Bài này mik giải đc rồi nha !

\(\left(2x-12\right)+\left(8x-19\right)-\left(9x+16\right)=4\)

\(\Leftrightarrow2x-12+8x-19-9x-16=4\)

\(\Leftrightarrow x=51\)Vậy x = 51 

Ta có: \(\left(2x-12\right)+\left(8x-19\right)-\left(9x+16\right)=4\)

     \(\Leftrightarrow2x-12+8x-19-9x-16=4\)

     \(\Leftrightarrow2x+8x-9x=4+12+16+19\)

     \(\Leftrightarrow x=51\)

Vậy............

\(P=\frac{2n-1}{n-1}=\frac{2\left(n-1\right)+1}{n-1}=\frac{1}{n-1}\)

hay \(n-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

Ta có: \(P=\frac{2n-1}{n-1}\)

    \(\Leftrightarrow P=\frac{2n-2+1}{n-1}\)

    \(\Leftrightarrow P=2+\frac{1}{n-1}\)

Để \(P\inℤ\)\(\Rightarrow\)\(2+\frac{1}{n-1}\inℤ\)mà \(2\)nguyên 

\(\Rightarrow\)\(1⋮n-1\)\(\Rightarrow\)\(n-1\inƯ\left(1\right)\in\left\{\pm1\right\}\)

+ \(n-1=1\)\(\Leftrightarrow\)\(n=2\)

+ \(n-1=-1\)\(\Leftrightarrow\)\(n=0\)

Vậy ......

Ta có: \(B=\frac{3^{2021}-3^{2019}}{3^{2021}+3^{2020}}\)

    \(\Leftrightarrow B=\frac{3^{2019}.\left(3^2-1\right)}{3^{2020}.\left(3+1\right)}\)

    \(\Leftrightarrow B=\frac{8}{3.4}\)

    \(\Leftrightarrow B=\frac{2}{3}\)

Vậy \(B=\frac{2}{3}\)

Ta có : \(B=\frac{3^{2021}-3^{2019}}{3^{2021}+3^{2020}}\)

\(=\frac{3^{2019}\left(3^2-1\right)}{3^{2020}\left(3+1\right)}\)

\(=\frac{3^{2019}.8}{3^{2020}.4}\)

\(=\frac{2}{3}\)

\(4\sqrt{33}\)và \(\sqrt{29}\sqrt{14}\)

Theo bài ra ta có : 

\(4\sqrt{33}=\sqrt{4^2.33}=\sqrt{16.33}=\sqrt{528}\)

theo công thức sau : \(A\sqrt{B}=\sqrt{A^2.B}\)

mà \(\sqrt{29}\sqrt{14}=\sqrt{29.14}=\sqrt{406}\)

theo công thức sau : \(\sqrt{A}\sqrt{B}=\sqrt{AB}\)

vì 528 > 406 \(\Rightarrow\sqrt{528}>\sqrt{406}\)

hay \(4\sqrt{33}>\sqrt{29}\sqrt{14}\)

Ta có : \(33< 36\)

\(\Rightarrow\sqrt{33}< \sqrt{36}=6\)

\(\Rightarrow4\sqrt{33}< 4.6=24\)

Ta lại có : \(14>9\)

\(\Rightarrow\sqrt{14}>\sqrt{9}=3\)

\(\Rightarrow29\sqrt{14}>29.3=87>24\)

Suy ra : \(4\sqrt{33}< 29\sqrt{14}\)

a.Ta có:

ˆBID=12ˆBIC=12(180o−ˆBCI−ˆIBC)=12(180o−12ˆBCA−12ˆABC)=12(180o−12(ˆBCA+ˆABC)=12(180o−12(180o−ˆBAC)=60oBID^=12BIC^=12(180o−BCI^−IBC^)=12(180o−12BCA^−12ABC^)=12(180o−12(BCA^+ABC^)=12(180o−12(180o−BAC^)=60o 

Lại có :

ˆNIB=ˆIBC+ˆICB=12ˆABC+12ˆACB=12(ˆABC+ˆACB=12(180o−ˆBAC)=60oNIB^=IBC^+ICB^=12ABC^+12ACB^=12(ABC^+ACB^=12(180o−BAC^)=60o

→ˆNIB=ˆBID→NIB^=BID^

→ΔNIB=ΔDIB(g.c.g)→ΔNIB=ΔDIB(g.c.g)

→BN=BD→BN=BD

b.Chứng minh tương tự câu a

→CD=CM→CD=CM

→BN+CM=BD+CD=BC→đpcm→BN+CM=BD+CD=BC→đpcm