Tham khảo:

Mỗi loài, mỗi giống, mỗi cá thể có hệ gen (vật chất di truyền) khác nhau. Nên dù ăn cùng một loại thức ăn khi tiêu hóa có thể thu được các loại axit amin giống nhau nhưng tổng hợp prôtêin theo hệ gen khác nhau do đó prôtêin của chúng khác nhau-> tính chất khác nhau. mà ta biết prôtêin có tính đặc thù rất cao vì số lượng, thành phần, trật tự sắp xếp các aa.

Thay x = 116, y = 16, z = 2 vào biểu thức, ta được:

\(116\left(2.116-2\right)+16\left(2-2.16\right)=116.230+16.\left(-30\right)=26200\)

\(1,\left(2x-5\right)^2=\dfrac{9}{16}\)

\(\Leftrightarrow\left|2x-5\right|=\dfrac{3}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=\dfrac{3}{4}\\2x-5=-\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{8}\\x=\dfrac{17}{8}\end{matrix}\right.\)

\(2,\left(1-2x\right)^3=-\dfrac{27}{64}\)

\(\Leftrightarrow1-2x=-\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{7}{8}\)

\(3,\left(3x+\dfrac{1}{2}\right)^4=0\)

\(\Leftrightarrow3x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x=-\dfrac{1}{6}\)

\(4,\left(x+\dfrac{3}{4}\right)^3+5=-3\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right)^3=-8\)

\(\Leftrightarrow x+\dfrac{3}{4}=-2\)

\(\Leftrightarrow x=-\dfrac{11}{4}\)

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

\(\Leftrightarrow\left|x-\dfrac{1}{5}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=\dfrac{1}{2}\\x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{3}{10}\end{matrix}\right.\)

\(6,2x=\dfrac{84}{163}\)

\(\Leftrightarrow x=\dfrac{42}{163}\)

\(a.2^n=8\)

\(\Leftrightarrow2^n=2^3\)

\(\Leftrightarrow n=3\)

\(b.5^{2n-3}=125\)

\(\Leftrightarrow5^{2n-3}=5^3\)

\(\Leftrightarrow2n-3=3\)

\(\Leftrightarrow n=3\)

\(c.8^n:4=128\)

\(\Leftrightarrow2^{3n}:2^2=128\)

\(\Leftrightarrow2^{3n-2}=2^7\)

\(\Leftrightarrow3n-2=7\)

\(\Leftrightarrow n=3\)

\(d.7^{n-2}.7^4=7^4\)

\(\Leftrightarrow7^{n-2}=1\)

\(\Leftrightarrow n-2=0\)

\(\Leftrightarrow n=2\)

\(Mg\rightarrow\left\{{}\begin{matrix}Mg+2HCl--->MgCl_2+H_2\uparrow\\Mg+H_2SO_4--->MgSO_4+H_2\uparrow\end{matrix}\right.\)

\(Al\rightarrow\left\{{}\begin{matrix}2Al+6HCl--->2AlCl_3+3H_2\uparrow\\2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3\downarrow+3H_2\uparrow\end{matrix}\right.\)

\(CaO\rightarrow\left\{{}\begin{matrix}CaO+2HCl--->CaCl_2+H_2O\\CaO+H_2SO_4--->CaSO_4\downarrow+H_2O\end{matrix}\right.\)

\(Al_2O_3\rightarrow\left\{{}\begin{matrix}Al_2O_3+6HCl--->2AlCl_3+3H_2O\\Al_2O_3+3H_2SO_4--->Al_2\left(SO_4\right)_3\downarrow+3H_2O\end{matrix}\right.\)

\(K_2O\rightarrow\left\{{}\begin{matrix}K_2O+2HCl--->2KCl+H_2O\\K_2O+H_2SO_4--->K_2SO_4+H_2O\end{matrix}\right.\)

\(KOH\rightarrow\left\{{}\begin{matrix}KOH+HCl--->KCl+H_2O\\2KOH+H_2SO_4--->K_2SO_4+2H_2O\end{matrix}\right.\)

\(Mg\left(OH\right)_2\rightarrow\left\{{}\begin{matrix}Mg\left(OH\right)_2+2HCl--->MgCl_2+2H_2O\\Mg\left(OH\right)_2+H_2SO_4--->MgSO_4+2H_2O\end{matrix}\right.\)

\(Al\left(OH\right)_3\rightarrow\left\{{}\begin{matrix}Al\left(OH\right)_3+3HCl--->AlCl_3+3H_2O\\2Al\left(OH\right)_3+3H_2SO_4--->Al_2\left(SO_4\right)_3\downarrow+6H_2O\end{matrix}\right.\)

\(CaCO_3\rightarrow\left\{{}\begin{matrix}CaCO_3+2HCl--->CaCl_2+CO_2\uparrow+H_2O\\CaCO_3+H_2SO_4--->CaSO_4\downarrow+CO_2\uparrow+H_2O\end{matrix}\right.\)

\(K_2S\rightarrow\left\{{}\begin{matrix}K_2S+2HCl--->2KCl+H_2S\uparrow\\K_2S+H_2SO_4--->K_2SO_4+H_2S\uparrow\end{matrix}\right.\)

\(5x\left(x-3\right)=3-x\)

\(\Leftrightarrow5x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(5x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)

\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{1}{4}=\dfrac{6}{12}+\dfrac{8}{12}+\dfrac{3}{12}=\dfrac{17}{12}\)

thiếu nghiệm e nhé

\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{4}\\x-\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

thiếu nghiệm e nhé