\(70,08dm^2=700800mm^2\)

\(0,0030hm^2=30m^2\)

10,07kg=10070g

5,06 tấn=5 tấn 600hg

|x|=|y|

mà x>0; y<0

nên x=-y

2x+y=-2y+y=-y

\(\left|x\right|=\left|y\right|\) và \(x>0;y< 0\)

\(\Rightarrow y=-x\)

\(\Rightarrow2x\pm x=x\)

Vậy \(2x+y=x\)

13

Khó quá bn ah. Tính máy tính ik😂😂😂

\(\dfrac{6^{13}-3\cdot6^{12}+3^{13}}{9^5\left(4^6+1\right)}\)

\(=\dfrac{2^{13}\cdot3^{13}-2^{12}\cdot3^{13}+3^{13}}{3^{10}\left(2^{12}+1\right)}\)

\(=\dfrac{3^{12}\left(2^{13}-2^{12}+1\right)}{3^{10}\left(2^{12}+1\right)}=\dfrac{3^2\cdot\left[2^{12}\left(2-1\right)+1\right]}{2^{12}+1}\)

\(=\dfrac{9\left(2^{12}+1\right)}{2^{12}+1}\)

=9

\(\left[\left(6x-12\right):3\right].32=64\)

\(\left(6x-12\right):3=64:32\)

\(\left(6x-12\right):3=2\)

\(6x-12=2.3\)

\(6x-12=6\)

\(6x=12+6\)

\(6x=18\)

\(x=18:3\)

\(x=6\)

\(\left[\left(6x-12\right):3\right]\cdot2^5=64\)

=>\(\left(2x-4\right)=\dfrac{64}{2^5}=2\)

=>2x=4+2=6

=>\(x=\dfrac{6}{2}=3\)

Ý bạn là sao nhỉ???

Bn viết sai hoặc là viết thiếu câu hỏi r.

Chớ mik đọc hổng hỉu j hớt á

Để \(\dfrac{n}{-3}\) là phân số dương thì n<0

mà n nguyên 

nên \(n\in Z^-\)

\(\dfrac{x+5}{97}+\dfrac{x+5}{98}+\dfrac{x+5}{99}=0\\ \Rightarrow\left(x+5\right).\left(\dfrac{1}{97}+\dfrac{1}{98}+\dfrac{1}{99}\right)=0\)
Vì \(\dfrac{1}{97}+\dfrac{1}{98}+\dfrac{1}{99}\ne0\) nên:
\(x+5=0\\ \Rightarrow x=-5\)
Vậy...

\(\dfrac{x+5}{97}+\dfrac{x+5}{98}+\dfrac{x+5}{99}=0\)

=>\(\left(x+5\right)\left(\dfrac{1}{97}+\dfrac{1}{98}+\dfrac{1}{99}\right)=0\)

=>x+5=0

=>x=-5

`12x + 6/7 : 2 = 9x + 8`

`=> 12x + 3/7 - 9x  =  8`

`=> 3x   =  8 -  3/7`

`=> 3x = 53/7`

`=> x = 53/7 : 3`

`=> x = 53/7 . 1/3`

`=> x=  53/21`

Vậy ..

\(12x+\dfrac{6}{7}\times\dfrac{1}{2}=9x+8\)

\(12x+\dfrac{3}{7}=9x+8\)

\(12x=9x+8-\dfrac{3}{7}\)

\(12x=9x+\dfrac{53}{7}\)

\(\Rightarrow\) \(\dfrac{53}{7}=3x\)

\(\Rightarrow\dfrac{53}{7}\div3=x\)

\(\Rightarrow\dfrac{53}{21}=x\)

Vậy \(x=\dfrac{53}{21}\)

ΔEHF vuông tại H

=>\(HE^2+HF^2=EF^2\)

=>\(HE=\sqrt{5^2-3^2}=4\left(cm\right)\)

Xét ΔHEG vuông tại H và ΔHFE vuông tại H có

\(\widehat{HEG}=\widehat{HFE}\left(=90^0-\widehat{G}\right)\)

Do đó: ΔHEG~ΔHFE

=>\(\dfrac{HE}{HF}=\dfrac{HG}{HE}\)

=>\(HE^2=HF\cdot HG\)

=>\(HG=\dfrac{4^2}{3}=\dfrac{16}{3}\left(cm\right)\)

ΔEHG vuông tại H

=>\(HE^2+HG^2=EG^2\)

=>\(EG=\sqrt{\left(\dfrac{16}{3}\right)^2+4^2}=\dfrac{8\sqrt{13}}{3}\left(cm\right)\)