Ta có:\(\left|5a-6b+300\right|^{2007}=0\Leftrightarrow5a-6b+300=0\Leftrightarrow5a=-300+6b\)

\(\left|2a-3b\right|^{2008}=0\Leftrightarrow2a-3b=0\Leftrightarrow2a=3b\Leftrightarrow a=\frac{3}{2}b\)

\(\Rightarrow5\cdot\frac{3}{2}b=-300+6b\)

\(\Rightarrow7,5b=-300+6b\)

\(\Rightarrow1,5b=-300\)

\(\Rightarrow b=-200\)

\(\Rightarrow a=-300\)

bai nay de ma ban .Ban **** minh di minh lam cho 

\(\left|5a-6b+300\right|^{2007}+\left(2a-3b\right)^{2008}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}5a-6b+300=0\\2a-3b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a-6b=-300\\2a-3b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-300\\b=-200\end{matrix}\right.\)

Ta có:

\(\left|5a-6b+300\right|^{2011}\ge0\forall a,b\)

\(\left(2a-3b\right)^{2010}\ge0\forall a,b\)

\(\Rightarrow\left|5a-6b+300\right|^{2011}+\left(2a-3b\right)^{2010}\ge0\forall a,b\)

Mặt khác: \(\left|5a-6b+300\right|^{2011}+\left(2a-3b\right)^{2010}=0\)

nên: \(\left\{{}\begin{matrix}5a-6b+300=0\\2a-3b=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5a-6b=-300\\2\cdot\left(2a-3b\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5a-6b=-300\\4a-6b=0\end{matrix}\right.\)

\(\Rightarrow5a-6b-\left(4a-6b\right)=-300-0\)

\(\Rightarrow5a-6b-4a+6b=-300\)

\(\Rightarrow a=-300\)

Khi đó: \(2\cdot\left(-300\right)-3b=0\)

\(\Rightarrow-3b=600\)

\(\Rightarrow b=-200\)

Vậy \(a=-300;b=-200\)

\(\text{#}Toru\)

\(\left|5a-6b+300\right|^{2011}>=0\forall a,b\)

\(\left(2a-3b\right)^{2010}>=0\forall a,b\)

Do đó: \(\left|5a-6b+300\right|^{2011}+\left(2a-3b\right)^{2010}>=0\forall a,b\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}5a-6b+300=0\\2a-3b=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5a-6b=-300\\2a-3b=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5a-6b=-300\\4a-6b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-300\\3b=2a\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=-300\\b=\dfrac{2}{3}a=\dfrac{2}{3}\cdot\left(-300\right)=-200\end{matrix}\right.\)

Lời giải:
Ta thấy: 
$|5a-6b+300|^{2011}\geq 0, \forall a,b$ (tính chất trị tuyệt đối) 

$(2a-3b)^{2022}\geq 0, \forall a,b$

Do đó để tổng của chúng bằng $0$ thì:

$|5a-6b+300|=2a-3b=0$

$\Leftrightarrow 5a-6b=-300(*)$ và $2a=3b$

$2a=3b\Rightarrow a=1,5b$. Thay vào đk $(*)$ thì:

$5.1,5b-5b=-300$

$\Rightarrow 7,5b-5b=-300$

$\Rightarrow 2,5b=-300\Rightarrow b=-120$

$a=1,5b=1,5(-120)=-180$

Ta có \(\hept{\begin{cases}\left|2a-3b+500\right|^{2021}\ge0\forall a;b\\\left(5a-6b\right)^{2020}\ge0\forall a;b\end{cases}}\Rightarrow\left|2a-3b+500\right|^{2021}+\left(5a-6b\right)^{2020}\ge0\forall a;b\)

Dấu "=" xảy ra <=> 

\(\hept{\begin{cases}2a-3b=500\\5a-6b=0\end{cases}}\Rightarrow\hept{\begin{cases}4a-6b=1000\\5a-6b=0\end{cases}}\Rightarrow\hept{\begin{cases}a=-1000\\b=-\frac{2500}{3}\end{cases}}\)

Vậy a = -1000 ; b = -2500/3 là giá trị cần tìm

Ko bít Tự làm

=>5a-6a+300=0 và 2a-3b=0

=>a=300 và 3b=2a=600

=>a=300; b=200