Bài tập cuối chương 1

Thay \(x=1;y=-1;z=2\) vào P

\(P=1.\left(-1\right)^2.2-2.1^2.\left(-1\right).2^2+3.\left(-1\right).2+1\\ =5\)

\(a,Q=\left(-2x^3y+7x^2y+3xy\right)+P=\left(-2x^3y+7x^2y+3xy\right)+\left(3x^2y-2xy^2-4xy+2\right)\\ =-2x^3y+7x^2y+3xy+3x^2y-3xy^2-4xy+2\\ =-2x^3y^2+10x^2y-3xy^2-xy+2\)

\(b,M=\left(3x^2y^2-5x^2y+8xy\right)-P\\ =\left(3x^2y^2-5x^2y+8xy\right)-\left(3x^2y-2xy^2-4xy+2\right)\\ =3x^2y^2-5x^2y+8xy-3x^2y^2+2xy^2+4xy-2\\ =-3x^2y+12xy-2\)

a) \(x^2y\left(5xy-2x^2y-y^2\right)\)

\(=5x^3y^2-2x^4y^2-x^2y^3\)

b) \(\left(x-2y\right)\left(2x^3+4xy\right)\)

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

a) \(18x^4y^3:12\left(-x\right)^3y\)

\(=\left(18:-12\right)\left(x^4:x^3\right)\left(y^3:y\right)\)

\(=-\dfrac{3}{2}xy^2\)

b) \(x^2y^2-2xy^3:\dfrac{1}{2}xy^2\)

\(=\dfrac{xy^2\left(x-2y\right)}{\dfrac{1}{2}xy^2}\)

\(=\dfrac{x-2y}{\dfrac{1}{2}}\)

\(=2x-4y\)

\(a,=\left(4x^2-25\right)-\left(6x^2+9x-4x-6\right)\\ =4x^2-25-6x^2-5x+6=-2x^2-5x-19\\ b,=4x^2-4x+1-4\left(x^2-4\right)\\ =4x^2-4x+1-4x^2+16\\ =-4x+17\)

\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)

\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left(x^2+2xy+y^2-3xy\right)\\ =\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

Thay \(x+y=3;xy=2\)

\(=3.\left(3^2-3.2\right)=9\)

a: \(=\dfrac{2x^2-1-x^2-3}{x-2}=\dfrac{x^2-4}{x-2}=x+2\)

b: \(=\dfrac{x\left(x-y\right)+y\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}\)

\(=\dfrac{x^2-xy+xy+y^2}{x^2-y^2}=\dfrac{x^2+y^2}{x^2-y^2}\)

c: \(=\dfrac{x+1-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x+1}\)

d: \(=\dfrac{\left(x+2\right)\cdot y-x\left(y-2\right)}{xy\left(x+y\right)}\)

\(=\dfrac{2y+2x}{xy\left(x+y\right)}=\dfrac{2}{xy}\)

e: \(=\dfrac{1}{x\left(2x-3\right)}-\dfrac{1}{\left(2x-3\right)\left(2x+3\right)}\)

\(=\dfrac{2x+3-x}{x\left(2x-3\right)\left(2x+3\right)}=\dfrac{x+3}{x\left(2x-3\right)\left(2x+3\right)}\)

g: \(=\dfrac{-2x+x+3-x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2}{x+3}\)

\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)

\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)

\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)

\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)

\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)

Giá 1kg thanh long sau khi giảm là: \(x - 1000\) (đồng)

Với số tiền đó, hôm qua người đó mua được số thanh long là: \(\dfrac{b}{a}\) (kg)

Với số tiền đó, hôm nay người đó mua được số thanh long là: \(\dfrac{b}{{a - 1000}}\)  (kg)

Hôm nay mua nhiều hơn hôm qua số kg là:

\(\dfrac{b}{{a - 1000}} - \dfrac{b}{a} = \dfrac{{ba}}{{\left( {a - 1000} \right)a}} - \dfrac{{b\left( {a - 1000} \right)}}{{\left( {a - 1000} \right)a}} = \dfrac{{ba - ba + 1000b}}{{a\left( {a - 1000} \right)}} = \dfrac{{1000b}}{{{a^2} - 1000a}}\) (kg)