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1, \(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)\(=\frac{4y.y}{11x^2.x^2}.\frac{-3x^2}{2.4y}\)\(=\frac{y}{11x^2}.\frac{-3}{2}=\frac{-3y}{22x^2}\)
2, \(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)\(=\frac{4x^2}{5y^2}.\frac{5y}{6x}.\frac{3y}{2x}\)\(=\frac{2x.2x}{5y.y}.\frac{5y}{3.2x}.\frac{3y}{2x}\)\(=\frac{2x}{y}.\frac{1}{3}.\frac{3y}{2x}\)
\(\frac{2x}{3y}.\frac{3y}{2x}=1\)
3, \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)\(=\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}\)\(=\frac{\left(x+2\right)}{3}.\frac{1}{2}=\frac{x+2}{6}\)
4, \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\left(-\frac{2\left(x-2\right)}{x+2}\right)=\frac{5}{4}.\frac{-2}{1}=-\frac{5}{2}\)
5, \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{3}{-\left(x-6\right)}=\frac{x+6}{2\left(x+5\right)}.\frac{-3}{1}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
6, \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}=\frac{\left(x-3y\right)\left(x+3y\right)}{\left(xy\right)^2}.\frac{3xy}{2\left(x-3y\right)}=\frac{x+3y}{xy}.\frac{3}{2}=\frac{3\left(x+3y\right)}{2xy}\)
7, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)
a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=24-11x\)
b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)
\(=8x^2y-6y^2-9x^2y+12y^2\)
\(=6y^2-x^2y\)
c) \(3y^2\left[\left(2x-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
\(=3y^2\cdot\left(2x-1+y+1\right)-y\cdot\left(1-y-y^2\right)+y\)
\(=6xy^2-3y^2+3y^3+3y^2-y+y^2+y^3+y\)
\(=4y^3+y^2+6xy^2\)
\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)
\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(a,3x^3y^3-15x^2y^2=3x^2y^2\left(xy-5\right)\)
\(b,5x^3y^2-25x^2y^3+40xy^4\)
\(=5xy^2\left(x^2-5xy+8y^2\right)\)
\(c,-4x^3y^2+6x^2y^2-8x^4y^3\)
\(=-2x^2y^2\left(2x-3+4x^2y\right)\)
\(d,a^3x^2y-\frac{5}{2}a^3x^4+\frac{2}{3}a^4x^2y\)
\(=a^3x^2\left(y-\frac{5}{2}x^2+\frac{2}{3}ay\right)\)
\(e,a\left(x+1\right)-b\left(x+1\right)=\left(x+1\right)\left(a-b\right)\)
\(f,2x\left(x-5y\right)+8y\left(5y-x\right)\)
\(=2x\left(x-5y\right)-8y\left(x-5y\right)=\left(x-5y\right)\left(2x-8y\right)\)
\(g,a\left(x^2+1\right)+b\left(-1-x^2\right)-c\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(a-b-c\right)\)
\(h,9\left(x-y\right)^2-27\left(y-x\right)^3\)
\(=9\left(x-y\right)^2+27\left(x-y\right)^3\)
\(=9\left(x-y\right)^2\left(1+3x-3y\right)\)
a,3x3y3−15x2y2=3x2y2(xy−5)a,3x3y3−15x2y2=3x2y2(xy−5)
b,5x3y2−25x2y3+40xy4b,5x3y2−25x2y3+40xy4
=5xy2(x2−5xy+8y2)=5xy2(x2−5xy+8y2)
c,−4x3y2+6x2y2−8x4y3c,−4x3y2+6x2y2−8x4y3
=−2x2y2(2x−3+4x2y)=−2x2y2(2x−3+4x2y)
d,a3x2y−52a3x4+23a4x2yd,a3x2y−52a3x4+23a4x2y
=a3x2(y−52x2+23ay)=a3x2(y−52x2+23ay)
e,a(x+1)−b(x+1)=(x+1)(a−b)e,a(x+1)−b(x+1)=(x+1)(a−b)
f,2x(x−5y)+8y(5y−x)f,2x(x−5y)+8y(5y−x)
=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)
g,a(x2+1)+b(−1−x2)−c(x2+1)g,a(x2+1)+b(−1−x2)−c(x2+1)
=(x2+1)(a−b−c)=(x2+1)(a−b−c)
h,9(x−y)2−27(y−x)3h,9(x−y)2−27(y−x)3
=9(x−y)2+27(x−y)3
a) \(4x^2\left(5x^3-2x+3\right)\)
\(=20x^5-8x^3+12x^2\)
b) \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)\)
\(=12y^5+2y^4-y^2\)
c) \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-14x^2+8x\)
d) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+22x-55-6x^2-23x-21\)
\(=-x-76\)
1, \(4x^2\left(5x^3-2x+3\right)=20x^5-8x^3+12x^2\)
2, \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)=12y^5+2y^4-y^2\)
