Bài 1:

a) \(x^2+10x+26+y^2+2y=(x^2+10x+25)+(y^2+2y+1)\)

..................................................= \(\left(x+5\right)^2+\left(y+1\right)^2\)

b) \(z^2-6z+5-t^2-4t=(z^2-6t+9)-(t^2+4t+4)\)

............................................= \(\left(z-3\right)^2-\left(t+2\right)^2\)

c) \(x^2-2xy+2y^2+2y+1=(x^2-2xy+y^2)+(y^2+2y+1)\)

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

d) \(4x^2-12x-y^2+2y+8=\left(4x^2-12x+9\right)-\left(y^2-2y+1\right)\)

.................................................= \(\left(2x-3\right)^2-\left(y-1\right)^2\)

Bài 2:

a) \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-16\)

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

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

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

x=2
y=3
z=4

\(\dfrac{3x-2y}{4}=\dfrac{4y-3z}{2}=\dfrac{2z-4x}{3}=\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\2z-4x=0\\4y-3z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\\ \Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x-2y+3z}{2-6+12}=\dfrac{8}{8}=1\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\\z=4\end{matrix}\right.\)

1) x² + 10x + 26 + y² + 2y 

= (x² + 10x + 25) + (y² + 2y + 1)

= (x² + 5x + 5x + 25) + (y² + y + y + 1)

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

= (x + 5)² + (y + 1)²

2) z² - 6z + 13 + t² + 4t 

= (z² - 6z + 9) + (t² + 4t + 4) 

= (z² - 3z - 3z + 9) + (t² + 2t + 2t + 4)

= z(z - 3) - 3(z - 3) + t(t + 2) + 2(t + 2)

= (z - 3)² + (t + 2)²

3) x² - 2xy + 2y² + 2y + 1

(x² - 2xy + y²) + (y² + 2y + 1)

= (x - xy - xy + y²) + (y² + y + y +1)

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

= (x - y)² + (y + 1)²

a: \(A=-4x^5y^3-2x^2y^3z^2-2y^4\)

b: \(B=-4x^5y^3-2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3=-4x^5y^3+\dfrac{1}{5}x^4y^3-\dfrac{8}{3}y^4\)

vào chtt tick nha bạn hiền

Nhóm 1:-5x\(^2\)yz;\(\dfrac{2}{3}\)x\(^2\)yz

Nhóm 2:3xy\(^2\)z;-\(\dfrac{2}{3}\)xy\(^2\)z

Nhóm 3:10x\(^2\)y\(^2\)z;\(\dfrac{5}{7}\)x\(^2\)y\(^2\)z

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

=>\(\dfrac{4\left(3x-2y\right)}{4.4}=\dfrac{3\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)

=>\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)

=>\(\dfrac{12x-8y}{16}=0\)

=>12x-8y=0

=>12x=8y

=>\(\dfrac{12x}{24}=\dfrac{8y}{24}\)

=>\(\dfrac{x}{2}=\dfrac{y}{3}\)(1)

Lại có \(\dfrac{8y-6z}{4}=0\)

=>8y-6z=0

=>8y=6z

=>\(\dfrac{8y}{24}=\dfrac{6z}{24}\)

=>\(\dfrac{y}{3}=\dfrac{z}{4}\)(2)

từ (1) và (2)=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)

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