a: \(A=\dfrac{x^5}{x^3}\cdot\dfrac{y^{-2}}{y}=x^2\cdot y^{-1}=\dfrac{x^2}{y}\)

b: \(B=\dfrac{x^2\cdot y^{-3}}{x^3\cdot y^{-12}}=\dfrac{x^2}{x^3}\cdot\dfrac{y^{-3}}{y^{-12}}=\dfrac{1}{x}\cdot y^{-3+12}=\dfrac{y^9}{x}\)

a) \(A=\dfrac{x^5y^{-2}}{x^3y}=\dfrac{x^5}{x^3}.\dfrac{1}{y^{2-1}}=x^{5-3}y^{-1}=x^2y^{-1}\).

b) \(B=\dfrac{x^2y^{-3}}{\left(x^{-1}y^4\right)^{-3}}=\dfrac{x^2y^{-3}}{x^3y^{-12}}=x^{2-3}y^{-3-\left(-12\right)}=\dfrac{1}{xy^9}\)

a)      Cách 1:

\(6(y - x) - 2(x - y)\)

\( = 6y - 6x - 2x + 2y\)

\( = 8y - 8x\)

Cách 2:

\(6(y - x) - 2(x - y)\\= 6(y-x)+2(y-x)\\=(6+2).(y-x)\\=8.(y-x)\\=8y-8x\)

b)      \(3{x^2} + x - 4x - 5{x^2}\)

\( = (3{x^2} - 5{x^2}) + (x - 4x)\)

\( =  - 2{x^2} - 3x\)

`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`

`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`

`=5x^2-5x^2-4x+6x+1+9+245`

`=2x+255`

`b)(x-2)(x^2+2x+4)-(25+x^3)`

`=x^3-8-x^3-25=-33`

Lời giải:

a. 

$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$

$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$

$=5x^2+2x+10-(5x^2-245)=2x+255$

b.

$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$

$=-8-25=-33$

a: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)\)

\(=4x^2-4x+1+x^2+6x+9-5x^2+245\)

\(=2x+255\)

b: \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+25\right)\)

\(=x^3-8-x^3-25\)

=-33

Bài 1:

\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)

Bài 2:

\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)

câu d của bài 2 là của bài 1 nha mình để nhầm chỗ huhu

\(a,\left(2x-1\right)^2-\left(x-3\right)\left(x+3\right)-1969\\ =4x^2-4x+1-x^2+9-1969\\ =3x^2-4x-1959\)

\(b,\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\\ =4x^2-9y^2-4x^2+4xy-y^2\\ =8y^2+4xy=4y\left(2y+x\right)\)

\(c,\left(x+3y\right)^2+\left(x+y\right)\left(x-y\right)+280\\ =x^2+6xy+9y^2+x^2-y^2+280\\ =2x^2+8y^2+6xy+280\)

a: \(\left(2x-1\right)^2-\left(x-3\right)\cdot\left(x+3\right)-1969\)

\(=4x^2-4x+1-x^2+9-1969\)

\(=3x^2-4x-1959\)

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

\(=4x^2-9y^2-4x^2+4xy-y^2\)

\(=-10y^2+4xy\)

a)\(\text{( 2 x − 1 )^2− ( x − 3 ) ( x + 3 ) − 1969}\)

\(\text{= 4x^2 − 4x + 1 − x^2 + 9 − 1969}\)

\(\text{=3x^2− 4 x − 1959}\)

b) \(\text{( 2 x − 3 y ) ( 2 x + 3 y ) − ( 2 x − y )^2}\)

=\(\text{= 4 x^2− 9 y^2− 4 x^2 + 4 x y − y^2}\)

\(\text{= -10 y^2+ 4 x y = -2 y ( 5 y -2 x )}\)

c)\(\text{( x + 3 y )^2 + ( x + y ) ( x − y ) + 280}\)

\(\text{= x^2 + 6 x y + 9 y^2 + x^2 − y^2 + 280}\)

\(\text{= 2 x^2 + 8 y^2 + 6 x y + 280}\)

a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

\(=2x^2-4xy+\dfrac{15}{4}y^2\)

b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)

\(=2x^2+2x+13-2x^2+2\)

=2x+15

a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)

b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)

\(=2x+15\)

a; \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

= \(x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

= \(2x^2-4xy+\dfrac{15}{4}y^2\)

b; \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

= \(x^2-4x+4+x^2+6x+9-2x^2+2\)

= \(2x+15\)

a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)

\(=14\sqrt{2x}+30\)

b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)

\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)

\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)

\(=\dfrac{11}{2}\sqrt{x}\)

c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)

\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)

d

a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)

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

\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)

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

\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)

c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)

\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)

\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)

\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)

d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)

\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)

\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)

\(D=0\)