Gọi \(p_1;n_1\) và \(p_2;n_2\) lần lượt là số hạt p và n của M và X .

Theo đề bài ta có: \(\left\{{}\begin{matrix}2\left(2p_1+n_1\right)+\left(2p_2+n_2\right)=140\left(1\right)\\4p_1+2p_2-\left(2n_1+n_2\right)=44\left(2\right)\\\left(p_1+n_1\right)-\left(p_2+n_2\right)=23\left(3\right)\\\left(2p_1+n_1\right)-\left(2p_2+n_2\right)=34\left(4\right)\end{matrix}\right.\)

Lấy (1) - (2) và 2.(3) - (4) ta được :

\(\left\{{}\begin{matrix}4n_1+2n_2=96\\n_1-n_2=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_1=20\\n_2=8\end{matrix}\right.\)Thay vào (1) và (3)

ta có : \(\left\{{}\begin{matrix}4p_1+2p_2=92\\p_1-p_2=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p_1=19\\p_2=8\end{matrix}\right.\)

Vậy M là \(p_1+n_1=19+20=39\) Kali ; X là : \(p_2+n_2=16\) Oxi

Vậy M₂X là K₂O

BĐT cần chứng minh tương đương :

\(\dfrac{a^8+b^8+c^8}{a^3b^3c^3}\ge\dfrac{ab+bc+ac}{abc}\)

\(\Leftrightarrow\dfrac{a^8+b^8+c^8}{a^2b^2c^2}\ge ab+bc+ac\)

\(\Leftrightarrow\dfrac{a^6}{b^2c^2}+\dfrac{b^6}{a^2c^2}+\dfrac{c^6}{a^2b^2}\ge ab+bc+ac\)

Do \(a^2+b^2+c^2\ge ab+bc+ac\)

Ta phải cm

\(\dfrac{a^6}{b^2c^2}+\dfrac{b^6}{a^2c^2}+\dfrac{c^6}{a^2b^2}\ge a^2+b^2+c^2\)(1)

Đặt : \(\left(a^2;b^2;c^2\right)=\left(x;y;z\right)\)

\(\Rightarrow\left(1\right)\Leftrightarrow\dfrac{x^3}{yz}+\dfrac{y^3}{xz}+\dfrac{z^3}{xy}\ge x+y+z\)

Áp dụng C.B.S

\(\Rightarrow\dfrac{x^3}{yz}+\dfrac{y^3}{xz}+\dfrac{z^3}{xy}=\dfrac{x^4}{xyz}+\dfrac{y^4}{xyz}+\dfrac{z^4}{xyz}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3xyz}\)

Theo Bunhiacopxki: \(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\)\(\Rightarrow\left(x^2+y^2+z^2\right)^2\ge\dfrac{\left(x+y+z\right)^4}{9}\)

Theo Cauchy : \(\Rightarrow3xyz\le\dfrac{\left(x+y+z\right)^3}{9}\)

\(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{3xyz}\ge\dfrac{\dfrac{\left(x+y+z\right)^4}{9}}{\dfrac{\left(x+y+z\right)^3}{9}}=x+y+z\)

\(\Rightarrow\)\(\Rightarrow\dfrac{x^3}{yz}+\dfrac{y^3}{xz}+\dfrac{z^3}{xy}\ge x+y+z\)

=> đpcm

Ta có: \(5x^2+6xy+5y^2=4\left(x+y\right)^2+\left(x-y\right)^2\ge4\left(x+y\right)^2\)

tương tự: \(5y^2+6yz+5z^2\ge4\left(y+z\right)^2\) ;\(5z^2+6xz+5z^2\ge4\left(x+z\right)^2\)

\(\Rightarrow P\ge\dfrac{2\left(x+y\right)}{x+y+2z}+\dfrac{2\left(y+z\right)}{y+z+2x}+\dfrac{2\left(x+z\right)}{x+z+2y}\)

\(\Leftrightarrow\dfrac{P}{2}\ge\dfrac{x+y}{x+y+2z}+\dfrac{y+z}{y+z+2x}+\dfrac{x+z}{x+z+2y}\)

\(\Leftrightarrow\dfrac{P}{2}\ge\dfrac{x+y}{\left(x+z\right)+\left(y+z\right)}+\dfrac{y+z}{\left(x+y\right)+\left(x+z\right)}+\dfrac{x+z}{\left(x+y\right)+\left(y+z\right)}\)Theo BDT Nesbit

