a) \(\left(x-3\right)\left(x-2\right)< 0\)

Ta có : \(x-2>x-3\)

\(\Rightarrow\left\{{}\begin{matrix}x-3< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x>2\end{matrix}\right.\Rightarrow2< x< 3\)

Vậy \(2< x< 3\)

b) \(3x+x^2=0\)

\(x\left(3+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\3+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{-3;0\right\}\)

a+1/2=c+2/4=c+1/2=>a=c=>3a=3c

b+2/3=c+2/4=c+1/2=>b=c+1/2-2/3=c-1/6=>2b=2c-1/3

3a-2b+c=3c-2c+1/3+c=2c+1/3=105

=>2c=314/3=>c=157/3

b=c-1/6=157/3-1/6=313/6

a=c=157/3

Dù kh hiểu gì Nhưng mình camon cậu ạ

\(\)\(A=2^0+2^1+2^2+2^3+...+2^{2012}\\ A=1+2+\left(2^2+2^3+2^4\right)+\left(2^5+2^6+2^7\right)+...+\left(2^{2010}+2^{2011}+2^{2012}\right)\\ A=3+2^2\cdot\left(1+2+2^2\right)+2^5\cdot\left(1+2+2^2\right)+...+2^{2010}\cdot\left(1+2+2^2\right)\\ A=3+2^2\cdot\left(1+2+4\right)+2^5\cdot\left(1+2+4\right)+...+2^{2010}\cdot\left(1+2+4\right)\\ A=3+2^2\cdot7+2^5\cdot7+...+2^{2010}\cdot7\\ A=3+7\cdot\left(2^2+2^5+...+2^{2010}\right)\\ \)

Cô giải rồi lên đây giải làm j nữa.

2. GTLN

có A= x - |x|

Xét x >= 0 thì A= x - x = 0 (1)

Xét x < 0 thì A=x - (-x) = 2x < 0 (2)

Từ (1) và (2) => A =< 0

Vậy GTLN của A bằng 0 khi x >= 0

Bài1:

\(C=x^2+3\text{|}y-2\text{|}-1\)

Với mọi x;ythì \(x^2>=0;3\text{|}y-2\text{|}>=0\)

=>\(x^2+3\text{|}y-2\text{|}>=0\)

Hay C>=0 với mọi x;y

Để C=0 thì \(x^2=0\) và \(\text{|}y-2\text{|}=0\)

=>\(x=0vày-2=0\)

=>\(x=0và.y=2\)

Vậy....

\(\left(x-3\right)^2+\left|y^2-9\right|=0\)

Vì \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\forall x\\\left|y^2-9\right|\ge0\forall y\end{matrix}\right.\)

để bt = 0 \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left|y^2-9\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y^2-9=0\Rightarrow y^2=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\\left[{}\begin{matrix}y=3\\y=-3\end{matrix}\right.\end{matrix}\right.\)

Vậy.....

\(\left(x-3\right)^2+\left|y^2-9\right|=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left|y^2-9\right|=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\y^2-9=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\y^2=9\left[{}\begin{matrix}y=3\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=3\\y=3hoặcy=-3\end{matrix}\right.\)

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

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}=\dfrac{a^2+b^2}{b^2+c^2}\)

\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a^2+b^2}{b^2+c^2}\)

\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a^2+b^2}{b^2+c^2}\)

\(\Rightarrow\dfrac{a}{c}=\dfrac{a^2+b^2}{b^2+c^2}\)

Vậy nếu \(\dfrac{a}{b}=\dfrac{b}{c}\) thì \(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\left(đpcm\right)\)

b) Vì 50 > 49 nên \(\sqrt{50}\) > \(\sqrt{49}\) = 7

Vì 2 > 1 nên \(\sqrt{2}\) > \(\sqrt{1}\) = 1

\(\Rightarrow\) \(\sqrt{50}\) + \(\sqrt{2}\) > 7 + 1 = 8 (1)

Ta nhận thấy: 50 + 2 = 52 < 64. \(\Rightarrow\) \(\sqrt{50+2}\) < \(\sqrt{64}\) = 8 (2)

Từ (1) và (2) suy ra ​​​\(\sqrt{50}\) + \(\sqrt{2}\) > \(\sqrt{50+2}\)​

Vậy,...

OK, tôi sẽ giúp bn.

a) Vì 26 > 25 nên \(\sqrt{26}\) > \(\sqrt{25}\) = 5

Vì 17 > 16 nên \(\sqrt{17}\) > \(\sqrt{16}\) = 4

\(\Rightarrow\) \(\sqrt{26}\) + \(\sqrt{17}\) > 5 + 4 = 9

Vậy, \(\sqrt{26}\) + \(\sqrt{17}\) > 9

\(B=\left[\dfrac{1}{100}-1^2\right]\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]\cdot...\cdot\left[\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2\right]\cdot...\cdot\left[\dfrac{1}{100}-\left(\dfrac{1}{120}\right)^2\right]\)

\(=\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{100}\right)\cdot...\cdot\left(\dfrac{1}{100}-\dfrac{1}{14400}\right)\)

=0

mk ko chép đề mà tách luôn nha

M = x²x² + x²x² + x²y² + x²y² + x²y² + y²y² + y²

= ( x²x² + x²y² ) + ( x²x² + x²y² ) + ( x²y² + y²y² ) + y²

= x²( x² + y² ) + x²( x² + y² ) + y²( x² + y² ) + y²

= ( x² + y² ) (x² + x² + y² ) + y²

= 1( x² + 1) + y²

= x² + y² +1 = 2

thanks bn