à ừ đấy, cj làm nhanh để rep tin nhắn của em àm quên mất

Ta có:

\(P\left(x\right)+Q\left(x\right)=x^5-2x^2+1\)

\(\Rightarrow Q\left(x\right)=P\left(x\right)-\left(x^5-2x^2+1\right)\)

\(=x^4-3x^2+\dfrac{1}{2}-x-x^5+2x^2-1\)

\(=-x^5+x^4-x^2-x-\dfrac{1}{2}\)

Vậy \(Q\left(x\right)=-5^2+x^4-x^2-x-\dfrac{1}{2}\)

\(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)

\(=-5,13:\left(5\dfrac{5}{28}-2\dfrac{13}{36}+1\dfrac{16}{63}\right)\)

\(=5,13:\left[\left(5-2+1\right)+\left(\dfrac{5}{28}-\dfrac{13}{36}+\dfrac{16}{63}\right)\right]\)

\(=5,13:\left(4+\dfrac{1}{14}\right)\)

\(=1,26\)

Nếu \(x>\dfrac{1}{2}\) , ta có:

\(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\Rightarrow\dfrac{7}{2}:\left(2x-1\right)=\dfrac{21}{22}\Rightarrow x=\dfrac{7}{3}\left(tm\right)\)

Nếu \(x< \dfrac{1}{2}\), ta có:

\(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\Rightarrow\dfrac{7}{2}:\left(1-2x\right)=\dfrac{21}{22}\Rightarrow-2x=\dfrac{8}{3}\Rightarrow x=-\dfrac{4}{3}\left(tm\right)\)

Vậy \(x=\dfrac{7}{3};x=\dfrac{4}{3}\)

Vì \(0\le a\le b\le c\le1\) nên:

\(\left(a-1\right)\left(b-1\right)\ge ab+1\ge a+b\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\left(1\right)\)

Tương tự: \(\dfrac{a}{bc+1}\le\dfrac{a}{b=c}\left(2\right);\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\left(3\right)\)

Do đó: \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\left(4\right)\)

Mà: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\le\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(5\right)\)

Từ (4) và (5) suy ra \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\left(đpcm\right)\)

\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

\(=\dfrac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)

\(=\dfrac{2^{12}.3^4.2}{2^{12}.3^5.4}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)

\(=\dfrac{1}{6}-\dfrac{-10}{3}\)

\(=\dfrac{7}{2}\)

Xét \(x=0\Rightarrow y=0,z=0\Rightarrow2y+4z=0\) (vô lý)

Suy ra \(x\ne0;y\ne0;z\ne0\)

Khi đó từ đề suy ra:

\(\dfrac{2y+4x}{xy}=\dfrac{4z+6y}{yz}=\dfrac{6x+2z}{zx}=\dfrac{2^2+4^2+6^2}{x^2+y^2+z^2}\)

\(\Rightarrow\dfrac{2}{x}+\dfrac{4}{y}=\dfrac{4}{y}+\dfrac{6}{z}=\dfrac{6}{z}+\dfrac{2}{x}=\dfrac{2^2+4^2+6^2}{x^2+y^2+z^2}\)

Đặt \(\dfrac{2}{x}=\dfrac{4}{y}=\dfrac{6}{z}=\dfrac{1}{k}\left(k\ne0\right)\) thì \(\dfrac{2^2+4^2+6^2}{x^2+y^2+z^2}=\dfrac{2}{k}\)

Suy ra: \(x=2k;y=4k;z=6k\) và \(x^2+y^2+z^2=28k\left(3\right)\)

Thay \(x=2k,y=4k,z=6k\) vào (3) ta được:

\(\left(2k\right)^2+\left(4k\right)^2+\left(6k\right)^2=28k\)

\(\Rightarrow56k^2-28k=0\Rightarrow\left[{}\begin{matrix}k=0\left(ktm\right)\\k=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)

Với \(k=\dfrac{1}{2}\Rightarrow x=1;y=2;z=3\)

Vậy \(x=1;y=2;z=3\)