\(\dfrac{x+y}{\left(x+z\right)+\left(y+z\right)}+\dfrac{y+z}{\left(x+y\right)+\left(x+z\right)}+\dfrac{x+z}{\left(x+y\right)+\left(y+z\right)}\ge\dfrac{3}{2}\)

Vậy \(\dfrac{P}{2}\ge\dfrac{3}{2}\Leftrightarrow P\ge3\)

Min P = 3 khi x = y = z

Áp dụng BĐT Cauchy :

\(\dfrac{ab^2}{a^2+b^2}\le\dfrac{ab^2}{2ab}=\dfrac{b}{2}\)\(\Leftrightarrow a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{b}{2}\)

\(\Leftrightarrow\dfrac{a^3}{a^2+b^2}\ge a-\dfrac{b}{2}\)

Tương tự : \(\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2}\) ; \(\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\)

Cộng ba BĐT lại theo vế theo vế

\(\Rightarrow\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge\left(a+b+c\right)-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\left(x+y\right)^2+\left(x-y\right)^2+\dfrac{3}{\left(x+y\right)^2}=\dfrac{85}{3}\\\left(x+y\right)+\left(x-y\right)+\dfrac{1}{x+y}=\dfrac{13}{3}\end{matrix}\right.\)

\(a=x+y\); \(b=x-y\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a^2+b^2+\dfrac{3}{a^2}=\dfrac{85}{3}\\a+b+\dfrac{1}{a}=\dfrac{13}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\left(a+\dfrac{1}{a}\right)^2-6+b^2=\dfrac{85}{3}\\a+\dfrac{1}{a}=\dfrac{13}{3}-b\end{matrix}\right.\)

\(\Rightarrow3\left(\dfrac{13}{3}-b\right)^2-6+b^2=\dfrac{85}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}b=1\\b=\dfrac{11}{2}\end{matrix}\right.\)đến đây tự làm nha

Áp dụng BĐT Cauchy

\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ac\right)\ge9abc\)

\(\Rightarrow\sqrt{\dfrac{\left(a+b+c\right)\left(ab+bc+ac\right)}{abc}}\ge3\)

\(\Rightarrow P\ge3+\dfrac{4bc}{\left(b+c\right)^2}\)

Ta cần tìm Min của \(3+\dfrac{4bc}{\left(b+c\right)^2}\)

Không mất tính tổng quát giả sử \(b\ge c\)

\(\Rightarrow b+c\le2b\)\(\Leftrightarrow\left(b+c\right)^2\le4b^2\Leftrightarrow\dfrac{4bc}{\left(b+c\right)^2}\ge\dfrac{c}{b}\)

\(b\ge c\Rightarrow\dfrac{c}{b}\ge1\)

Vậy \(3+\dfrac{4bc}{\left(b+c\right)^2}\ge4\)

Dấu đẳng thức xảy ra khi a = b = c

\(\dfrac{x^2}{y^2}+\dfrac{y^2}{z^2}\ge\dfrac{2x}{z}\); \(\dfrac{y^2}{z^2}+\dfrac{z^2}{x^2}\ge\dfrac{2y}{x}\); \(\dfrac{x^2}{y^2}+\dfrac{z^2}{x^2}\ge\dfrac{2z}{y}\)

Cộng ba vế bđt sau đó chia 2 ta được đpcm

Giả sử \(x_0\) là nghiệm chung của hai phương trình

\(\Rightarrow\left\{{}\begin{matrix}x^2_0+ax_0+b=0\left(1\right)\\x^2_0+cx_0+d=0\left(2\right)\end{matrix}\right.\)

Lấy (1) - (2)

\(\Rightarrow x_0\left(a-c\right)=-\left(b-d\right)\)

\(\Leftrightarrow x_0^2\left(a-c\right)^2=\left(b-d\right)^2\) (*)

Lấy (1) nhân c; (2) nhân a

\(\Rightarrow\left\{{}\begin{matrix}cx^2_0+acx_0+bc=0\left(3\right)\\ax^2_0+acx_0+ad=0\left(4\right)\end{matrix}\right.\)

Lấy (4) -(3)

\(\Rightarrow\left(a-c\right)x^2_0+ad-bc=0\Leftrightarrow\left(a-c\right)x^2_0=-\left(ad-bc\right)\)

\(\Leftrightarrow\left(a-c\right)^2x^2_0=-\left(a-c\right)\left(ad-bc\right)\)(**)

Từ (*) và (**) ta được

\(\left(b-d\right)^2=-\left(a-c\right)\left(ad-bc\right)\Leftrightarrow\left(b-d\right)^2+\left(a-c\right)\left(ad-bc\right)\)