Ta có : \(\left|x-y\right|+\left(y-3\right)^2\le0\) ( 2 )

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

\(\Rightarrow\left|x-y\right|+\left(y-3\right)^2\ge0\) ( 1 )

- Thấy từ 1 và 2 chỉ xảy ra : \(\left|x-y\right|+\left(y-3\right)^2=0\)

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

\(\Rightarrow x=y=3\)

Vậy ...

\(\Leftrightarrow x-5=-4\)

\(\Leftrightarrow x=-4+5=1\)

Vậy ...

- Thời gian họ đi từ A đến B là : 3,75 giờ .

Mà họ nghỉ trên đường 25 phút .

=> Thời gian đi xe là : 10/3 giờ .

Vậy quãng đường AB dài là : \(S=vt=\dfrac{10}{3}.57=190\left(km\right)\)

Bài 5 :

a, ĐKXĐ ; \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có : \(P=1:\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=1:\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=1:\left(\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=1:\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

b, - Xét \(P-3=\dfrac{x+\sqrt{x}+1-3\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)

\(\Rightarrow P>3\)

\(A=4x^2+12xy+9y^2\)

\(B=25x^2-10xy+y^2\)

\(C=8x^3+12x^2y^2+6xy^4+y^6\)

\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)

\(E=x^3-27y^3\)

\(F=x^6-27\)

\(a,\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)

\(=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{132}.\dfrac{132}{103}=\dfrac{115}{103}\)

\(b,\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2-5^0\)

\(=\dfrac{1}{4}-4-1=-\dfrac{19}{4}\)

\(c,12\dfrac{1}{3}-\dfrac{5}{7}:\left(24-23\dfrac{5}{7}\right)\)

\(=\dfrac{37}{3}-\dfrac{5}{7}.\dfrac{2}{7}=\dfrac{37}{3}-\dfrac{10}{49}=\dfrac{1783}{147}\)

Ta có : \(x^2-4x+4+y^2=0\)

\(\Leftrightarrow\left(x-2\right)^2+y^2=0\)

Thấy : \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\y^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^2+y^2\ge0\)

- Dấu " = " xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\y=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

- Xem lại đề có thiếu số cuối không nha bạn ;-;