Đưa các biểu thức sau thành phân thức:
a) với và
b) với
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a: \(\frac{y}{4}-2+\frac{15}{4y}\)
\(=\frac{y-8}{4}+\frac{15}{4y}=\frac{y^2-8y+15}{4y}=\frac{\left(y-5\right)\left(y-3\right)}{4y}\)
\(\frac{y}{2}+\frac{6}{y}-\frac72=\frac{y-7}{2}+\frac{6}{y}=\frac{y\left(y-7\right)+12}{2y}\)
\(=\frac{y^2-7y+12}{2y}=\frac{\left(y-3\right)\left(y-4\right)}{2y}\)
Ta có: \(M=\frac{\frac{y}{4}-2+\frac{15}{4y}}{\frac{y}{2}+\frac{6}{y}-\frac72}\)
\(=\frac{\left(y-5\right)\left(y-3\right)}{4y}:\frac{\left(y-3\right)\left(y-4\right)}{2y}\)
\(=\frac{\left(y-5\right)\left(y-3\right)}{4y}\cdot\frac{2y}{\left(y-3\right)\left(y-4\right)}=\frac{y-5}{2\cdot\left(y-4\right)}\)
b: \(N=\frac{3b-\frac{1}{9b^2}}{1+\frac{1}{3b}+\frac{1}{9b^2}}\)
\(=\frac{27b^3-1}{9b^2}:\frac{9b^2+3b+1}{9b^2}\)
\(=\frac{\left(3b-1\right)\left(9b^2+3b+1\right)}{9b^2}\cdot\frac{9b^2}{9b^2+3b+1}\)
=3b-1
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a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)
b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)
a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)
b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)
=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)
B1:
[(m+n)+(2m-3n)]^2
= (m+n)^2 + 2(m+n)(2m-3n) + (2m-3n)^2
= m^2 +2mn +n^2 + 4m^2 - 6mn + 4mn - 6n^2 + 4m^2 - 12mn + 9n^2
= 9m^2 - 12mn + 4n^2
B2,3
bn lm theo hdt ( a +b + c) ^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc nha
a) Ta có M = y 2 − 8 y + 15 4 y : y 2 − 7 y + 12 2 y = y − 5 2 ( y − 4 )
b) Ta có N = 27 b 3 − 1 9 b 2 : 9 b 2 + 3 b + 1 9 b 2 = 3 b − 1