biểu thức trên = : (( x+y+z)-(x+y))²   ( theo hằng đẳng thức số 20

theo hằng đẳng thức số 2

(x + y +z)² -2(x + y +z)+(x+y)²

=x² +y² + z² +2xy + 2yz+2xz  -2x² -2xy -2y² -2xy-2xz-2yz+x²+2xy+y²

= z² 

(x+y+z)² - 2(x+y+z)(x+y)+(x+y)²

= (x+y+z+x+y)²

Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath.

x^n-1(x+y)-y(x^n-1+y^n-1)

= x^n-1.x+x^n-1.y-y.x^n-1-y.y^n-1

= x^n-1+1+(x^n-1y-x^n-1y)-y^n-1+1

= xⁿ-yⁿ

x^n-1.x+x^n-1.y-y.x^n-1-y.y^n-1

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)

\(P=\dfrac{4k^2-2k.3k+9k^2}{4k^2+2k.3k+9k^2}=\dfrac{13k^2-6k^2}{13k^2+6k^2}=\dfrac{7k^2}{19k^2}=\dfrac{7}{19}\)

\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)

ĐK: \(3x\ne\pm y;x\ne0\)

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

= \(\dfrac{3x\left(3x-y\right)-x\left(3x+y\right)+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{6x^2-4xy+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{2x\left(3x-2y+1\right)}{\left(3x-y\right)\left(3x+y\right)}\)

Thay x = 1; y=2, ta có:

A = \(\dfrac{2.1\left(3.1-2.2+1\right)}{\left(3.1-2\right)\left(3.1+2\right)}=0\)

\(=x^0+x^{-1}y-x^{-1}y-y^0\)

\(=1+0-1=0\)

\(a,=\left[\left(x+2\right)-\left(x-3\right)\right]^2=\left(x+2-x+3\right)^2=5^2=25\)

\(b=x^2-5\)

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