Killua Zoldyck sai nhé xem tui làm nà

\(=\frac{\left(88985+1\right)\times2003+-678}{88985\times2003+1325}\)

\(=\frac{88985\times2003+2003-678}{88985\times2003+1325}\)

\(=\frac{88985\times2003+1325}{88985\times2003+1325}\)

\(=1\)

Bài giải:

\(\frac{88986\times2003-678}{88985\times2003+1325}=\frac{88986-678}{88985+1325}\)

\(=\frac{88308}{90310}\)

Đáp số: 88308/90310

\(\frac{\left(88985+1\right)\cdot2003-678}{88985\cdot2003+1325}=\frac{88985\cdot2003+2003-678}{88985\cdot2003+1325}=\frac{88985\cdot2003+1325}{88985\cdot2003+1325}=1\)

Bài 2: 

a: =47,4+47,4x4=47,4x5=237

b: =(0,26+0,74)+(0,41+0,59)=1+1=2

Tính bằng cách thuận tiện ( đề `b` )

`13/50+74/100+41/100+59/100`

`=>` `26/100+74/100+41/100+59/100`

`=>` `(26/100+74/100) + ( 41/100 + 59/100)`

`=>` `1+1`

`=>` `2`

tính nhanh:

`a,` \(2 × 4 × 8 × 0 , 5 × 0 , 125 × 0 , 25 × ( 0 , 4321 + 0 , 5679 )\)

`=>`\( 2 × 4 × 8 × 0 , 5 × 0 , 125 × 0 , 25 × 1\)

`=>`\(( 2 × 0 , 5 ) × ( 4 × 0 , 25 ) × ( 8 × 0 , 125 ) × 1\)

`=>`\( 1 × 1 × 1 × 1=1\)

Sao bài của bạn lạ giữ cậu phải xóa chứ 

Hi°•☆▪¤○●□■♤♡♢♧

đây là đề \(\frac{88986\times2003-678}{88985\times2003+1325}\)
 

để bạn buồn cười quá

\(\frac{1-678}{1+1325}=\frac{-677}{1326}\)

chúc bạn học tốt nhé

:D

\(\frac{88986.2003-678}{88986.2003+1325}\)

\(=\frac{88986.2003+1325-2003}{88986.2003+1325}\)

\(=1-\frac{2003}{88986.2003+1325}\)

... bn tự tính nghen!!!

678  x 678 + 302 x 678 + 678 + 678  : 2

=678  x 678 + 302 x 678 + 678 + 678  x 0,5

=678 x (678+302+1+0,5)

=678 x 981,5

= 665457

665457 nha bn

6783 + 1325 = 8108

1 + 434566433635 = 434566433636

45 + 23 =68

K CHO MK NHA THANK YOU VERY MUCH

6783+1325=8108

1+434566433635=434566433636

45+23=68

6783 + 1325 = 8108

1 + 434566433635 = 434566433636

45 + 23 = 68 

\(4\cdot\left(-\dfrac{1}{2}\right)^{^3}+2\cdot\left(-\dfrac{1}{2}\right)^{^2}+3\cdot\left(-\dfrac{1}{2}\right)+1=\left(-\dfrac{1}{2}\right)\left[4\cdot\left(-\dfrac{1}{2}\right)^{^2}+2\left(-\dfrac{1}{2}\right)+3\right]+1=\left(-\dfrac{1}{2}\right)\left(4\cdot\dfrac{1}{4}-1+3\right)+1=\left(-\dfrac{1}{2}\right)\left(1-1+3\right)+1=3\left(-\dfrac{1}{2}\right)+1=-\dfrac{3}{2}+1=-\dfrac{1}{2}\)

Vì \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0,\left(y+0,4\right)^{100}\ge0,\left(z-3\right)^{678}\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0,4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-0,4\\z=3\end{matrix}\right.\)

Vậy \(\left(x,y,z\right)=\left(\dfrac{1}{5};-0,4;3\right)\)

Vì \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)

\(\left(y+0,4\right)^{100}\ge\forall y\)

\(\left(z-3\right)^{678}\ge0\forall z\)

\(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\)

mà \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)

Dấu ''='' xảy ra khi \(x=\dfrac{1}{5};y=-0,4;z=3\)