3, \(\left(5x^2-4x\right)\left(x-2\right)=5x^3-10x^2-4x^2+8x=5x^3-14x^2+8x\)
4, \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+33x-10x-55-\left(6x^2+14x+9x+21\right)\)
\(=6x^2+23x-55-6x^2-23x-21=-76\)
Answer:
\(2x^3+4x^2y+2xy^2\)
\(= 2 x ( x ² + 2 x y + y ² )\)
\(= 2 x ( x + y ) ² \)
\( − 3 x ^4 y − 6 x ^3 y ^2 − 3 x ^2 y ^3 \)
\(=-3x^2y(x^2+2xy+y^2)\)
\(=-3x^2y(x+y)^2\)
\(4x^5y^2+8x^4y^3+4x^3y^4\)
\(=4x^3y^2.x^2+4x^3y^2.2xy+4x^3y^2.y^2\)
\(=4x^3y^2.(x^2+2xy+y^2)\)
\(=4x^3y^2.(x+y)^2\)
1)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)
d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)
e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)
f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)
\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)
2)
a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )
d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
Bài 1:
a)
\(A=x^2+y^2-xy-3y+2016=(x^2-xy+\frac{y^2}{4})+(\frac{3y^2}{4}-3y+3)+2013\)
\(=(x-\frac{y}{2})^2+3(\frac{y}{2}-1)^2+2013\)
\(\geq 2013\)
Vậy GTNN của $A$ là $2013$. Giá trị này đạt được khi \(\left\{\begin{matrix} x-\frac{y}{2}=0\\ \frac{y}{2}-1=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} y=2\\ x=1\end{matrix}\right.\)
b)
\(B=2x^2+5y^2+4xy-6+5x-9\)
\(=5(y^2+\frac{4}{5}xy+\frac{4}{25}x^2)+\frac{6}{5}x^2+5x-15\)
\(=5(y+\frac{2}{5}x)^2+\frac{6}{5}(x^2+\frac{25}{6}x+\frac{25^2}{12^2})-\frac{485}{24}\)
\(=5(y+\frac{2}{5}x)^2+\frac{6}{5}(x+\frac{25}{12})^2-\frac{485}{24}\geq \frac{-485}{24}\)
Vậy GTNN của $B$ là $\frac{-485}{24}$
Giá trị này đạt được khi \(\left\{\begin{matrix} y+\frac{2}{5}x=0\\ x+\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-\frac{25}{12}\\ y=\frac{5}{6}\end{matrix}\right.\)
c)
\(C=x^2+xy+y^2-3x-3y+2018\)
\(=\frac{4x^2+4xy+4y^2-12x-12y+8072}{4}=\frac{(4x^2+4xy+y^2)+3y^2-12x-12y+8072}{4}\)
\(=\frac{(2x+y)^2-6(2x+y)+3y^2-6y+8072}{4}\)
\(=\frac{(2x+y)^2-6(2x+y)+9+3(y^2-2y+1)+8060}{4}=\frac{(2x+y-3)^2+3(y-1)^2+8060}{4}\)
\(\geq \frac{8060}{4}=2015\)
Vậy $C_{\min}=2015$. Giá trị đạt được khi \(\left\{\begin{matrix} 2x+y-3=0\\ y-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)
Bài 2:
a)
\(-A=x^2+4y^2-2x+4y-5=(x^2-2x+1)+(4y^2+4y+1)-7\)
\(=(x-1)^2+(2y+1)^2-7\geq -7\)
\(\Rightarrow A\leq 7\)
Vậy GTLN của $A$ là $7$.
Giá trị này đạt được khi \(\left\{\begin{matrix} x-1=0\\ 2y+1=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=\frac{-1}{2}\end{matrix}\right.\)
b)
ĐKĐB \(\Leftrightarrow B+2x^2+10y^2-6xy-4x+3y-2=0\)
\(\Leftrightarrow 2x^2-2x(3y+2)+(10y^2+3y-2+B)=0\)
Coi đây là PT bậc 2 ẩn $x$. Vì dấu "=" tồn tại nên PT luôn có nghiệm
\(\Rightarrow \Delta'=(3y+2)^2-2(10y^2+3y-2+B)\geq 0\)
\(\Leftrightarrow B\leq \frac{-11y^2+6y+8}{2}=\frac{\frac{97}{11}-11(y-\frac{3}{11})^2}{2}\leq \frac{97}{22}\)
Vậy $B_{\max}=\frac{97}{22}$
a, \(2x^2+3\left(x+1\right)\left(x-1\right)-5x\left(x+1\right)\)
\(=2x^2+3\left(x^2-1\right)-5x^2-5x\)
\(=2x^2+3x^2-3-5x^2-5x\)
\(=\left(2x^2+3x^2-5x^2\right)-3-5x\)
\(=-\left(5x+3\right)\)
b, \(\left(4x+3y\right)\left(2x-5y\right)-\left(2x+6y\right)\left(3x-5y\right)\)
\(=8x^2-20xy+6xy-\left(15y^2-6x^2-10xy-18xy-30y^2\right)\)
\(=8x^2-20xy+6xy-15y^2+6x^2+10xy+18xy+30y^2\)
\(=\left(8x^2+6x^2\right)+\left(-20xy+6xy+10xy+18xy\right)+\left(-15y^2+30y^2\right)\)
\(=14x^2+14xy+15y^2\)
\(=14x.\left(x+y\right)+15y^2\)
Chúc bạn học tốt!!!
a) \(2x^2+3.\left(x+1\right).\left(x-1\right)-5x\left(x+1\right)\)
= \(2x^2+3.\left(x^2-1\right)-5x.\left(x+1\right)\)
= \(2x^2+3x^2-3-5x^2-5x\)
= \(-5x-3\)