\(\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\\ =\dfrac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\dfrac{3\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\\ =\dfrac{24\cdot24+24\cdot23-23\cdot1}{24+24\cdot23+23\cdot23}\cdot\dfrac{1}{3}\\ =\dfrac{23\left(24-1\right)+24\cdot24}{24\left(1+23\right)+23\cdot23}\cdot\dfrac{1}{3}=\dfrac{23\cdot23+24\cdot24}{24\cdot24+23\cdot23}\cdot\dfrac{1}{3}\\ =1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)

\(9^2=\left(3^2\right)^2=3^4\) mà bn :))

\(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot\left(3^2\right)^2}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-\left(5^2\right)^5\cdot\left(7^2\right)^2}{\left(5^3\cdot7\right)^3+5^9\cdot\left(2\cdot7\right)^3}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\\ =\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\\ =\dfrac{2}{9}-\dfrac{-6}{1+8}=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)

Áp dụng t/c của dãy tỉ số bằng nhau , ta có 

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+1+x+z+x+y-3}{x+y+z}\\ =\dfrac{2x+2y+2z}{x+y+z}=2\)

\(\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\end{matrix}\right. \Rightarrow\left\{{}\begin{matrix}x+y+z+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\end{matrix}\right.\)(*)

+, Từ đầu bài `=>(y+z+1)/x=1/(x+y+z)`

mà `(y+z+1)/x=2`

nên `1/(x+y+z)=2`

`=>x+y+z=1/2`

Thay `x+y+z=1/2` vào `(**)` , ta đc :

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\)

thay `x=1/2;y=5/6;z=-5/6` vào `A`, ta đc :

\(A=2022\cdot\dfrac{1}{2}+\left(\dfrac{5}{6}+\dfrac{-5}{6}\right)^{2023}\\ =1011+0=1011\)

`a)`

Có `Delta ABC` cân tại `A=>hat(B_1)=hat(C_1);AB=AC`

Có `hat(B_1)+hat(ABM)=180^0` ( kề bù )

`hat(C_1)+hat(ACN)=180^0` (kề bù)

mà `hat(B_1)=hat(C_1)(cmt)`

nên `hat(ABM)=hat(ACN)`

Xét `Delta ABM` và `Delta ACN` có :

`AB=C(cmt)`

`hat(ABM)=hat(ACN)(cmt)`

`BM=CN(GT)`

`=>Delta ABM=Delta ACN(c.g.c)(đpcm)`

`b)` 

Có `Delta ABM=Delta ACN(cmt)=>hat(A_1)=hat(A_2)` ( 2 góc t/ứng )

Xét `Delta AHB` và `Delta AKC` có :

`hat(AHB)=hat(AHC)(=90^0)`

`AB=AC(cmt)`

`hat(A_1)=hat(A_2)(cmt)`

`=>Delta AHB=Delta AKC(c.h-g.n)(đpcm)`