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a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)
a: ĐKXĐ: x<>2; x<>-2
b: \(A=\dfrac{3x\left(x-2\right)+2x+6}{2\left(x-2\right)\left(x+2\right)}=\dfrac{3x^2-6x+2x+6}{2\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3x^2+4x+6}{2\left(x-2\right)\left(x+2\right)}\)
c: Khi x=-3 thì \(A=\dfrac{3\cdot\left(-3\right)^2-4\cdot3+6}{2\left(-3-2\right)\left(-3+2\right)}=\dfrac{21}{10}\)
Đề bài sai rồi bạn ! Mình sửa :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)
b) \(P=\left(\frac{x-1}{x+1}-\frac{x+1}{x-1}\right):\frac{2x}{3x-3}\)
\(\Leftrightarrow P=\frac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{x^2-2x+1-x^2-2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-6}{x+1}\)
c) Để P nhận giá trị nguyên
\(\Leftrightarrow\frac{-6}{x+1}\inℤ\)
\(\Leftrightarrow x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;-3;1;-4;2;-7;5\right\}\)
Ta loại các giá trị ktm
\(\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)
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
Vậy A\(Z=>x\in\)( 0;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))