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1) a)
=\(\left(4-1+8\right)x^2=11x^2\)
b) =\(\left(\dfrac{1}{2}-\dfrac{3}{4}+1\right)x^2y^2=\dfrac{3}{4}x^2y^2\)
c) =(3-7+4-6)y=5y 2) a) ...=\(\left[\left(\dfrac{-2}{3}y^3\right)-\dfrac{1}{2}y^3\right]+3y^2-y^2\\ =\left[\left(\dfrac{-2}{3}-\dfrac{1}{2}\right)y^3\right]+\left(3-1\right)y^2=\dfrac{-7}{6}y^3+2y^2\) b) ...=\(\left(5x^3-x^3\right)-\left(3x^2+4x^2\right)+\left(x-x\right)=4x^3-7x^2\) 3) a)A=\(\left(5.\dfrac{1}{2}\right).\left(x.x^2.x\right)\left(y^2.y^2\right)=\dfrac{5}{2}x^4y^4\) b)Vậy Đơn thức A có bậc 8; hệ số là \(\dfrac{5}{2}\); phần biến là \(x^4y^4\) c)Khi x=1;y=-1 thì A=\(\dfrac{5}{2}.1^4.\left(-1\right)^4=\dfrac{5}{2}\)
Lời giải:
$\frac{x}{y}=\frac{2}{3}\Rightarrow \frac{x}{2}=\frac{y}{3}$. Đặt $\frac{x}{2}=\frac{y}{3}=k$ thì:
$x=2k; y=3k$
Khi đó: $3x-2y=3.2k-3.2k=0$. Mẫu số không thể bằng $0$ nên $A$ không xác định. Bạn xem lại.
$B=\frac{2(2k)^2-2k.3k+3(3k)^2}{3(2k)^2+2.2k.3k+(3k)^2}=\frac{29k^2}{33k^2}=\frac{29}{33}$
\(a,Đặt\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\\ A=\dfrac{2x-3y}{x-5y}=\dfrac{2\cdot2k-3\cdot3k}{2k-5\cdot3k}\\ =\dfrac{4k-9k}{2k-15k} \\ =\dfrac{5k}{13k}\\ =\dfrac{5}{13}\)
\(b,Thayx-y=7vàoB,tacó:\\ B=\dfrac{2x+7}{3x-y}+\dfrac{2y-7}{3y-x}\\ =\dfrac{2x+x-y}{3x-y}+\dfrac{2y-x+y}{3y-x}\\ =\dfrac{3x-y}{3x-y}+\dfrac{3y-x}{3y-x}\\ =1+1\\ =2\)
\(c,Đặt\dfrac{x}{3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\\ C=\dfrac{5x^2+3y^2}{10x^2-3y^2}\\ =\dfrac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\\ =\dfrac{45k^2+75k^2}{90k^2-75k^2}\\ =\dfrac{120k^2}{15k^2}\\ =8\)
\(d,\dfrac{a}{b}=\dfrac{5}{7}\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}=k\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=7k\end{matrix}\right.\\ D=\dfrac{5a-b}{3a-2b}\\ =\dfrac{5\cdot5k-7k}{3\cdot5k-2\cdot7k}\\ =\dfrac{25k-7k}{15k-14k}\\ =\dfrac{18k}{k}=18\)
\(e,Thayx-y=5vàoE,tacó:\\ E=\dfrac{3x-5}{2x+y}-\dfrac{4y+5}{x+3y}\\ =\dfrac{3x-x+y}{2x+y}-\dfrac{4y+x-y}{x+3y}\\ =\dfrac{2x+y}{2x+y}-\dfrac{3y+x}{x+3y}\\ =1-1=0\)
Thay \(x=\dfrac{1}{2};y=-1\) vào B, ta được:
\(B=\left[\left(\dfrac{1}{2}\right)^3-4\cdot\left(\dfrac{1}{2}\right)^2\cdot\left(-1\right)+3\cdot\left(-1\right)^2-4\right]:\left[3\cdot\left(\dfrac{1}{2}\right)^3-3\cdot\left(-1\right)^2-3\cdot\left(-1\right)\right]\)
\(=\left(\dfrac{1}{8}+4\cdot\dfrac{1}{4}+3\cdot1-4\right):\left(3\cdot\dfrac{1}{8}-3\cdot1+3\right)\)
\(=\left(\dfrac{1}{8}+1+3-4\right):\left(\dfrac{3}{8}-3+3\right)\)
\(=\dfrac{1}{8}\cdot\dfrac{8}{3}=\dfrac{1}{3}\)