là dấu phần nguyên không phải dấu ngoặc thường
\(a,x\ne2;x\ne-2;x\ne0\)
\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(=\frac{1}{2-x}\)
\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\)
\(A=\frac{3}{x+4}-\frac{x\left(x-1\right)}{x+4}\times\frac{2x-5}{x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)
\(=\frac{3\left(x+4\right)}{\left(x+4\right)^2}-\frac{x\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)
\(=\frac{3x+12}{\left(x+4\right)^2}-\frac{\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{17}{\left(x+4\right)^2}\)
\(=\frac{\left(3x+12\right)\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{2x^2-7x+5}{\left(x+4\right)^2\left(x-2\right)}-\frac{17\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}\)
\(=\frac{3x^2+6x-24-2x^2+7x-5-17x+34}{\left(x+4\right)^2\left(x-2\right)}\)
\(=\frac{x^2-4x+5}{\left(x+4\right)^2\left(x-2\right)}=\frac{x^2-4x+5}{x^3+6x^2-32}\)
b) \(18A=1\)
<=> \(18\times\frac{x^2-4x+5}{x^3+6x^2-32}=1\)( ĐK : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\))
<=> \(\frac{x^2-4x+5}{x^3+6x^2-32}=\frac{1}{18}\)
<=> 18( x2 - 4x + 5 ) = x3 + 6x2 - 32
<=> 18x2 - 72x + 90 = x3 + 6x2 - 32
<=> x3 + 6x2 - 32 - 18x2 + 72x - 90 = 0
<=> x3 - 12x2 + 72x - 122 = 0
Rồi đến đây chịu á :)
ĐKXĐ: \(x\ne0;x\ne\pm2\)
a, \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left[\frac{3x^2}{3x\left(x-2\right)\left(x+2\right)}-\frac{6x\left(x+2\right)}{3x\left(x-2\right)\left(x+2\right)}+\frac{3x\left(x-2\right)}{3x\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(=\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{-18x}{3x\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}\)
\(=\frac{-3x}{3x\left(x-2\right)}=\frac{-1}{x-2}\)
b, Ta có: \(\left|x\right|=\frac{1}{2}\Rightarrow x=\pm\frac{1}{2}\)
Với \(x=\frac{1}{2}\) thì \(A=\frac{-1}{\frac{1}{2}-2}=\frac{-1}{\frac{-3}{2}}=\frac{2}{3}\)
Với \(x=\frac{-1}{2}\)thì \(A=\frac{-1}{\frac{-1}{2}-2}=\frac{-1}{\frac{-5}{2}}=\frac{2}{5}\)
c, Để A=2 <=> \(\frac{-1}{x-2}=2\Leftrightarrow-1=2x-4\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Vậy x=3/2 thì A=2
d, Để A<0 <=> \(\frac{-1}{x-2}< 0\Leftrightarrow x-2>0\Leftrightarrow x>2\)
Vậy với x>2 thì A<0
e, Để A thuộc Z <=> x-2 thuộc Ư(-1)={1;-1}
Ta có: x-2=1 => x=3 (t/m)
x-2=-1 => x=1 (t/m)
Vậy x thuộc {3;1} thì A thuộc Z
a) \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)(ĐKXĐ: x khác 0; + 2)
\(A=\left(\frac{x^2}{x\left(x^2-4\right)}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right)\)
\(A=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\right):\frac{6}{x+2}\)
\(A=\frac{-6x}{x\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}=\frac{-x}{x\left(x-2\right)}=\frac{1}{2-x}.\)
Vậy \(A=\frac{1}{2-x}.\)
b) \(\left|x\right|=\frac{1}{2}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\). Nếu \(x=\frac{1}{2}\)thì \(A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}.\)
Nếu \(x=-\frac{1}{2}\)thì \(A=\frac{1}{2+\frac{1}{2}}=\frac{2}{5}.\)Vậy ...
c) Để A=2 thì \(\frac{1}{2-x}=2\Rightarrow4-2x=1\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}.\)Vậy ...
d) Để A<0 thì \(\frac{1}{2-x}< 0\Rightarrow2-x< 0\Leftrightarrow x>2.\)Vậy ...
e) Để A thuộc Z thì \(\frac{1}{2-x}\in Z\Rightarrow1⋮2-x\). Mà 2-x thuộc Z (Do x thuộc Z)
Nên \(2-x\in\left\{1;-1\right\}\Rightarrow x\in\left\{1;3\right\}.\)(t/m ĐKXĐ)
Vậy x=1 hay x=3 thì A nguyên.
a) \(-ĐKXĐ:x\ne\pm2;1\)
Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)
b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)
\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)
Vậy với mọi x thỏa mãn x>1 thì A > 0
c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
Vậy x = -1;-2