cho biểu thức \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{8}{x^2-1}\right):\left(\frac{1}{x-1}-\frac{7x+3}{1-x^2}\right),\left(x\pm1,x\pm\frac{1}{2}\right)\)
1 rút gọn biểu thức A
2 tìm giá trị của x nguyên để A nhận giá trị nguyên
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13 tháng 12 2020
a, \(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right):\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}=\frac{x+1}{x-1}\)
b, Thay x = -2 ta được :
\(\frac{x+1}{x-1}=\frac{-2+1}{-2-1}=\frac{1}{3}\)
Vậy A nhận giá trị 1/3
13 tháng 12 2020
\(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right)\div\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\div\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{x+1}{x-1}\)
Với x = -2 (tmđk) => \(A=\frac{-2+1}{-2-1}=\frac{-1}{-3}=\frac{1}{3}\)
10 tháng 1 2020
a) P = \(\left(\frac{x}{x-1}+\frac{1}{x^2-x}\right):\left(\frac{1}{x+1}+\frac{2}{x^2+1}\right)\)
=> P = \(\left(\frac{x^2}{\left(x-1\right)x}+\frac{1}{x\left(x-1\right)}\right):\left(\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{2}{\left(x+1\right)\left(x-1\right)}\right)\)
=> P = \(\left(\frac{x^2+1}{x\left(x-1\right)}\right):\left(\frac{x-1+2}{\left(x+1\right)\left(x-1\right)}\right)\)
=> P = \(\frac{x^2+1}{x\left(x-1\right)}:\frac{x+1}{\left(x+1\right)\left(x-1\right)}\)
=> P = \(\frac{x^2+1}{x\left(x-1\right)}\cdot\left(x-1\right)\)
=> P = \(\frac{x^2+1}{x}\)
b) ĐKXĐ: x \(\ne\)0; x \(\ne\)\(\pm\)1
Để P > -1
=> \(\frac{x^2+1}{x}>-1\)
=> \(\frac{x^2+1}{x}+1>0\)
=> \(\frac{x^2+1+x}{x}>0\)
Do x2 + x + 1 > 0 \(\forall\)x (vì x2 + x + 1 = x2 + x + 1/4 + 3/4 = (x + 1/2)2 + 3/4 > 0 : giải thích)
=> x > 0
Vậy để P > -1 <=> x > 0 và x \(\ne\)1
10 tháng 1 2020
a)
\(P=\left(\frac{x}{x-1}+\frac{1}{x^2-x}\right):\left(\frac{1}{x+1}+\frac{1}{x^2+1}\right)\)
\(P=\left(\frac{x}{x-1}+\frac{1}{x\left(x-1\right)}\right):\left(\frac{1}{x+1}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(P=\left(\frac{x^2}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}\right):\left(\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(P=\frac{x^2+1}{x\left(x-1\right)}:\frac{x-1}{\left(x+1\right)\left(x-1\right)}\)
\(P=\frac{x^2+1}{x\left(x-1\right)}:\frac{1}{x+1}\)
?????????????????? Đề
tự làm nốt k hiểu đề cho sai à
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{8}{x^2-1}\right):\left(\frac{1}{x-1}-\frac{7x+3}{1-x^2}\right)\)
\(A=\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x+1\right)\left(x-1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}\right]:\left[\frac{x+1}{\left(x+1\right)\left(x-1\right)}-\frac{3-7x}{\left(x+1\right)\left(x-1\right)}\right]\)
\(A=\left[\frac{x^2+2x+1-x^2+2x-1+8}{\left(x+1\right)\left(x-1\right)}\right]:\frac{x+1-3+7x}{\left(x+1\right)\left(x-1\right)}\)
\(A=\frac{4x+8}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{8x-2}\)
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tìm giá trị x nguyên để A nguyên đi