Số chia hết cho 2 là: 428

Hok tot!

số 428 chia hết cho 2

\(T=\frac{\left(xy+z\right)\left(yz+x\right)\left(zx+y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)

\(T=\frac{\left(xy+1-x-y\right)\left(yz+1-y-z\right)\left(zx+1-z-x\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)

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

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

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

\(T=\frac{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)

\(T=1\). Vậy \(T\) không phụ thuộc vào \(x,y,z\).

\(A=\frac{2x+\sqrt{2x+2}}{\sqrt{x}}-5.\)

\(A=\frac{2x+\sqrt{2x+2}}{\sqrt{x}}-\frac{5}{1}\)

\(A=\frac{2x+\sqrt{2x+2}}{\sqrt{x}}-\frac{\sqrt{x.5}}{\sqrt{x.1}}\)

\(A=\frac{2x+\sqrt{2x+2}}{\sqrt{x}}-\frac{5\sqrt{x}}{\sqrt{x}}\)

\(A=\frac{2x+\sqrt{2x+2}-5\sqrt{x}}{\sqrt{x}}\)

428 nha bạn

TL 

số chia hết cho 2 là : 428 

~HT~ 

x=60 nhé

học tốt

x là 60

Ht

^^

TL:

6,2 x 5,4 + 6,2 x 4,6

= 6,2 x (5,4 + 4,6)

= 6,2 x 10

= 62

HT

TL:

6,2 x 5,4 + 6,2 x 4,6

= 6,2 x ( 5,4 + 4,6 )

= 6,2 x 10

= 62 

HT

TL

1 000 000

@minhnguvn

100 x 100 x 100 = 1000000

HT

@LeBaoPhuong

I/ Trắc nghiệm

1) C

2) C

3) B

4) D

5) B

6) A