4a² + 9b² - 20a + 6b + 26 = 0 <=> ( 2a - 5 )² + ( 3b + 1 )² = 0 <=> a = 5/2 ; b = -1/3

5a² + b² - 2a + 4ab + 1 = 0 <=> ( 2a + b )² + ( a - 1 )² = 0 <=> a = 1 ; b = -2

1) Ta có 4a² + 9b² - 20a + 6b + 26 = 0

<=> (4a² - 20a + 25) + (9b² + 6b + 1) = 0

<=> (2a - 5)² + (3b + 1)² = 0

<=> \(\hept{\begin{cases}2a-5=0\\3b+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{5}{2}\\b=-\frac{1}{3}\end{cases}}\)

Vậy a = 5/2 ; b = -1/3

2) Ta có 5a² + b² - 2a + 4ab + 1 = 0

<=> (4a² + 4ab + b²) + (a² - 2a + 1) = 0

<=> (2a + b)² + (a - 1)² = 0

<=> \(\hept{\begin{cases}2a+b=0\\a-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}b=-2\\a=1\end{cases}}\)

Vậy b = -2 ;  a = 1

2.

\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)

ĐKXĐ là :

\(a\ne0;-3;-2\)

Vs a = 1 ta có:

=> P=3

1.

\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)

Trả lời đi mn

Áp dụng bất đẳng thức Cauchy - Schwarz ta có:

\(\left(4a^2+9b^2\right)\left(2^2+2^2\right)\ge\left(2a.1-3b.2\right)^2=\left(4a-6b\right)^2=1\)

\(\Rightarrow4a^2+9b^2\ge\dfrac{1}{8}\).

Đẳng thức xảy ra khi \(a=\dfrac{1}{8};b=\dfrac{-1}{12}\).

a   \(2a>b;2a>0\Rightarrow2a+2a>b+0\Rightarrow4a>b\)

b   \(4a^2+b^2=5ab\Rightarrow4a^2+b^2-5ab=0\Rightarrow\left(4a^2-4ab\right)-\left(ab-b^2\right)=0\)

\(\Rightarrow4a\left(a-b\right)-b\left(a-b\right)=0\Rightarrow\left(4a-b\right)\left(a-b\right)=0\Rightarrow\hept{\begin{cases}4a-b=0\Rightarrow4a=b\\a-b=0\Rightarrow a=b\end{cases}}\)

c  \(20=4\cdot5>11\)mà \(2\cdot5=10>11\)đâu 

sai đề r

\(9a^2+12ab+4b^2+a^2+4a+4+b^2-6b+9=0\)

\(\Leftrightarrow\left(3a+2b\right)^2+\left(a+2\right)^2+\left(b-3\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}3a+2b=0\\a+2=0\\b-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-2\\b=3\end{matrix}\right.\)

\(M=\left(-4+3\right)^{2020}=1\)