\(\dfrac{x}{2}=\dfrac{y}{5}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{25}=\dfrac{10x^2-y^2}{10\cdot4-25}=\dfrac{x^2+y^2}{4+25}\\ \Leftrightarrow A=\dfrac{40-25}{29}=\dfrac{15}{29}\)

bạn làm đc ko????

Có 3x = -2y

=> \(\frac{x}{-2}=\frac{y}{3}\)

=>\(\frac{x^2}{\left(-2\right)^2}=\frac{y^2}{3^2}=\frac{x.y}{\left(-2\right).3}=\frac{-5}{-6}=\frac{5}{6}\)

=> x² = 5. (-2)² : 6 = \(\frac{10}{3}\)=> x = \(\sqrt{\frac{10}{3}}\)hoặc x = \(-\sqrt{\frac{10}{3}}\)

y² = 5.3³ : 6 = \(\frac{15}{2}\)=> y = \(\sqrt{\frac{15}{2}}\)hoặc y = \(-\sqrt{\frac{15}{2}}\)

KL: (x; y) = \(\left(\sqrt{\frac{10}{3}};\sqrt{\frac{15}{2}}\right)\); \(\left(-\sqrt{\frac{10}{3}};-\sqrt{\frac{15}{2}}\right)\)

2:

a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)

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

b: \(=\left(x^2-4y^2\right)-\left(x-2y\right)\)

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

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

c: \(=\left(y^2+10y+25\right)-9z^2\)

\(=\left(y+5\right)^2-\left(3z\right)^2\)

\(=\left(y+5+3z\right)\left(y+5-3z\right)\)

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

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

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

1:

a: \(x\left(3-4x\right)+5\left(3-4x\right)=\left(3-4x\right)\left(x+5\right)\)

b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)

\(=2y\left(5y-6\right)+4\left(5y-6\right)\)

\(=2\left(5y-6\right)\left(y+2\right)\)

c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)

\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)

\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)

d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)

\(=2y\left(x+y\right)\left[3\left(x-y\right)-4\left(x+y\right)\right]\)

\(=2y\left(x+y\right)\left(3x-3y-4x-4y\right)\)

\(=2y\left(x+y\right)\left(-x-7y\right)\)

Bài 1

a) x(3 - 4x) + 5(3 - 4x)

= (3 - 4x)(x + 5)

b) 2y(5y - 6) - 4(6- 5y)

= 2y(5y - 6) + 4(5y - 6)

= (5y - 6)(2y + 4)

= 2(5y - 6)(y + 2)

c) 27(x - 2)³ - 3x(2 - x)²

= 27(x - 2)³ - 3x(x - 2)²

= 3(x - 2)²[9(x - 2) - x]

= 3(x - 2)²(9x - 18 - x)

= 3(x - 2)²(8x - 18)

= 6(x - 2)²(4x - 9)

d) 6y(x² - y²) - 8y(x + y)²

= 6y(x - y)(x + y) - 8y(x + y)²

= 2y(x + y)[3(x - y) - 4(x + y)]

= 2y(x + y)(3x - 3y - 4x - 4y)

= 2y(x + y)(-x - 7y)

= -2y(x + y)(x + 7y)

A=\(\left(\frac{2}{5}xy+3x^2y^3-4xy^3\right)-\left(-x^2y^3+xy^3+\frac{2}{5}xy\right)+5\)

A=\(\frac{2}{5}xy+3x^2y^3-4xy^3+x^2y^3-xy^3-\frac{2}{5}xy+5\)

A=\(\left(\frac{2}{5}xy-\frac{2}{5}xy\right)+\left(3x^2y^3+x^2y^3\right)+\left(-4xy^3-xy^3\right)+5\)

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

Thay \(x=\frac{1}{2}\) và \(y=-1\) vào đa thức A, ta được:

\(4.\left(\frac{1}{2}\right)^2.\left(-1\right)^3-5.\left(\frac{1}{2}\right).\left(-1\right)^3+5\)

=\(4.\frac{1}{4}.\left(-1\right)-5.\frac{1}{2}.\left(-1\right)+5\)

=\(\left(-1\right)-\left(-\frac{5}{2}\right)+5\)

= \(\frac{3}{2}+5=\frac{13}{2}\)

Vậy giá trị của đa thức A tại \(x=\frac{1}{2}\) và \(y=-1\) là \(\frac{13}{2}\)

Nhớ tick cho mình nha!

a, xy + x + y = 4

=> xy + x = 4 - y

=> x(y+1) = 5 - y - 1

=> x(y+1) = 5 - (y+1)

=> x(y+1) + (y+1) = 5

=> (x+1)(y+1) = 5

Vì x, y \(\in\) Z => x+1 \(\in\) Z và y + 1 \(\in\) Z

Mà 5 = 1.5 = (-1)(-5)

Ta có bảng :

Vậy các cặp số (x; y) là : (0; 4);(4; 0);(-2; -6);( -6; -2)

b, x - y + xy = 3

=> xy + x = 3 + y

=> x(y+1) = 2 + 1 + y

=> x(y+1) - (y+1) = 2

=> (x-1)(y+1) = 2

Vì x, y \(\in\) Z => x-1 \(\in\) Z và y+1 \(\in\) Z

Mà 2 = 1.2 = (-1)(-2)

Ta có bảng :

Vậy các cặp số (x; y) là (2; 1);(3; 0);(0; -3);(-1; -2)

a) xy + x+y = 4

x.(y + 1) + y = 4

x.(y + 1) + (y + 1) = 4 +1

(x + 1).(y + 1) = 5

Vì x,y\(\in\)Z và x,y \(\in\) Ư(5)

nên ta xét bảng sau:

Vậy (x ; y) là (0;4),(-2;-6),(4;0),(-6;-2)