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Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
\(Q=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
\(\Leftrightarrow\) \(Q=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(2-x\right)}+\frac{5}{\left(x+3\right)\left(2-x\right)}+\frac{-1}{\left(x+3\right)\left(2-x\right)}\)
\(\Rightarrow\) \(Q=\left(x-2\right)\left(x+2\right)+5-1\)
\(\Leftrightarrow\) \(Q=x^2-4+5-1\)
\(\Leftrightarrow\) \(Q=x^2\)
Thay \(Q=\frac{-3}{4}\) ta được:
\(x^2=\frac{-3}{4}\)
Vì \(\frac{-3}{4}>0\forall x\)
\(\Rightarrow\) Pt vô nghiệm
Vậy không có giả trị nào của x thỏa mãn \(Q=\frac{-3}{4}\)
Chúc bn học tốt!!
dkxd \(\hept{\begin{cases}\\\end{cases}}x-2=0;x+2=0\Leftrightarrow\hept{\begin{cases}\\\end{cases}x=+2;x=-2}\)
b/ \(\frac{x^2}{x^2-4}-\frac{x}{x+2}-\frac{2}{x-2}=\frac{x^2}{\left(x-2\right).\left(x+2\right)}-\frac{x.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}-\frac{2.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}\)
\(\frac{x^2-x^2-2x-2x+4}{\left(x-2\right).\left(x+2\right)}=\frac{4}{\left(x-2\right)\left(x+2\right)}\)
tới khúc này bí rồi ^^
a,ĐKXĐ của A là:\(x\ne+2;-2\)
b,\(\frac{x^2-x^2+2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4}{\left(x+2\right)\left(x-2\right)}\)
c,Để A\(\in\)Z=> (x+2)(x-2)\(\inƯ\)(4) hay \(x^2-4\inƯ\)(4)=\(\left(4;-4;2;-2;1;-1\right)\)
Ta có bảng
| \(x^2-4\) | x |
| 4 | \(\sqrt{8}\) |
| -4 | 0 |
| 2 | \(\sqrt{6}\) |
| -2 | \(\sqrt{2}\) |
| 1 | \(\sqrt{5}\) |
Vậy A\(Z=>x\in\)( 0;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))
a) \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)(với \(x\ne\pm2;x\ne-1\))
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{-\left(6-5x\right)}{x^2-4}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{x+2}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{1}{x-2}:\frac{x+1}{x-2}=\frac{1}{x-2}\cdot\frac{x-2}{x+1}=\frac{1}{x+1}\)
b) Với \(M=\frac{1}{4}\)ta có :
\(M=\frac{1}{x+1}\Rightarrow\frac{1}{4}=\frac{1}{x+1}\)
\(\Rightarrow1\left(x+1\right)=4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
a, \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}\)
\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}=\frac{1}{x-2}.\frac{x-2}{x+1}=\frac{1}{x+1}\)
b, Ta có : M = 1/4 hay \(\frac{1}{x+1}=\frac{1}{4}\Leftrightarrow4=x+1\Leftrightarrow x=3\)
Công thức tổng quát:
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Do đó:
\(A=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x-4}+\frac{1}{\left(x-1\right)\left(x+10\right)}\)
Bạn tự làm tiếp nhé.
a) \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
\(A=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{4x^2-8x+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)
\(a,A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
\(=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)
\(=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x^2-8x+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)