\(A=\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\)

\(\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\right)\)

\(=\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{4}{4}=1\)

M N P Q 8 12 10

Xét tam giác MNP có NQ là tia phân giác ^MNP nên 

\(\frac{NM}{NP}=\frac{MQ}{QP}\)mà \(MQ=MP-QP=5-QP\)(1) 

hay \(\frac{8}{12}=\frac{5-QP}{QP}\Rightarrow8QP=60-12QP\)

\(\Leftrightarrow20QP=60\Leftrightarrow QP=3\)cm 

suy ra (1) \(MQ=5-3=2\)cm 

Vậy QP = 3 cm ; MQ = 2cm 

Ta có NQ là ta phân giác 

\(\Rightarrow\)MQ=PQ mà MQ+PQ=MP =10 cm

\(\Rightarrow\)MQ=PQ=10:2=5(CM)

Vậy ...........

\(-4x^2-3x+5=0\)

\(\Leftrightarrow4x^2+3x-5=0\)

\(\Leftrightarrow4\left(x^2+\frac{3}{4}x-\frac{5}{4}\right)=0\)

\(\Leftrightarrow x^2+\frac{3}{4}x-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x^2+2.x.\frac{3}{8}+\frac{9}{64}\right)-\frac{89}{64}=0\)

\(\Leftrightarrow\left(x+\frac{3}{8}\right)^2=\frac{89}{64}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{8}=\frac{\sqrt{89}}{8}\\x+\frac{3}{8}=\frac{-\sqrt{89}}{8}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{89}-3}{8}\\x=\frac{-\sqrt{89}-3}{8}\end{cases}}\)

Vậy phương trình có tập nghiệm : \(S=\left\{\frac{\pm\sqrt{89}-3}{8}\right\}\)

\(x^4-y^2+2x^3+2x^2+x+3=0\)

\(\Leftrightarrow\left(x^4+2x^3+x^2\right)+\left(x^2+x\right)=y^2-3\)

\(\Leftrightarrow\left(x^2+x\right)^2+\left(x^2+x\right)=y^2-3\)

\(\Leftrightarrow\left(2x^2+2x+1\right)^2=4y^2-11\)

\(\Leftrightarrow\left(2x^2+2x+1\right)^2-\left(2y\right)^2=11\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x^2+2x+1\right)^2=25\\\left(2y\right)^2=26\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1;x=-2\\y=\pm3\end{cases}}\)

oke bạn

oke bạn