x=5

y=\(\frac{1}{4}\)

x=5

\(\frac{1+4y}{18}=\frac{1+5y}{24}\Rightarrow24+96y=18+90y\)

\(\Rightarrow6+6y=0\Leftrightarrow6\left(1+y\right)=0\)Vậy y = -1

Thay y = -1 ta có :

\(\frac{1-5}{24}=\frac{1-6}{6x}\Leftrightarrow\frac{-5}{30}=-\frac{5}{6x}\left(\frac{-4}{24}=-\frac{5}{30}=\frac{1-5}{24}\right)\)

Vậy 6x = 30 hay x = 5 

\(\frac{a^2-2ab}{a^2b}.P=\frac{a^2b-4b^3}{3ab^2}\)

\(P=\frac{a^2b-4b^3}{3ab^2}:\frac{a^2-2ab}{a^2b}\)

\(P=\frac{a^2b-4b^3}{3ab^2}.\frac{a^2b}{a^2-2ab}\)

\(P=\frac{b\left(a^2-4b^2\right)}{3ab^2}.\frac{a^2b}{a\left(a-2b\right)}\)

\(P=\frac{b\left(a-2b\right)\left(a+2b\right)}{3ab^2}.\frac{a^2b}{a\left(a-2b\right)}\)

\(P=\frac{b\left(a+2b\right)}{3b}.\frac{a}{a}\)

\(P=\frac{a+2b}{3}\)

P=\(\frac{a^2b.b\left(a^2-4b^2\right)}{3ab^2.a\left(a-2b\right)}=\frac{a^2b^2\left(a-2b\right)\left(a+2b\right)}{3a^2b^2\left(a-2b\right)}\)

=> P=\(\frac{a+2b}{3}\)

Ta có: 1+2y/18=1+4y/24  

=>  24(1+2y)=18(1+4y)   

  =>24+48y=18+72y 

   =>24-18=72y-48y   

  =>6=24y   

=>  y=1/4

    thay y=1/4 vào đề ta có: 

(1+ 1/2)/18=1+1/24=(1+3/2)/6x  

  =>1/12=(5/2)/6x  

  => 12(5/2)=6x 

   =>30=6x  

=>x=5                                  

Vậy x=5 

   y=1/4

a) Biến đổi VT . Mẫu chung là ( a + 2b )( a - 2b )

\(VT=\frac{a+2b-6b-2\left(a-2b\right)}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 1 )

Biến đổi VP 

\(-\frac{1}{2a}\left(\frac{a^2+4b^2}{a^2-4b^2}+1\right)=-\frac{1}{2a}\cdot\frac{a^2+4b^2+a^2-4b^2}{a^2-4b^2}\)

\(=-\frac{1}{2a}\cdot\frac{2a^2}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP ( đpcm )

b) \(a^3+b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)^3\)

<=> \(b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)^3=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)-a^3\)( * )

Biến đổi VT của ( * ) ta có :

\(VT=\left[b+\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right]\left[b^2-\frac{b^2\left(2a^3+b^3\right)}{a^3-b^3}+\frac{b^2\left(2a^3+b^3\right)^2}{\left(a^3-b^3\right)^2}\right]\)

\(=\frac{3a^3b}{a^3-b^3}\cdot\frac{3a^6b^2+3a^3b^5+3b^8}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 1 )

\(VP=\left[\frac{a\left(a^3+2b^3\right)}{a^3-b^3}-a\right]\left[\frac{a^2\left(a^3+2b^3\right)^2}{\left(a^3-b^3\right)^2}+\frac{a^2\left(a^3+2b^3\right)}{a^3-b^3}+a^2\right]\)

\(=\frac{3ab^3}{a^3-b^3}\cdot\frac{3a^8+3a^5b^3+3a^2b^6}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP => ( * ) đúng 

=> Hằng đẳng thức đúng 

Đặt \(\left(\frac{1}{2a+1};\frac{1}{2b+1};\frac{1}{2c+1}\right)=\left(x;y;z\right)\Rightarrow x+y+z\ge1\)

Mặt khác do \(a;b;c>0\Rightarrow x;y;z< 1\)

Ta có: \(P=\frac{x}{3-2x}+\frac{y}{3-2y}+\frac{z}{3-2z}\)

Ta có đánh giá: \(\frac{x}{3-2x}\ge\frac{27x-2}{49}\) \(\forall x\in\left(0;1\right)\)

\(\Leftrightarrow9x^2-6x+1\ge0\Leftrightarrow\left(3x-1\right)^2\ge0\) (luôn đúng)

Thiết lập tương tự và cộng lại:

\(P\ge\frac{27\left(x+y+z\right)-6}{49}\ge\frac{21}{49}=\frac{3}{7}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=1\)