a) 321475 + 423507 x 2

= 321475 + 847014 = 1168489

843275 - 123568 x 5

= 843275 - 617840 = 225435

b) 1306 x 8 + 24573

= 10448 + 24573 = 35021

609 x 9 - 4845

= 5481 - 4845 = 636

a) Ta có: P(x)+Q(x)

\(=x^3+3x^2+3x-2-x^3-x^2-5x+2\)

\(=2x^2-2x\)

Ta có: P(x)-Q(x)

\(=x^3+3x^2+3x-2+x^3+x^2+5x-2\)

\(=2x^3+4x^2+8x-4\)

b) Đặt H(x)=0

\(\Leftrightarrow2x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

3 x 50 = 150

3 x 200 = 600

2 x 4000 = 8000

2 x 80 =160

4 x 300 = 1200

3 x 7000 = 21000

9 x 70 = 630

5 x 300 = 1500

6 x 2000 = 12000 

3 x 50 = 150          

3 x 200 = 600                                

2 x 4 000  = 8000         

2 x 80 = 160

4 x 300 = 1200

3 x 7 000 = 21000

9 x 70 = 630

5 x 300 = 1500

6 x 2 000 = 12000 

3 x 50 = 150

3 x 200 = 600

2 x 4 000 = 8 000

2 x 80 = 160

4 x 300 = 1200

3 x 7000 = 21 000

9 x 70 = 630

5 x 300 = 1500

6 x 2000 = 12000

a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=\left(x-y\right)\left(x-y+2\right)+37\)(1)

Thay x-y=7 vào biểu thức (1), ta được:

\(A=7\cdot\left(7+2\right)+37=7\cdot9+37=100\)

Vậy: Khi x-y=7 thì A=100

b) Ta có: \(x+y=2\)

\(\Leftrightarrow\left(x+y\right)^2=4\)

\(\Leftrightarrow x^2+y^2+2xy=4\)

\(\Leftrightarrow2xy+10=4\)

\(\Leftrightarrow2xy=-6\)

\(\Leftrightarrow xy=-3\)

Ta có: \(A=x^3+y^3\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)\)(2)

Thay x+y=2; \(x^2+y^2=10\) và xy=-3 vào biểu thức (2), ta được:

\(A=2\cdot\left(10+3\right)=2\cdot13=26\)

Vậy: Khi x+y=2 và \(x^2+y^2=10\) thì A=26

\(\Rightarrow A=x^2+2x+y^2-2y-2xy+37=x^2-2xy+y^2+2\left(x-y\right)+37=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2\cdot7+37=100\)

\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[x^2+y^2-\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}\right]=2\cdot\left[10+3\right]=2\cdot13=26\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)=-\dfrac{z}{y}\cdot\dfrac{-x}{z}\cdot-\dfrac{y}{x}=-1\)

a)P(x)+Q(x)=4x^3+x^2-x+5+2x^2+4x-1
                   =x^2+2x^2+4x^3-x+5+4x-1
                   =3x^2+4x^3-x+5+4x-1

\(a,A=\dfrac{x^2-x-2}{x^2-1}+\dfrac{1}{x-1}-\dfrac{1}{x+1}\)

\(\Rightarrow A=\dfrac{x^2-x-2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x^2-x-2x+x+1-x+1}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x^2-3x+2}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x^2-2x-x+2}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow A=\dfrac{x-2}{x+1}\)

\(b,A=\dfrac{3}{4}\\ \Rightarrow\dfrac{x-2}{x+1}=\dfrac{3}{4}\\ \Rightarrow4\left(x-2\right)=3\left(x+1\right)\\ \Rightarrow4x-8=3x+3\\ \Rightarrow4x-8-3x-3=0\\ \Rightarrow x-11=0\\ \Rightarrow x=11\)

\(c,\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

Thay x=5 vào A ta có:

\(A=\dfrac{x-2}{x+1}=\dfrac{5-2}{5+1}=\dfrac{3}{6}=\dfrac{1}{2}\)

Thay x=1 vào A ta có:

\(A=\dfrac{x-2}{x+1}=\dfrac{1-2}{1+1}=\dfrac{-1}{2}\)

a: Khi x=2 và y=-3 thì \(x^2+2y=2^2+2\cdot\left(-3\right)=4-6=-2\)

b: \(A=x^2+2xy+y^2=\left(x+y\right)^2\)

Khi x=4 và y=6 thì \(A=\left(4+6\right)^2=10^2=100\)

c: \(P=x^2-4xy+4y^2=\left(x-2y\right)^2\)

Khi x=1 và y=1/2 thì \(P=\left(1-2\cdot\dfrac{1}{2}\right)^2=\left(1-1\right)^2=0\)