Để \(B=\frac{1}{2\left(n-1\right)^2+3}\) đạt GTLN <=> \(2\left(n-1\right)^2+3\) đạt GTNN

Vì \(\left(n-1\right)^2\ge0\) \(\forall\) \(n\) \(\Rightarrow2\left(n-1\right)^2\ge0\) \(\forall\) \(n\)

\(\Rightarrow2\left(n-1\right)^2+3\ge3\) \(\forall\) \(n\)

Dấu "=" xảy ra <=> \(2\left(n-1\right)^2=0\Rightarrow n=1\)

Vật GTLN của B là \(\frac{1}{3}\) tại n = 1

taco

|x+1|\(\ge\)0

|x+2|\(\ge\)0

|x+3|\(\ge\)0

\(\Rightarrow\)4x\(\ge\)0

\(\Rightarrow\)x\(\ge\)0

tu do ta co        x+1+x+2+x+3=4x

3x+6=4x

4x-3x=6

x=6

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=0\)

\(\Leftrightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\frac{329}{5}=0\)

\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{5}\ne0\Rightarrow x+329=0\Rightarrow x=-329\)

Vậy x = - 329

Ta có:

\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\)\(\left(\frac{x+349}{5}-4\right)=0\)

\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

do số trong ngoặc khác 0 nên

pt <=> x+329=0

<=> x=-329

a) ta có: \(AB^2+AC^2=24^2+32^2=40^2=BC^2\)

=> theo Pitago đảo thì tam giác ABC vuông tại A

b) Ta có: MC=AC-AM=32-7=25

\(\Delta ABM\)vuông tại A có: \(AM^2+AB^2=MB^2\)=> MB=\(\sqrt{AM^2+AB^2}=\sqrt{7^2+24^2}=25\)

Do đó: MB=MC => \(\Delta MBC\)cân tại M

=> \(\widehat{MBC}=\widehat{MCB}\)

Mặt khác \(\widehat{AMB}\)là góc ngoài \(\Delta MBC\)nên: \(\widehat{AMB}\)=\(\widehat{MBC}+\widehat{MCB}=2\widehat{MCB}\)(ĐPCM)

=\(\frac{2^{12}.3^5+2^{12}.3^4}{2^{12}.3^6+2^{12}.3^3}\)

=\(\frac{2^{12}\left(3^5+3^4\right)}{2^{12}\left(3^6+3^3\right)}\)

\(=\frac{324}{756}\)

=\(\frac{3}{7}\)

\(\frac{AB}{AC}=\frac{5}{12}\Leftrightarrow\frac{AB}{5}=\frac{AC}{12}=\frac{AC-AB}{12-5}=\frac{14}{7}=2\)

=> \(\hept{\begin{cases}AB=2.5=10\\AC=2.12=24\end{cases}}\)

Áp dụng Pitago => \(BC=\sqrt{AB^2+AC^2}=\sqrt{10^2+24^2}=26\)

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