tính bằng cách hợp lí
\(1+8.\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right)-9^8\)
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`1789 + 1789 + 1789 xx 3`
`=1789xx1 + 1789xx1 + 1789 xx 3`
`=1789 xx ( 1 + 1 + 3)`
`=1789 xx 5`
`= 8945`
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
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`
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 \(AE=15cm\) !!!
\(1+8.\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^2-1\right).\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^8-1\right)\left(3^8+1\right)-9^8\)
\(=1+\left(3^8\right)^2-1-9^8=\left(3^2\right)^8-9^8=9^8-9^8=0\)