\(B=\dfrac{3}{4}xy^2-\dfrac{1}{3}x^2y-\dfrac{5}{6}xy^2+2x^2y=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y\)

Bậc:3

Thay x=-1, y=1 vào B ta có:

\(B=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y=-\dfrac{1}{12}.\left(-1\right).1^2+\dfrac{5}{3}.\left(-1\right)^2.1=\dfrac{1}{12}+\dfrac{5}{3}=\dfrac{7}{4}\)

Câu 3:

a: A(x)=x^3+3x^2-4x-12

B(x)=x^3-3x^2+4x+18

A(x)+B(x)

=x^3+3x^2-4x-12+x^3-3x^2+4x+18

=2x^3+6

A(x)-B(x)

=x^3+3x^2-4x-12-x^3+3x^2-4x-18

=6x^2-8x-30

b: A(-2)=(-8)+3*4-4*(-2)-12

=-20+3*4+4*2=0

=>x=-2 là nghiệm của A(x)

B(-2)=(-8)-3*(-2)^2+4*(-2)+18=-10

=>x=-2 ko là nghiệm của B(x)

a) Ta có bảng sau:

Vậy: 

b) Ta có bảng sau:

Vậy: ...

`@` `\text {Ans}`

`\downarrow`

`a)`

`(x-1)(y+4) = 5`

`=> (x-1)(y+4) \in \text {Ư(5)} = +-1; +-5`

Ta có bảng sau:

Vậy, ta có các cặp `x,y` thỏa mãn `{2; -9}; {6; -5}; {0; 1}; {-4; -8}`

a) (x-1)(y+4)=5

⇒ x-1 và y+4 ϵ {-1;1;-5;5}

⇒ (x;y) ϵ {(0;-5);(-2;1);(-4;-5);(6;-3)

b) (2x+3)(y-2)=11

⇒ 2x+3 và y-2 ϵ {-1;1;-11;11}

⇒ (x;y) ϵ {(-2;-9);(-1;13);(-7;1);(4;3)}

c) xy+2x+y=12

⇒ x(y+2)+y+2-2=12

⇒ (x+1)(y+2)=14

⇒ x+1 và y+2 ϵ {-1;1;-2;2;-7;7;-14;14}

⇒ (x;y) ϵ {(-2;-16);(0;12);(-3;-9);(1;5);(-8;-4);(6;0);(-15;-3);(13;-1)}

d) xy-x-3y=4

⇒ y(x-3)-(x-3)-3=4

⇒ (x-3)(y-1)=7

⇒ x-3 và y-1 ϵ {-1;1;-7;7}

⇒ (x;y) ϵ {(2;-6);(4;8);(-4;0);(10;2)}

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4:

Bài 2 :

a) \(ĐKXĐ:\hept{\begin{cases}x;y>0\\x\ne y\end{cases}}\)

b) \(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\right):\frac{x\sqrt{xy}+y\sqrt{xy}}{\sqrt{xy}\left(y-x\right)}\)

\(\Leftrightarrow A=\frac{x-\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}:\frac{x+y}{y-x}\)

\(\Leftrightarrow A=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}\cdot\frac{y-x}{x+y}\)

\(\Leftrightarrow A=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(y-x\right)}{x+y}\)

c) Thay \(x=4+2\sqrt{3},y=4-2\sqrt{3}\)vào A, ta được :

   \(A=\frac{\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\left(4-2\sqrt{3}-4-2\sqrt{3}\right)}{4+2\sqrt{3}+4-2\sqrt{3}}\)

\(\Leftrightarrow A=\frac{\left(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\right).\left(-4\sqrt{3}\right)}{8}\)

\(\Leftrightarrow A=\frac{\left(1+\sqrt{3}-\sqrt{3}+1\right).\left(-4\sqrt{3}\right)}{8}=\frac{-8\sqrt{3}}{8}=-\sqrt{3}\)

Vậy ....

Bài 1:

\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}=\frac{2\sqrt{2\cdot4}-\sqrt{3\cdot4}}{\sqrt{2\cdot9}-\sqrt{16\cdot3}}-\frac{\sqrt{5}+\sqrt{9\cdot3}}{\sqrt{30}-\sqrt{2}}\)

\(=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}-\sqrt{2}}=\frac{\left(4\sqrt{2}-2\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)-\left(\sqrt{5}+3\sqrt{3}\right)\left(3\sqrt{2}-4\sqrt{3}\right)}{\left(3\sqrt{2}-4\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)}\)

\(=\frac{4\sqrt{60}-8-2\sqrt{90}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{3\sqrt{60}-6-4\sqrt{90}+4\sqrt{6}}\)

\(=\frac{8\sqrt{15}-8-6\sqrt{10}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{6\sqrt{15}-6-12\sqrt{10}+4\sqrt{6}}\)

\(=\frac{12\sqrt{15}-2\sqrt{10}-7\sqrt{6}+28}{6\sqrt{15}-12\sqrt{10}+4\sqrt{6}-6}\)

Bài 2:

a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

hay \(x=\dfrac{2}{7}\)

b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow x^3=-8\)

hay x=-2

Bài 1: 

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

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

\(=xy\)

=1

b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)

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

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

\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)