`Answer:`

`a^2.(b-c)+b^2.(c-a)+c^2.(a-b)`

`=a^2.(b-c)+b^2[(c-b)-(a-b)]+c^2.(a-b)`

`=a^2.(b-c)+b^2.(c-b)+b^2.(a-b)+c^2.(a-b)`

`=(b-c)(a^2-b^2)-(a-b)(b^2-c^2)`

`=(b-c)(a-b)(a+b)-(a-b)(b-c)(b+c)`

`=(a-b)(b-c)(a+b-b-c)`

`=(a-b)(b-c)(a-c)`

\(a^2.\left(b-c\right)+b^2,\left(c-a\right)+c^2.\left(a-b\right)\)

\(=a^2.\left(b-c\right)-b^2.\left(a-c\right)+c^2.\left(a-b\right)\)

\(=a^2.\left(b-c\right)-b^2.\left[\left(a-b\right)+\left(b-c\right)\right]+c^2.\left(a-b\right)\)

\(=a^2.\left(b-c\right)-b^2.\left(a-b\right)-b^2.\left(b-c\right)+c^2.\left(a-b\right)\)

\(=\left(b-c\right).\left(a^2-b^2\right)+\left(a-b\right).\left(c^2-b^2\right)\)

\(=\left(b-c\right).\left(a-b\right).\left(a+b\right)-\left(b-c\right).\left(b+c\right).\left(a-b\right)\)

\(=\left(b-c\right).\left(a-b\right).\left(a+b-b-c\right)\)

\(=\left(b-c\right).\left(a-b\right).\left(a-c\right)\)

vì \(x+y+z=0\)   \(\Rightarrow\)\(x+y=-z\)

\(\Rightarrow\left(x+y\right)^3=-z^3\)\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)

\(\Rightarrow x^3+y^3+z^3=-3xy.\left(x+y\right)\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)   do  \(x+y=-z\)

\(\Rightarrow\left(x^3+y^3+z^3\right).\left(x^2+y^2+z^2\right)=3xyz.\left(x^2+y^2+z^2\right)\)

\(\Rightarrow3xyz.\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3.\left(y^2+z^2\right)+y^3.\left(x^2+z^2\right)+z^3.\left(x^2+y^2\right)\)

lại có: \(x^2+y^2=\left(x+y\right)^2-2xy=z^2-2xy\)

tương tự thì:  \(y^2+z^2=x^2-2yz\)

\(z^2+x^2=y^2-2xz\)

vì vậy nên \(3xyz.\left(x^2+y^2+z^2\right)=x^5+y^5+z^5+x^3.\left(x^2-2yz\right)+y^3.\left(y^2-2xz\right)+z^3.\left(z^2-2xy\right)\)

\(\Rightarrow3xyz.\left(x^2+y^2+z^2\right)=2x^5+2y^5+2z^5-2xyz.\left(x^2+y^2+z^2\right)\)

\(\Rightarrow5xyz.\left(x^2+y^2+z^2\right)=2.\left(x^5+y^5+z^5\right)\) 

đpcm

VÌ \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(\Rightarrow\left(x+y\right)^5=-z^5\)

\(\Leftrightarrow x^5+y^{^5}+5\left(x^4y+xy^4+2x^3y^2+2x^2y^3\right)=-z^5\)

\(\Leftrightarrow x^5+y^{^5}+z^5+5xy\left(x^3+y^3+2x^3y^2+2x^2y^3\right)=0\)

\(\Leftrightarrow x^5+y^{^5}+z^5+5xy\left(x+y\right)+\left(x^2-xy+y^2+2xy\right)=0\)

\(\Leftrightarrow x^5+y^{^5}+z^5-5xyz\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow x^5+y^{^5}+z^5=5xyz\left(x^2+xy+y^2\right)\)

\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(2x^2+2xy+2y^2\right)\)

\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+\left(x+y\right)^2+y^2\right)\)

\(\Leftrightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)Vì (x+y=-z)

HT

327 + 443 = 770

770 nhé e 

HT 

ah hỏi kb j !

TL

12cm

Vẽ hình tròn đường kính 12cm

HT

đây hok tot

Đổi: 25% = 0,25

        30% = 0,3

ta có: x - 0,25 + 0,3 = 19,5

                   x - 0,25 = 19,5 - 0,3

                   x - 0,25 = 19,2

                             x = 19,2 + 0,25

                             x = 19,45

Vậy STN đó là: 19,45

k cho tớ nhá:))

gọi số đó là x,ta có:

\(x-25+30=19,5\)

\(x+5=19,5\)

\(x=19,5-5\)

\(x=14,5\)

`Answer:`

Bài 1:

Ta có: `AB<AC<BC` (Vì `23cm<25cm<30cm`)

`=>\hat{C}<\hat{B}<\hat{A}`

Bài 2:

Vì chu vi của `\triangleABC=AB+BC+CA=20cm`

Mà `\triangleABC` cân ở `A=>AB=AC`

Xét `\triangleABC:` 

`AB=AC=\frac{20-6}{2}=\frac{14}{2}=7cm`

`=>AB=AC>BC`

`=>\hat{C}=\hat{B}>\hat{A}`

`Answer:`

\(A=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{10000}\right)\)

\(\Rightarrow A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)

\(\Rightarrow A=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{99.101}{100^2}\)

\(\Rightarrow A=\frac{\left(1.2.3.4.5.....9\right).\left(3.4.5.....101\right)}{\left(2.3.4.....100\right).\left(2.3.4.5.....100\right)}\)

\(\Rightarrow A=\frac{1}{100}.\frac{1}{2}.101\)

\(\Rightarrow A=\frac{101}{200}\)

`Answer:`

Ta có `hat{zOt}+\hat{yOz}=90^o`

\(\Rightarrow\frac{1}{2}.Oz+\widehat{yOz}=90^o\)

\(\Rightarrow\frac{1}{2}.4\widehat{yOz}+\widehat{yOz}=90^o\)

\(\Rightarrow\widehat{yOz}.3=90^o\)

\(\Rightarrow\widehat{yOz}=30^o\)

`=>\hat{xOz}=120^o` (Vì `\hat{xOz}=4\hat{yOz}`

Vậy `\hat{xOy}=\hat{yOz}+\hat{xOz}=120^o+30^o=150^o`