\(\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}.3+9}-2\sqrt{3}\)

\(\sqrt{\left(2\sqrt{3}-3\right)^2}-2\sqrt{3}\)

\(2\sqrt{3}-3-2\sqrt{3}\)

= -3

Bình phương 2 cái đấy ta có

x⁶ - 6x⁴y² + 9x²y⁴ = 100

y⁶ - 6x²y⁴ + 9x⁴y² = 900

Cộng vế theo vế được

x⁶ + 3x²y⁴ + 3x⁴y² + y⁶ = 1000

<=> (x² + y²)³ = 1000

<=> x² + y² = 10

MN ƠI GIÚP MK NHA

bài 2

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{a+c}{b}=>a=b=c\)

=>x=2 => A=(4-2+1)^2016 =3^2016

=(\(\dfrac{99}{2}+1+\dfrac{98}{3}+1+...+\dfrac{1}{100}+1\)):(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\)) -2

=(\(\dfrac{101}{2}+\dfrac{101}{3}+...\dfrac{101}{100}\)):(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\)) -2

=101(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\)):(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\))-2

=101 -2 =99

-_-

\(\left\{{}\begin{matrix}\sqrt{a}=x\ge0\\\sqrt{b}=y\ge0\end{matrix}\right.\) \(\Rightarrow x+y=1\)

\(P=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=x^2+y^2-xy\)

\(P=\left(x+y\right)^2-3xy=1-3xy\)

Do \(\left\{{}\begin{matrix}x\ge0\\y\ge0\end{matrix}\right.\) \(\Rightarrow xy\ge0\Rightarrow P\le1\Rightarrow P_{max}=1\) khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\) hay \(\left(a;b\right)=\left(1;0\right);\left(0;1\right)\)

Lại có \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\Rightarrow P\ge1-3.\frac{1}{4}=\frac{1}{4}\)

\(\Rightarrow P_{min}=\frac{1}{4}\) khi \(x=y=\frac{1}{2}\) hay \(a=b=\frac{1}{4}\)

\(\Rightarrow m^2+n^2=1+\frac{1}{16}=\frac{17}{16}\)