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a, \(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
=\(\left(\frac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}\)
\(=\frac{-3}{x-2}\)
b. Thay : x=-4
=>-3/x-2=-3/(-4)-2=1/2
câu a quy đồng mẫu lên: x^2-4=(x+2)(x-2). câu b thì thay vào. câu c toán 7 tự làm
\(A=\left(\frac{2x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{5-x^2}{x+2}\right)\) ĐKXĐ : \(x\ne\pm2\)
\(A=\left(\frac{2x}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4}{x+2}+\frac{5-x^2}{x+2}\right)\)
\(A=\left(\frac{2x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+5-x^2}{x+2}\right)\)
\(A=\frac{x-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1}\)
\(A=\frac{x-6}{x-2}\)
Câu 1 :
a, \(\frac{3}{x+3}-\frac{x-6}{x^2+3x}=\frac{3x-x+6}{x\left(x+3\right)}=\frac{2x+6}{x\left(x+3\right)}=\frac{2}{x}\)
b, \(\frac{2x^2-x}{x-1}+\frac{x+1}{1-x}+\frac{2-x^2}{x-1}=\frac{2x^2-x-x-1+2-x^2}{x-1}\)
\(=\frac{x^2-2x+1}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
Bài 2 :
a, Với \(x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x+x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=\frac{-3}{x-2}\)
b, Thay x = -4 vào biểu thức trên ta được :
\(-\frac{3}{-4-2}=-\frac{3}{-6}=\frac{1}{2}\)
c, Để A \(\inℤ\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
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;
\(a^5+b^5+c^5⋮30\left(ĐCCM\right)\)
\(ĐK:\hept{\begin{cases}x+2\ne0\\x-2\ne0\end{cases}\Rightarrow x\ne\pm2}\)
a) \(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(A=\left[\frac{x}{\left(x-2\right).\left(x+2\right)}+\frac{x-2}{\left(x-2\right).\left(x+2\right)}-\frac{2x+4}{\left(x-2\right).\left(x+2\right)}\right]:\left(\frac{2}{x+2}\right)\)
\(A=\frac{x+x-2-2x-4}{\left(x-2\right).\left(x+2\right)}\cdot\frac{x+2}{2}=\frac{-6}{\left(x-2\right).\left(x+2\right)}\cdot\frac{\left(x+2\right)}{2}=\frac{-6}{2.\left(x-2\right)}=-\frac{3}{x-2}\)
b) \(A=-\frac{3}{x-2}=\frac{-3}{-4-2}=\frac{-3}{-6}=\frac{1}{2}\)
c) để A thuộc Z => 3 chia hết cho x-2 =>.....(tự làm nha bn)