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\(a,x\ne2;x\ne-2;x\ne0\)
\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(=\frac{1}{2-x}\)
\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)
Bài 2 :
a) Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)
b) \(A=\left(\frac{1}{x-2}-\frac{1}{x+2}\right)\cdot\frac{x^2-4x+4}{4}\)
\(A=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\left(\frac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\frac{4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\frac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)
\(A=\frac{x-2}{x+2}\)
c) Thay x = 4 ta có :
\(A=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}\)
Vậy.........
\(4x^2y^3.\frac{2}{4}x^3y=4x^2y^3.\frac{1}{2}x^3y=2x^5y^4\)
\(\left(5x-2\right)\left(25x^2+10x+4\right)\)
\(=\left(5x-2\right)\left[\left(5x\right)^2+5x.2+2^2\right]\)
\(=\left(5x\right)^3-2^3\)
\(=125x^3-8\)
a) \(ĐKXĐ:x\ne1\)
b) \(\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right):\left(1-\frac{2x}{x^2+1}\right)\)
\(=\left(\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right):\frac{x^2+1-2x}{x^2+1}\)
\(=\left(\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\right):\frac{\left(x-1\right)^2}{x^2+1}\)
\(=\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}.\frac{x^2+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)^2}{\left(x-1\right)^3}\)
\(=\frac{1}{x-1}\)
c) Với \(\forall x\)(\(x\ne1\)) thì biểu thức được xác định .
P/s : Theo mik câu c nên chuyển thành : Tìm x để biểu thức đạt giá trị nguyên.
Tại thấy câu c k khác j câu a !
a) Điều kiện: \(x\ne0;x\ne1\)
b) \(A=\left(\frac{x}{x-1}-\frac{1}{x^2-x}\right):\frac{x^2+2x+1}{x}\)
\(A=\left(\frac{x}{x-1}-\frac{1}{x.\left(x-1\right)}\right):\frac{\left(x+1\right)^2}{x}\)
\(A=\left(\frac{x^2}{\left(x-1\right).x}-\frac{1}{x.\left(x-1\right)}\right):\frac{\left(x+1\right)^2}{x}\)
\(A=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right).x}.\frac{x}{\left(x+1\right)^2}\)
\(A=\frac{x+1}{x}.\frac{x}{\left(x+1\right)^2}=\frac{1}{x+1}\)
c) Thay: \(x=2\)vào \(\frac{1}{x+1}\)ta có: \(A=\frac{1}{2+1}=\frac{1}{3}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
b)
\(A=\left(\frac{x}{x-1}-\frac{1}{x^2-x}\right):\frac{x^2+2x+1}{x}\)
\(A=\left(\frac{x}{x-1}-\frac{1}{x\left(x-1\right)}\right)\cdot\frac{x}{x^2+2x+1}\)
\(A=\left(\frac{x\cdot x}{x\left(x-1\right)}-\frac{1}{x\left(x-1\right)}\right)\cdot\frac{x}{\left(x+1\right)^2}\)
\(A=\frac{x^2-1}{x\left(x-1\right)}\cdot\frac{x}{\left(x+1\right)^2}=\frac{\left(x^2-1\right)\cdot x}{x\left(x-1\right)\left(x+1\right)^2}=\frac{\left(x+1\right)\left(x-1\right)\cdot x}{x\left(x-1\right)\left(x+1\right)^2}=\frac{1}{x+1}\)
c) \(A=\frac{1}{x+1}=\frac{1}{2+1}=\frac{1}{3}\)
Vậy \(A=\frac{1}{3}\)
a, ĐKXĐ : x khác -1 và 1
b, A = 2x^2+4x+2/(x-1).(x+1) . (x-1)/10
= 2.(x^2+2x+1)/10.(x+1)
= (x+1)^2/5.(x+1)
= x+1/5
k mk nha
a, ĐKXĐ: \(x\ne\pm1\)
b, \(A=\left(\frac{2x}{x-1}+\frac{4x}{x^2-1}-\frac{2}{x+1}\right)\frac{x-1}{10}\)
\(A=\left(\frac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{4x}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right)\frac{x-1}{10}\)
\(A=\frac{2x^2+2x+4x-2x+2}{\left(x-1\right)\left(x+1\right)}.\frac{x-1}{10}\)
\(A=\frac{2x^2+4x+2}{10\left(x+1\right)}\)
\(A=\frac{2\left(x+1\right)^2}{10\left(x+1\right)}\)
\(A=\frac{\left(x+1\right)}{5}\)
Bài 1:
ĐKXĐ: \(x\ne\left\{-1;1\right\}\)
\(P=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right).\frac{4\left(x^2-1\right)}{5}\)
\(P=\left(\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x+3\right)}{2\left(x-1\right)\left(x+1\right)}\right).\frac{4\left(x^2-1\right)}{5}\)
\(P=\left(\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x^2-1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(P=\frac{10.4.\left(x^2-1\right)}{2\left(x^2-1\right).5}=\frac{40}{10}=4\)
Bài 2:
ĐK: \(x\ne\left\{-2;2;\right\}\)
\(A=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)
\(A=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right).\frac{x+2}{6}\)
\(A=\left(\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)}{6}\)
\(A=\frac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}=\frac{-1}{x-2}\)
b/ \(\left|x\right|=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}A=\frac{-1}{\frac{1}{2}-2}=\frac{2}{3}\\A=\frac{-1}{-\frac{1}{2}-2}=\frac{2}{5}\end{matrix}\right.\)
c/ \(A< 0\Rightarrow\frac{-1}{x-2}< 0\Rightarrow\frac{1}{x-2}>0\Rightarrow x-2>0\Rightarrow x>2\)
\(\)
Mong sau này sẽ được cậu giúp đỡ thật nhiều :)