a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

GT => (a+1)(b+1)(c+1)=(a+1)+(b+1)+(c+1)

Đặt \(\frac{1}{a+1}=x,\frac{1}{1+b}=y,\frac{1}{c+1}=z\), ta cần tìm min của\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}\)với xy+yz+zx=1

\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\Leftrightarrow\frac{2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)Mà  (x+y)(y+z)(z+x) >= 8/9 (x+y+z)(xy+yz+xz) >= \(\frac{8\sqrt{3}}{9}\) nên \(M\)=< \(\frac{3\sqrt{3}}{4}\),dấu bằng xảy ra khi a=b=c=\(\sqrt{3}-1\)

Theo giả thiết, ta có: \(abc+ab+bc+ca=2\)

\(\Leftrightarrow abc+ab+bc+ca+a+b+c+1=a+b+c+3\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(a+1\right)+\left(b+1\right)+\left(c+1\right)\)

\(\Leftrightarrow\frac{1}{\left(a+1\right)\left(b+1\right)}+\frac{1}{\left(b+1\right)\left(c+1\right)}+\frac{1}{\left(c+1\right)\left(a+1\right)}=1\)

Đặt \(\left(a+1;b+1;c+1\right)\rightarrow\left(\frac{\sqrt{3}}{x};\frac{\sqrt{3}}{y};\frac{\sqrt{3}}{z}\right)\). Khi đó giả thiết bài toán được viết lại thành xy + yz + zx = 3 

Ta có: \(M=\Sigma_{cyc}\frac{a+1}{a^2+2a+2}=\Sigma_{cyc}\frac{a+1}{\left(a+1\right)^2+1}\)\(=\Sigma_{cyc}\frac{1}{a+1+\frac{1}{a+1}}=\Sigma_{cyc}\frac{1}{\frac{\sqrt{3}}{x}+\frac{x}{\sqrt{3}}}\)

\(=\sqrt{3}\left(\frac{x}{x^2+3}+\frac{y}{y^2+3}+\frac{z}{z^2+3}\right)\)

\(=\sqrt{3}\text{​​}\Sigma_{cyc}\left(\frac{x}{x^2+xy+yz+zx}\right)=\sqrt{3}\Sigma_{cyc}\frac{x}{\left(x+y\right)\left(x+z\right)}\)

\(\le\frac{\sqrt{3}}{4}\Sigma_{cyc}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)=\frac{3\sqrt{3}}{4}\)

Đẳng thức xảy ra khi \(x=y=z=1\)hay \(a=b=c=\sqrt{3}-1\)

đây nha bạn

Áp dụng bổ đề quen thuộc \(x^3+y^3\ge xy\left(x+y\right)\), ta được: \(\frac{1}{2a^3+b^3+c^3+2}=\frac{1}{\left(a^3+b^3\right)+\left(a^3+c^3\right)+2}\le\frac{1}{ab\left(a+b\right)+ac\left(a+c\right)+2}\)\(=\frac{bc}{ab^2c\left(a+b\right)+abc^2\left(a+c\right)+2bc}=\frac{bc}{b\left(a+b\right)+c\left(a+c\right)+2bc}\)\(\le\frac{bc}{ab+ac+4bc}=\frac{bc}{b\left(a+c\right)+c\left(a+b\right)+2bc}\)\(\le\frac{1}{9}\left(\frac{bc}{b\left(a+c\right)}+\frac{bc}{c\left(a+b\right)}+\frac{bc}{2bc}\right)=\frac{1}{9}\left(\frac{c}{a+c}+\frac{b}{a+b}+\frac{1}{2}\right)\)(1)

Tương tự, ta có: \(\frac{1}{a^3+2b^3+c^3+2}\le\frac{1}{9}\left(\frac{c}{b+c}+\frac{a}{a+b}+\frac{1}{2}\right)\)(2); \(\frac{1}{a^3+b^3+2c^3+2}\le\frac{1}{9}\left(\frac{b}{b+c}+\frac{a}{a+c}+\frac{1}{2}\right)\)(3)

Cộng theo vế ba bất đẳng thức (1), (2), (3), ta được: \(P\le\frac{1}{9}\left(1+1+1+\frac{3}{2}\right)=\frac{1}{2}\)

Vậy giá trị lớn nhất của P là \(\frac{1}{2}\)đạt được khi x = y = z = 1

A=-3..check mk nhá

Có: 2a² + 2b² = 5ab => 2(a² + b²) = 5ab => a² + b² = \(\frac{5}{2}\)ab 

\(A=\frac{2b}{a-b}+1=\frac{2b+a-b}{a-b}=\frac{a+b}{a-b}=\frac{\left(a+b\right)^2}{\left(a-b\right)^2}=\frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{\frac{5}{2}ab+2ab}{\frac{5}{2}ab-2ab}=\frac{\frac{9}{2}ab}{\frac{1}{2}ab}=9\)

Vậy A = 9

Nhận xét: vì b>a>0 nên a-b < 0  và 2b>a-b nên A<0.

Bình phương hai vế: 

A² = 9. nên A = 3; A = -3. Đáp số: A = -3

a) Có:

 \(a+b+c=0\\\Leftrightarrow\left(a+b+c\right)^2=0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\\ \Leftrightarrow2ab+2bc+2ca=-1\\ \Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\\ \Leftrightarrow\left(ab+bc+ca\right)^2=\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}-0=\dfrac{1}{4} \)

câu (b) cho đa thức P (x) = cái gì?

Chào thanh niên lâu lắm mới on à

Thay a=-3b vào M 

\(DK.a\ne0;b\ne0\)

\(M_b=\frac{2a+b}{a-b}-\frac{2a-b}{a+2b}=\frac{-6b+b}{-3b-b}-\frac{-6b-b}{-3b+2b}=\frac{5}{4}-\frac{-7}{-1}=-\frac{23}{4}\)

Ta có :\(\left(a-b\right)^2\ge0\forall a;b\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow2a^2+2b^2\ge a^2+b^2+2ab\Leftrightarrow2a^2+2b^2\ge\left(a+b\right)^2\)

Suy ra \(\frac{2011}{2a^2+2b^2+2008}\le\frac{2011}{\left(a+b\right)^2+2008}=\frac{2011}{4+2008}=\frac{2011}{2012}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=1\)