Quy đổi hh thành Fe và O

Đặt nFe=a(mol);nO=b(mol); nSo2=c(mol).

=> \(56a+16b=69.6\)(1)

Bảo toàn e => \(3a-2b-2c=0\)(2)

Kết tủa Fe(OH)3: \(n_{Fe\left(OH\right)_3}=0.7\left(mol\right)\)

=> \(n_{H^+_{\left(du\right)}}=1,55.2-0,7.3=1\left(mol\right)\)

=> \(n_{H_2SO_4\left(pu\right)}=2-0,5=1,5\left(mol\right)\)

=> \(b+2c=1,5\) (3)

Từ (1)(2) (3)

=> \(\left\{{}\begin{matrix}a=0,9\left(mol\right)\\b=1,2\left(mol\right)\\c=0,15\left(mol\right)\end{matrix}\right.\)=> \(V=3,36\left(l\right)\)

Với \(-1\le sinx,cosx\le1\)

ta có \(sin^{10}x\le sin^2x;cos^{18}x\le cos^2x\)

=> \(VT\le sin^2x+cos^2x=1\)

=> đẳng thức xảy ra khi \(\left[{}\begin{matrix}sinx=0\\cosx=0\end{matrix}\right.\)\(\)=> \(x=\frac{k\pi}{2}\left(k\in Z\right)\)

\(\lim\limits_{x\rightarrow-3}\frac{\sqrt{2x+10}-\sqrt[3]{x+11}}{x^3+27}=\lim\limits_{x\rightarrow-3}\frac{\sqrt{2x+10}-2+2-\sqrt[3]{x+11}}{x^3+27}=\lim\limits_{x\rightarrow-3}\frac{\frac{2\left(x+3\right)}{\sqrt{2x+10}+2}+\frac{-3-x}{4+2\sqrt[3]{x+11}+\sqrt[3]{\left(x+11\right)^2}}}{\left(x+3\right)\left(x^2-3x+9\right)}\)

=> \(\lim\limits_{x\rightarrow-3}S=\lim\limits_{x\rightarrow-3}\frac{\frac{2}{\sqrt{2x+10}+2}-\frac{1}{4+2\sqrt[3]{x+11}+\sqrt[3]{\left(x+11\right)^2}}}{x^2-3x+9}=\frac{5}{324}\)

a, Ta có \(\frac{a+b}{a+1}=\frac{\left(a+b\right)\left(a+1\right)-a\left(a+b\right)}{a+1}=a+b-\frac{a\left(a+b\right)}{a+1}\)

Mà \(\frac{1}{a+1}\le\frac{a+1}{4a}\)

=> \(\frac{a+b}{1+a}\ge a+b-\frac{\left(a+1\right)\left(a+b\right)}{4}=\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}a^2-\frac{1}{4}ab\)

Khi đó

\(Vt\ge\frac{3}{2}\left(a+b+c\right)-\frac{1}{4}\left(a^2+b^2+c^2\right)-\frac{1}{4}\left(ab+bc+ac\right)\)

=> \(VT\ge\frac{9}{2}-\frac{1}{4}\left(9-2ab-2bc-2ac\right)-\frac{1}{4}\left(ab+bc+ac\right)\)

=> \(VT\ge\frac{9}{4}+\frac{1}{4}\left(ab+bc+ac\right)\)

Lại có \(ab+bc+ac\le\frac{1}{3}\left(a+b+c\right)^2=3\)

=> \(VT\ge ab+bc+ac\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

b,Ta có \(\frac{a}{b\left(a+b^2\right)}=\frac{a+b^2-b^2}{b\left(a+b^2\right)}=\frac{1}{b}-\frac{b}{a+b^2}\)

Mà \(a+b^2\ge2b\sqrt{a}\)

=> \(\frac{a}{b\left(a+b^2\right)}\ge\frac{1}{b}-\frac{1}{2\sqrt{a}}\)

Lại có \(\frac{1}{\sqrt{a.1}}\le\frac{1}{2}\left(\frac{1}{a}+1\right)\)

=> \(\frac{a}{b\left(a+b^2\right)}\ge\frac{1}{b}-\frac{1}{4}.\left(\frac{1}{a}+1\right)\)

Khi đó

\(VT\ge\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)

Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=3\)

=> \(VT\ge\frac{9}{4}-\frac{3}{4}=\frac{3}{2}\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

a, \(x^2+\sqrt{x+2021}=2021\) ĐK \(x\ge-2021\)

<=> \(x^2-2021=-\sqrt{x+2021}\)

Đặt \(\sqrt{x+2021}=a\left(a\ge0\right)\)

=> \(\left\{{}\begin{matrix}x^2-2021=-a\\a^2-2021=x\end{matrix}\right.\)

=> \(\left(x-a\right)\left(x+a\right)+a+x=0\)

<=> \(\left[{}\begin{matrix}x+a=0\\x-a+1=0\end{matrix}\right.\)

+ \(x+a=0\)

=> \(\sqrt{x+2021}=-x\)

=> \(\left\{{}\begin{matrix}x\le0\\x^2-x-2021=0\end{matrix}\right.\)=> \(x=\frac{1-7\sqrt{165}}{2}\)

+ \(x-a+1=0\)

=> \(x+1=\sqrt{x+2021}\)

=> \(\left\{{}\begin{matrix}x\ge-1\\x^2+x-2020\end{matrix}\right.\)=> \(x=\frac{-1+\sqrt{8081}}{2}\)

Vậy \(S=\left\{\frac{-1+\sqrt{8081}}{2};\frac{1-7\sqrt{165}}{2}\right\}\)

ĐK \(x\le\frac{-5-\sqrt{41}}{8}\)hoặc \(x\ge\frac{1+\sqrt{5}}{2}\)

Nhân liên hợp 2 vế ta có:

=> \(\left(4x^2+5x-1-4x^2+4x+4\right)=3\left(3x+1\right)\left(\sqrt{4x^2+5x-1}+2\sqrt{x^2-x-1}\right)\)<=> \(3\left(3x+1\right)=3\left(3x+1\right)\left(\sqrt{4x^2+5x-1}+2\sqrt{x^2-x-1}\right)\)

<=>\(\left[{}\begin{matrix}x=-\frac{1}{3}\left(koTMĐKXĐ\right)\\\sqrt{4x^2+5x-1}+2\sqrt{x^2-x-1}=1\left(2\right)\end{matrix}\right.\)

Kết hợp (2) với PT ban đầu ta có:

=> \(2\sqrt{4x^2+5x-1}=9x+4\)

=> \(\left\{{}\begin{matrix}x\ge-\frac{4}{9}\\4\left(4x^2+5x-1\right)=81x^2+72x+16\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x\ge-\frac{4}{9}\\65x^2+52x+20=0\end{matrix}\right.\)

=> PT vô nghiệm

Vậy PT vô nghiệm

Ta có \(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2=x+y+z\left(1\right)\)

Áp dụng bất đẳng thức buniacoxki ta có

\(\left(x^3+y^3+z^3\right)\left(x+y+z\right)\ge\left(x^2+y^2+z^2\right)^2\)

Kết hợp với (1)=> \(x^3+y^3+z^3\ge x^2+y^2+z^2\left(2\right)\)

\(\left(x^4+y^4+z^4\right)\left(x^2+y^2+z^2\right)\ge\left(a^3+b^3+c^3\right)^2\)

Kết hợp với (2)=> \(x^4+y^4+z^4\ge x^3+y^3+z^3\)(ĐPCM)

Dấu bằng xảy ra khi x=y=z=1