\(1+8.\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)

\(=1+\left(3^2-1\right).\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)

\(=1+\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)

\(=1+\left(3^8-1\right)\left(3^8+1\right)-9^8\)

\(=1+\left(3^8\right)^2-1-9^8=\left(3^2\right)^8-9^8=9^8-9^8=0\)

`1789 + 1789 + 1789 xx 3`

`=1789xx1 + 1789xx1 + 1789 xx 3`

`=1789 xx ( 1 + 1 + 3)`

`=1789 xx 5`

`= 8945`

1789+1789+1789x3

= 1789 x 3 x 3

= 5367 x 3 

= 16101

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

\(\Rightarrow M=-4y^3+8xy^2-5x^2y+x^3-8xy^2+4y^3\)

\(\Rightarrow M=x^3-5x^2y\)

Ta thay \(x=1,y=-3598\) vào M:

\(M=1^3-5.1^2.\left(-3598\right)=17991\)

Tổng `2` xe đầu chứa được :

`45 + 53 = 88(tấn-hàng)`

Trung bình `2` xe đầu chở được :

`88:2=44(tấn-hàng)`

Xe thứ `3` chở được :

`44+6=50(tấn - hàng)`

Trung bình hai xe đầu chở được số tấn hàng là:

                ( 45+53) : 2= 49 ( tấn )

Xe thứ ba chở được số tấn hàng là:

                      49 + 6 = 55 ( tấn)

                                        đáp số:...................

`258 - (x - 212) = 26`

`x-212 = 258 - 26`

`x-212=232`

`x=232+212`

`x=444`

258 - (x - 212) = 26

        x - 212 = 258 - 26

       x - 212  = 232

       x = 232 + 212

       x = 444

`0,25 = 1/4`

Tổng số phần bằng nhau:

`1+4=5(phần)`

Số lớn:

`125:5xx4=100`

Số bé:

`125-100=25`

=)) giải dùm.Cinn ciuu

5 tấn 33 yến = 533 yến

17 tấn 79 kg = 17079 kg

32 yến 6 kg = 326 kg

5 tấn 33 yến =  533 yến

17 tấn 79kg = 17079 kg

32 yến 6kg = 326 kg

học tốt nha :)

`x - ( 45 + 56)= 21`

`x - 101 = 21`

`x=21+101`

`x=122`

__________________

`6xx x xx 5 = 240`

`x xx ( 6 xx 5)  =240`

`x xx 30= 240`

`x=240 : 30`

`x=8`

___________________

`173 + (x - 49) = 219`

`x - 49 = 219 - 173`

`x-49= 46`

`x=46+49`

`x=95`

a. X - (45 + 56) = 21

X - 101 = 21

X = 21 + 101

X = 122

b. 6xXx5 = 240

30xX = 240

X = 240 \(\div\)30 

X = 8

c. 173 + (X - 49) = 219

X - 49 = 219 - 173

X - 49 = 46

X = 46 + 49

X = 95

c) This makes no sense. What is quadrilateral ABCE while B, E, C are collinear? May be you mean ABCD but you make some mistake when typing.

ADCE is a parallelogram, therefore, \(CE//AD\) or \(BC//AD\)

Thus, we can easily prove that ABCD is a trapezoid.

Consider the triangle ABE, its height AH is also a median. This means ABE is an isosceles triangle, whose bisector is also AH. Thus, \(\widehat{BAH}=\widehat{EAH}\)

Clearly, we get \(\widehat{BAH}=\widehat{ECA}\left(=90^o-\widehat{B}\right)\), so, \(\widehat{EAH}=\widehat{ECA}\left(=\widehat{BAH}\right)\)

On the other hand, \(EC//AD\Rightarrow\widehat{ECA}=\widehat{CAD}\) (2 staggered angles)

Thus, \(\widehat{EAH}=\widehat{CAD}\left(=\widehat{ECA}\right)\) or \(\widehat{EAH}+\widehat{EAC}=\widehat{CAD}+\widehat{EAC}\) or \(\widehat{CAH}=\widehat{DAE}\)

ADCE is a parallelogram, therefore, \(\widehat{DAE}=\widehat{DCE}\), so, \(\widehat{CAH}=\widehat{DCE}\left(=\widehat{DAE}\right)\) or \(\widehat{CAH}=\widehat{DCB}\)

We also have \(\widehat{CAH}=\widehat{B}\left(=90^o-\widehat{ACB}\right)\), so \(\widehat{B}=\widehat{DCB}\left(=\widehat{CAH}\right)\)

Consider the trapezoid ABCD (AD//BC), it has \(\widehat{B}=\widehat{DCB}\), therefore, ABCD is an isosceles trapezoid.

d) It's pretty easy! Since ABE is an isosceles triangle, \(AE=AB\). Guess what? We've already had \(AB=15cm\). So, simply, we get \(AE=15cm\) !!!