Bài 1:

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

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

a: ĐKXĐ: x<>y; x<>1/2; x<>-2

b:

\(\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}\)

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

\(=\frac{-x^2-y^2}{y\left(x^2+y^2\right)}=\frac{-1}{y}\)

\(\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)

\(=\frac{2\left(x^2+1\right)-x\left(2x-1\right)}{2\left(2x-1\right)}\cdot\frac{-\left(2x-1\right)}{x+2}\)

\(=\frac{2x^2+2-2x^2+x}{2}\cdot\frac{-1}{x+2}=\frac{x+2}{-2\left(x+2\right)}=\frac{-1}{2}\)

Ta có: \(A=\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}+\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)

\(=\frac{-1}{y}+\frac{-1}{2}=\frac{-y-2}{2y}\)

Câu 1: x^2 + y^2 = 25

=>   x^2 + y^2 +2xy - 2xy = 25

=>   \(\left(x+y\right)^2\)\(-2xy=25\)

=> \(1-2xy=25\)=>  \(-2xy=-24\) =>   xy = 12

Câu 10: (a^2 + b^2)^2 - 4a^2.b^2 = a^4 +2a^2.b^2 + b^2 - 4a^2.b^2 = a^4 - 2a^2.b^2 + b^2 = (a^2 - b^2)^2

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

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

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

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

\(=\left(2x\right)^3\)

\(=8x^3\)

\(---\)

\(C=8\left(x+2y\right)^3-6\left(x+2y\right)^2x+12\left(x+2y\right)x^2-8x^3\) (sửa đề)

\(=\left[2\left(x+2y\right)\right]^3-3\cdot\left(x+2y\right)^2\cdot2x+3\cdot\left(x+2y\right)\cdot\left(2x\right)^2-\left(2x\right)^3\)

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

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

\(=\left(4y\right)^3\)

\(=64y^3\)

\(---\)

\(D=\left(x-y\right)^3-3\cdot\dfrac{\left(x-y\right)^2}{2}\cdot y+3\cdot\dfrac{\left(x-y\right)}{4}\cdot y^2-\dfrac{y^3}{8}\)

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

\(=\left[\left(x-y\right)-\dfrac{y}{2}\right]^3\)

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

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

#\(Toru\)

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) M = (x² + 3xy - 3x³) + (2y³ - xy + 3x³)

= x² + 3xy - 3x³ + 2y³ - xy + 3x³

= x² + (3xy - xy) + (-3x³ + 3x³) + 2y³

= x² + 2xy + 2y³

Tại x = 5 và y = 4

M = 5² + 2.5.4 + 2.4³

= 25 + 40 + 2.64

= 65 + 128

= 193

b) N = x²(x + y) - y(x² - y²)

= x³ + x²y - x²y + y³

= x³ + (x²y - x²y) + y³

= x³ + y³

Tại x = -6 và y = 8

N = (-6)³ + 8³

= -216 + 512

= 296

c) P = x² + 1/2 x + 1/16

= (x + 1/2)²

Tại x = 3/4 ta có:

P = (3/4 + 1/2)² = (5/4)² = 25/16