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Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
ĐKXĐ: \(x\ne2;x\ne-2\)
a) \(A=\left(\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2}\right).\left(\frac{x}{2}-1\right)\)
\(=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{2x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right).\frac{x-2}{2}\)
\(=\frac{4x}{\left(x-2\right)\left(x+2\right)}.\frac{x-2}{2}\)
\(=\frac{2x}{x+2}\)
b) \(A=1\Leftrightarrow\frac{2x}{x+2}=1\)
\(\Rightarrow2x=x+2\)
\(\Leftrightarrow x=2\left(loại\right)\)
\(\)Vậy không có giá trị nào của x để A = 1.
ĐKXĐ \(x\ne0;x\ne1;x\ne-1\)
\(A=\frac{\left(x+1+1-x\right)}{\left(1-x^2\right)-\frac{5-x}{1-x^2}}:\frac{\left(1-2x\right)}{x^2-1}\)
\(A=\frac{\left(x-3\right)}{\left(1-x^2\right)}:\frac{\left(1-2x\right)}{\left(x^2-1\right)}\)
\(A=\frac{\left(3-x\right)}{\left(x^2-1\right)}:\frac{\left(1-2x\right)}{\left(x^2-1\right)}\)
\(A=\frac{\left(3x-2\right)}{1-2x}\)
\(a,ĐKXĐ:x\ne\pm1;x\ne\frac{1}{2}\)
\(A=\left(\frac{1}{x-1}+\frac{2}{x+1}-\frac{5-x}{1-x^{^2}}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{1}{x-1}+\frac{2}{x+1}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}:\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+4}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\frac{2x+4}{1-2x}\)
\(b,Vớix\ne\pm1;x\ne\frac{1}{2}\)ta có \(A=\frac{2x+4}{1-2x}=\frac{-1\left(1-2x\right)+5}{1-2x}=-1+\frac{5}{1-2x}\)
Với x thuộc Z để A nguyên thì \(5⋮1-2x\Rightarrow1-2x\inƯ\left\{5\right\}=\left\{\pm1;\pm5\right\}\)
Với 1-2x=1 => x= 0(TMĐKXĐ)
với 1-2x=-1 => x=1(loại)
với 1-2x=5 => x=-2(tmđkxđ)
với 1-2x=-5 => x=3(tmđkxđ)
Vậy với \(x\in\left\{0;-2;-3\right\}\)thì A nguyên
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/ \(A< 0\Rightarrow\frac{-1}{x-2}< 0\Rightarrow\frac{1}{x-2}>0\Rightarrow x-2>0\Rightarrow x>2\)
\(\)
a, Rút gọn :
\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x-10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{1\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\frac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}-\frac{2x-10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{x-5+2x+10-2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\frac{x+15}{\left(x+5\right)\left(x-5\right)}\)
Đức Hiệp Tùng Giúp tôi giải toán 
FA Trả lời 3 Đánh dấu 3 phút trước (13:18)
Kb đi buồn quá
Toán lớp 1
ĐKXĐ: x\(\ne\)1, x\(\ne\)-1
MTC (x-1)(x+1)
\(\Leftrightarrow\)(\(\frac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)+ \(\frac{2\left(x-1\right)}{MTC}\)-\(\frac{-\left(5-x\right)}{MTC}\)) : \(\frac{1-2x}{MTC}\)
\(\Rightarrow\)\(\left[-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)\right]:\left(1-2x\right)\)
\(\Leftrightarrow\frac{-x-1+2x-2+5-x}{1-2x}\)
=\(\frac{-2x+2x+2}{1-2x}\)
=\(\frac{2}{1-2x}\)
b. mình chỉ biết \(x< \frac{1}{2}\) thôi chứ ko biết làm sao
hình như là giải Bất phương trình \(\frac{2}{1-2x}>0\)