\(\frac{a+b-c}{c}=\frac{a+b}{c}-1\)

\(\frac{a+c-b}{b}=\frac{a+c}{b}-1\)

\(\frac{b+c-a}{a}=\frac{b+c}{a}-1\)

\(\Rightarrow\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}\)

\(\Rightarrow\)\(M=\left(\frac{a+b}{c}\right)^3=\left(\frac{a+c}{b}\right)^3=\left(\frac{b+c}{a}\right)^3\)

Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\left(k\in Q\right)\)

\(\Rightarrow\hept{\begin{cases}x=2k+1\\y=3k+2\\z=4k+3\end{cases}}\)Thay vào đẳng thức \(2x+3y-z=95\)ta được:

\(2.\left(2k+1\right)+3.\left(3k+2\right)-\left(4k+3\right)=95\)

\(\Rightarrow4k+2+9k+6-4k-3=95\)

\(\Rightarrow9k+5=95\)

\(\Rightarrow9k=90\)

\(\Rightarrow k=10\)

Suy ra\(\hept{\begin{cases}x=21\\y=32\\z=43\end{cases}}\)

Vậy \(x=21;y=32;z=43\)

CHÚC BN HOK TỐT HA

Đề nghị bạn kiểm tra lại đề mình thấy khi

\(\frac{a}{b}=\frac{1}{2};\frac{b}{c}=\frac{2}{4};\frac{c}{d}=\frac{4}{8}\)

thì thay vào bị sai

Yasuo gánh team 15' gg

A = 1 + 2 + 2^2 + 2^3 + ... + 2^100

2A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^101

2A - A = A = ( 2 + 2^2 + 2^3 + 2^4 + ... + 2^101 ) - ( 1 + 2 + 2^2 + 2^3 + ... + 2^100 )

A = 2^101 - 1

Vì 2^101 - 1 < 2^101 nên A < B hay B > A 

Ta có:

\(A=2^0+2^1+2^2+2^3+...+2^{100}\)

\(A=1+2+2^2+2^3+...+2^{100}\)

\(2A=2+2^2+2^3+2^4+...+2^{101}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{101}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(A=2^{101}-1\left(1\right)\)

\(B=2^{101}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\)suy ra:\(A< B\)

Vậy \(A< B\)

CHÚC BN HOK TỐT NHA