Ta có :

\(\dfrac{ab+bc+ca}{xy+yz+zx}=\dfrac{ab}{xy}=\dfrac{bc}{yz}=\dfrac{ca}{zx}\)

\(\Rightarrow\dfrac{a}{x}.\dfrac{b}{y}=\dfrac{b}{y}.\dfrac{c}{z}=\dfrac{c}{z}.\dfrac{a}{x}\)

Mà \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

\(\Rightarrow\dfrac{a^2}{x^2}=\dfrac{b^2}{y^2}=\dfrac{c^2}{z^2}=\dfrac{a^2+b^2+c^2}{x^2+y^2+z^2}\left(\text{Đ}PCM\right)\)

\(ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)\)

\(=a^3b-ab^3+b^3c-bc^3+c^3a-ca^3\)

\(=a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(=a^3\left(b-a+a-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(=\left(a-b\right)\left(c^3-a^3\right)+\left(c-a\right)\left(b^3-a^3\right)\)

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

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

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

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

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

\(16x^2+8x+100=\left(4x\right)^2+2.4x.1+1^2+99\\ =\left(4x+1\right)^2+99>=99>0\left(DPCM\right)\)

\(16x^2+8x+100>0\)

\(\Leftrightarrow\left(4x\right)^2+2.4x.1+1+99>0\)

\(\Leftrightarrow\left(4x+1\right)^2+99>0\left(\forall x\in R\right)\)

\(x^2-9x+20=\left(x-\dfrac{9}{2}\right)^2-\left(\dfrac{9}{2}\right)^2+20=\left(x-\dfrac{9}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)

\(\widehat{A}+\widehat{B}+\widehat{C}=180^o=>60^o+2.\widehat{C}+\widehat{C}=180^o\\ < =>3.\widehat{C}=120^o\\ < =>\widehat{C}=40^o\\=>\widehat{B}=2.40^o=80^o\)

Giải giúp em với ạ huhu

\(\left(2x+1\right)^2-4.\left(x+2\right)=9\)

\(\Leftrightarrow4x^2+4x+1-4x-8=9\)

\(\Leftrightarrow4x^2=16\)

\(\Leftrightarrow x^2=4\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

vậy...............

\(\left(2x+1\right)^2-4\left(x+2\right)=9\)

\(\Leftrightarrow4x^2+4x+1-4x-8=9\)

\(\Leftrightarrow4x^2-7-9=0\)

\(\Leftrightarrow4x^2-16=0\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

     4⁴⁹ + 8³³ + 16²⁵

= (2²)⁴⁹ + (2³)³³ + (2⁴)²⁵

= 2⁹⁸ +  2⁹⁹ + 2¹⁰⁰

= 2⁹⁸ . ( 1 + 2 + 2²) 

= 2⁹⁸. 7

7 ⋮ 7 ⇔2⁹⁸. 7 =  4⁴⁹ + 8³³ + 16²⁵ ⋮7 (đpcm)

CM :\(4^{49}+8^{33}+16^{25}⋮7\)

\(4^{49}=64^{16}.4\)

ta có: 64 : 7 dư 1 nên 64¹⁶ chia 7 sẽ dư 1¹⁶=1

\(\Rightarrow64^{16}.4\div7\) sẽ dư 4

\(8\div7\) dư 1 nên \(8^{33}\div7\) cũng sẽ dư 1

\(16^{25}=4^{50}=64^{16}.16\)

lập luận tương tự như trên bạn sẽ có

\(16^{25}\div7\)dư 16 tức là chia 7 dư 2

từ đó ta có:

\(4^{49}+8^{33}+16^{25}=64^{16}.4+8^{33}+64^{16}.16\div7\) sẽ dư 7 tức là nó chia hết cho 7(đpcm)