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a, \(\dfrac{3a^2b-4ab^2}{5ab}=\dfrac{ab\left(3a-4b\right)}{5ab}=\dfrac{3a-4b}{5}\)
b, \(\dfrac{3x^3y^2-5x^2y^3+4x^3y^3}{x^2y^2}=\dfrac{x^2y^2\left(3x-5y+4xy\right)}{x^2y^2}\)
\(=3x-5y+4xy\)
c, \(\dfrac{2a^5b^4+3a^4b^3}{-3a^4b^5}=\dfrac{a^4b^3\left(2ab+3\right)}{-3a^4b^5}=\dfrac{2ab+3}{-3b^2}\)
d, \(\dfrac{-a^5b^4+3a^6b^2}{4a^4b^2}=\dfrac{-a^4b^2\left(ab^2+3a^2\right)}{4a^4b^2}=\dfrac{-\left(ab^2+3a^2\right)}{4}\)
Chúc bạn học tốt!!!
a. \(\left(3a^2b-4ab^3\right):5ab=3a^2b:5ab-4ab^3:5ab=\dfrac{3}{5}a-\dfrac{4}{5}b^2\)
b. \(\left(3x^3y^2-5x^2y^3+4x^3y^3\right):x^2y^2=3x^3y^2:x^2y^2-5x^2y^3:x^2y^2+4x^3y^3:x^2y^2=3x-5y+4xy\)
c. \(\left(2a^5b^4+3a^4b^3\right):\left(-3a^4b^5\right)=2a^5b^4:\left(-3a^4b^5\right)+3a^4b^3:\left(-3a^4b^5\right)=-\dfrac{2a}{3b}-\dfrac{1}{b^2}\)
d. \(\left(-a^5b^4+3a^6b^2\right):4a^4b^2=\left(-a^5b^4\right):4a^4b^2+3a^6b^2:4a^4b^2=-\dfrac{1ab^2}{4}+\dfrac{3a^2}{4}\)
Bài 1:
\(A=a^4-2a^3+3a^2-4a+5\)
\(=(a^4-2a^3+a^2)+2a^2-4a+5\)
\(=(a^4-2a^3+a^2)+2(a^2-2a+1)+3\)
\(=(a^2-a)^2+2(a-1)^2+3\)
\(=a^2(a-1)^2+2(a-1)^2+3=(a-1)^2(a^2+2)+3\)
Vì \((a-1)^2\geq 0,\forall a\in\mathbb{R}; a^2+2>0, \forall a\)
\(\Rightarrow A=(a-1)^2(a^2+2)+3\geq 0+3=3\)
Vậy \(A_{\min}=3\Leftrightarrow (a-1)^2=0\Leftrightarrow a=1\)
Bài 2:
a)
\(M=3xyz+x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)\)
\(=3xyz+x^2y+x^2z+yx^2+yz^2+zx^2+zy^2\)
\(=(x^2y+xy^2+xyz)+(y^2z+yz^2+xyz)+(zx^2+z^2x+xyz)\)
\(=xy(x+y+z)+yz(y+z+x)+xz(z+x+y)\)
\(=(x+y+z)(xy+yz+xz)\)
b)
\(Q=(a+b+c)^3-a^3-b^3-c^3\)
\(=[(a+b)+c]^3-a^3-b^3-c^3\)
\(=(a+b)^3+c^3+3(a+b)^2c+3(a+b)c^2-a^3-b^3-c^3\)
\(=a^3+b^3+3ab^2+3a^2b+c^3+3(a+b)^2c+3(a+b)c^2-a^3-b^3-c^3\)
\(=3ab(a+b)+3(a+b)c(a+b+c)\)
\(=3(a+b)[ab+c(a+b+c)]=3(a+b)[a(b+c)+c(b+c)]\)
\(=3(a+b)(b+c)(a+c)\)
a) (a2 - 1)3- ( a4 + a2+1)(a2-1)
= (a2 - 1)3 - (a2 - 1)3 =0
b) (a4 - 3a2+ 9)(a2+3) - (3+a2)3
= (3+a2)3 - (3+a2)3
=0
a) Ta có: \(\dfrac{m^2+2m+1}{m^2-1}\)
\(=\dfrac{\left(m+1\right)^2}{\left(m+1\right)\left(m-1\right)}\)
\(=\dfrac{m+1}{m-1}\)
b) Ta có: \(\dfrac{2a^4+3a^3+2a+3}{\left(a^2-a+1\right)\left(4a+6\right)}\)
\(=\dfrac{a^3\left(2a+3\right)+\left(2a+3\right)}{\left(a^2-a+1\right)\left(4a+6\right)}\)
\(=\dfrac{\left(2a+1\right)\left(a+1\right)\left(a^2-a+1\right)}{2\left(a^2-a+1\right)\left(2a+3\right)}\)
\(=\dfrac{a+1}{2}\)