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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.
câu 1
a)\(ĐKXĐ:x^3-8\ne0=>x\ne2\)
b)\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2-2x+4\right)}=\frac{3}{x-2}\left(#\right)\)
Thay \(x=\frac{4001}{2000}\)zô \(\left(#\right)\)ta được
\(\frac{3}{\frac{4001}{2000}-2}=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}=\frac{3}{\frac{1}{2000}}=6000\)
B1:dài quá :vv
B2:\(Q=\frac{x^2}{x^4+x^2+1}=\frac{x^2}{x^4+2x^2+1-x^2}=\frac{x^2}{\left(x^2+1\right)-x^2}=\frac{x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
\(=\frac{x}{x^2-x+1}.\frac{x}{x^2+x+1}=\frac{2}{3}.\frac{x}{x^2+x+1}\)
\(\frac{x}{x^2-x+1}=\frac{2}{3}\Rightarrow\frac{x^2-x+1}{x}=\frac{3}{2}\Rightarrow\frac{x^2-x+1}{x}+2=\frac{3}{2}+2\Rightarrow\frac{x^2+x+1}{x}=\frac{7}{2}\)
\(\Rightarrow\frac{x}{x^2+x+1}=\frac{2}{7}\Rightarrow Q=\frac{2}{3}.\frac{2}{7}=\frac{4}{21}\)
3.
Ta có: \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)=a\left(a+1\right)\left(a-1\right)\left(a^2+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)+5a\left(a-1\right)\left(a+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)\)
Do a(a-1)(a+1)(a-2)(a+2) là tích của 5 số hạng liên tiếp nên chia hết cho 2,3 và 5
Lại có a(a-1)(a+1) là tích của 3 số hạng liên tiếp nên chia hết cho 2,3 suy ra 5a(a-1)(a+1) chia hết cho 2,3,5
Từ đó:a(a-1)(a+1)(a-1)(a+2)+5a(a-1)(a+1) chia hết cho 2,3,5 hay a(a-1)(a+1)(a-2)(a+2)+5a(a-1)(a+1) chia hết cho 30 \(\Leftrightarrow a^5-a\) chia hết cho 30
Tương tự ta có\(b^5-b\) chia hết cho 30, \(c^5-c\) chia hết cho 30
Do đó:\(a^5-a+b^5-b+c^5-c⋮30\)
\(\Leftrightarrow a^5+b^5+c^5-\left(a+b+c\right)⋮30\)
Mà a+b+c=0 nên;
\(a^5+b^5+c^5⋮30\left(ĐCCM\right)\)
a, \(A=\frac{\left(x+2\right)^2}{x}\left(1-\frac{x^2}{x+2}\right)=\frac{\left(x+2\right)^2}{x}\left(\frac{x+2-x^2}{x+2}\right)\)
\(=\frac{-\left(x+2\right)^2\left(x-2\right)\left(x+1\right)}{x\left(x+2\right)}=\frac{-\left(x\pm2\right)\left(x+1\right)}{x}\)
c, Theo bài ra ta có : \(C=\frac{A}{B}\)hay \(\frac{\frac{-\left(x\pm2\right)\left(x+1\right)}{x}}{\frac{4}{\left(x-2\right)^2}}=\frac{\frac{-\left(x+2\right)\left(x+1\right)}{x}}{\frac{4}{x-2}}\)
d, Theo bài ra ta có :
\(C>0\)hay \(\frac{\frac{-\left(x+2\right)\left(x+1\right)}{x}}{\frac{4}{x-2}}>0\)
\(\Leftrightarrow\frac{-\left(x+2\right)\left(x+1\right)}{x}.\frac{x-2}{4}>0\)
\(\Leftrightarrow-\left(x+2\right)\left(x+1\right)>0\Leftrightarrow\left(x+2\right)\left(x+1\right)>0\)
\(\Leftrightarrow x>-2;x>-1\Rightarrow x>-1\)
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
a, ĐKXĐ: \(x\ne\pm3\)
\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)
b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)
c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)
\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)
a, ĐỂ A xác định :
\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)
\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)
\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)
\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)
\(A=\frac{x-4}{x+3}\)
b
\(P=\frac{x-4}{x^2+x-12}=\frac{x-4}{x^2+4x-3x-12}=\frac{x-4}{\left(x-3\right)\left(x+4\right)}\)
ĐKXĐ: \(\left(x-3\right)\left(x+4\right)\ne0\) => \(\hept{\begin{cases}x-3\ne0\\x+4\ne0\end{cases}}\)=> \(\hept{\begin{cases}x\ne3\\x\ne-4\end{cases}}\)
a, x2 - 3x = 0
=> x(x - 3) = 0
=> \(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Mà \(\hept{\begin{cases}x\ne3\\x\ne-4\end{cases}}\)
=> x = 0
=> \(P=\frac{0-4}{\left(0-3\right)\left(0+4\right)}=\frac{-4}{\left(-3\right).4}=\frac{1}{3}\)
b, Với \(\hept{\begin{cases}x\ne3\\x\ne-4\end{cases}}\)
\(P.\left(x+4\right)=\frac{2}{3}\)
=> \(\frac{x-4}{\left(x-3\right)\left(x+4\right)}.\left(x+4\right)=\frac{2}{3}\)
=> \(\frac{x-4}{x-3}=\frac{2}{3}\)
=> \(2\left(x-3\right)=3\left(x-4\right)\)
=> 2x - 6 = 3x - 12
=> -x = -6
=> x = 6 (TM ĐKXĐ)
c, Với \(\hept{\begin{cases}x\ne3\\x\ne-4\end{cases}}\)
\(P\left(x-3\right)\)có giá trị nguyên
=> \(\frac{x-4}{\left(x-3\right)\left(x+4\right)}.\left(x-3\right)\)nguyên
=> \(\frac{x-4}{x+4}\)nguyên
=> x - 4 chia hết cho x + 4
<=> x + 4 - 8 chia hết cho x + 4
Có x + 4 chia hết cho x + 4
=> 8 chia hết cho x + 4
=> x + 4 thuộc Ư(8)
=> x + 4 thuộc {1; -1; 2; -2; 4; -4; 8; -8}
=> x thuộc {-3; -5; -2; -6; 0; -8; 4; -12}