cho tam giác EFG=tam giác HIK. biết cạnh EF=7cm,IK=7cm,GE=7cm. Chu vi EFG=?cm
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét ΔMNP có :
PM = PN ( gt )
⇒ ΔMNP cân.
⇒ ^PMN = ^PNM ( t/c Δcân )
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\sqrt{\frac{25}{36}}-\left|-\frac{1}{24}\right|+\left(1\frac{1}{2}\right)^5:\left(\frac{3}{2}\right)^2\)
\(=\frac{5}{6}-\frac{1}{24}+\left(\frac{3}{2}\right)^5:\left(\frac{3}{2}\right)\)
\(=\frac{20}{24}-\frac{1}{24}+\left(\frac{3}{2}\right)^4\)
\(=\frac{19}{24}+\frac{81}{16}\)
\(=\frac{38}{48}+\frac{243}{48}\)
\(=\frac{281}{48}\)
\(c,\left(1-\frac{2}{5}\right)^2+\left|-\frac{3}{5}\right|+\sqrt{\left(1,6-0,7\right)^2}\)
\(=\left(\frac{5}{5}-\frac{2}{5}\right)^2+\frac{3}{5}+\sqrt{\left(0,9\right)^2}\)
\\(=\left(\frac{3}{5}\right)^2+\frac{3}{5}+0,9\)
\(=\frac{9}{25}+\frac{3}{5}+\frac{9}{10}\)
\(=\frac{18}{50}+\frac{30}{50}+\frac{45}{50}\)
\(=\frac{93}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
TL :
21 cm
~HT~
Chu vi của hình tam giác EFG = 7 + 7 + 7 = 21 cm
Vì cạnh IK cũng bằng cạnh FG
- HT -