TL :

Cách đổi tên :

Bước 1 : Vô Luyenthi24.vn

Giải thích các bước giải:

a2+b2+c2=ab+bc+caa2+b2+c2=ab+bc+ca

⇔2a2+2b2+2c2−2ab−2bc−2ca=0⇔2a2+2b2+2c2−2ab−2bc−2ca=0

⇔(a2−2ab+b2)+(b2−2bc+c2)+(c2−2ac+a2)=0⇔(a2−2ab+b2)+(b2−2bc+c2)+(c2−2ac+a2)=0

⇔(a−b)2+(b−c)2+(c−a)2=0⇔(a−b)2+(b−c)2+(c−a)2=0

Vì (a−b)2; (b−c)2; (c−a)2 ≥0Vì (a−b)2; (b−c)2; (c−a)2 ≥0

nên ⎧⎨⎩a−b=0b−c=0c−a=0nên {a−b=0b−c=0c−a=0

Xét ΔMNP có : 

PM = PN ( gt ) 

⇒ ΔMNP cân.

⇒ ^PMN = ^PNM ( t/c Δcân )

b) 15/12x - 3/7x = 6/5x - 12

<=> 1/20x            = -13/14

<=>    x                = -13/14 : 1/20

<=>    x                = -130/7

a. \(\left|x+\frac{1}{5}\right|-4=-2\) 

\(\left|x+\frac{1}{5}\right|=\left(-2\right)+4\)

\(\left|x+\frac{1}{5}\right|=2\)

\(\Rightarrow x+\frac{1}{5}=2\)    ;  \(x+\frac{1}{5}=-2\)

\(x=2-\frac{1}{5}\)             \(x=\left(-2\right)-\frac{1}{5}\)

\(x=\frac{9}{5}\)                      \(x=\frac{-11}{5}\)

Vậy \(x=\frac{9}{5};x=\frac{-11}{5}\)

#Y/n

\(\widehat{LMN}=143^o\)

#Y/n

TL

LMN= 143 Độ nghen

Xin k

Hok tốt
 

\(a,\sqrt{\frac{25}{36}}-\left|-\frac{1}{24}\right|+\left(1\frac{1}{2}\right)^5:\left(\frac{3}{2}\right)^2\)

\(=\frac{5}{6}-\frac{1}{24}+\left(\frac{3}{2}\right)^5:\left(\frac{3}{2}\right)\)

\(=\frac{20}{24}-\frac{1}{24}+\left(\frac{3}{2}\right)^4\)

\(=\frac{19}{24}+\frac{81}{16}\)

\(=\frac{38}{48}+\frac{243}{48}\)

\(=\frac{281}{48}\)

\(c,\left(1-\frac{2}{5}\right)^2+\left|-\frac{3}{5}\right|+\sqrt{\left(1,6-0,7\right)^2}\)

\(=\left(\frac{5}{5}-\frac{2}{5}\right)^2+\frac{3}{5}+\sqrt{\left(0,9\right)^2}\)

\\(=\left(\frac{3}{5}\right)^2+\frac{3}{5}+0,9\)

\(=\frac{9}{25}+\frac{3}{5}+\frac{9}{10}\)

\(=\frac{18}{50}+\frac{30}{50}+\frac{45}{50}\)

\(=\frac{93}{50}\)

?????????????????????????

TL :

\(\frac{5^9.16^2}{10^4.4^3}\)= 781,25

HT

come on

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2021.2022}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...\frac{1}{2021}-\frac{1}{2022}=1-\frac{1}{2022}=\frac{2021}{2022}\)

#Y/n

Mik nghĩ bạn viết đề sai nên ms làm vậy.