3 1/2 x X=1/5
co ca phep tinh
Những câu hỏi liên quan
3/5 + 1/2 + 3/10 =
tinh ca phep tinh
\(\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{3}{10}=\dfrac{6}{10}+\dfrac{5}{10}+\dfrac{3}{10}=\dfrac{14}{10}=\dfrac{7}{5}\)
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\(\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{3}{10}\\ =\dfrac{6}{10}+\dfrac{5}{10}+\dfrac{3}{10}\\ =\dfrac{14}{10}\\ =\dfrac{7}{5}\)
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thuc hien phep tinh:1/(x-1)-(x^3-x)/(x^2+1).(1/(x^2-2x+1)+1/(1-x^2))
thuc hien phep tinh:
(x+2)(1+x-x^2+x^3-x^4)-(1-x)(1+x+x^2+x^3+x^4)
thuc hien phep tinh:
(x+2)(1+x-x^2+x^3-x^4)-(1-x)(1+x+x^2+x^3+x^4)
thuc hien phep tinh:3*x^2*(6*x^(n-3)+1)-2*x^n*(9*x^(n-3)-1)
thuc hien phep tinh:
(x+2)(1+x-x^2+x^3-x^4)-(1-x)(1+x+x^2+x^3+x^4)
thuc hien phep tinh :
a, (x+1)(1+x-x^2+x^3-x^4)-(x-1)(1+x+x^2+x^3+x^4)
b, (2b^2-2-5b+6b^3)(3+3b^2-b)
thuc hien phep tinh :
a, (x+1)(1+x-x^2+x^3-x^4)-(x-1)(1+x+x^2+x^3+x^4)
b, (2b^2-2-5b+6b^3)(3+3b^2-b)
thuc hien phep tinh
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{-\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+x+3x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x+3}{\left(x-1\right)^2}\)
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\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)
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