giúp mik nốt những câu còn lại được không 

12-12=0

\(\left\{{}\begin{matrix}\sqrt{3}x-\sqrt{2}y=1\\\sqrt{2}x+\sqrt{3}y=\sqrt{3}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\sqrt{3}x-\sqrt{2}y=1\left(1\right)\\x=\frac{\sqrt{3}-\sqrt{3}y}{\sqrt{2}}\end{matrix}\right.\)

Thay x vào (1) có:

\(\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{3}y\right)}{\sqrt{2}}-\sqrt{2}y=1\) <=> \(\frac{3-3y-2y}{\sqrt{2}}=1\) <=> \(3-5y=\sqrt{2}\)

<=> \(5y=3-\sqrt{2}\) <=> \(y=\frac{3-\sqrt{2}}{5}\)

=> x=\(\frac{\sqrt{3}-\sqrt{3}.\frac{3-\sqrt{2}}{5}}{\sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}\left(1-\frac{3-\sqrt{2}}{5}\right)=\frac{\sqrt{3}}{\sqrt{2}}.\frac{2+\sqrt{2}}{5}=\frac{\sqrt{3}}{\sqrt{2}}.\frac{\sqrt{2}\left(1+\sqrt{2}\right)}{5}=\frac{\sqrt{3}\left(1+\sqrt{2}\right)}{5}\)

C1: Áp dụng bđt Côsi:

\(B=a+a+\frac{1}{8a^2}+\frac{7}{8a^2}\ge3\sqrt[3]{a.a.\frac{1}{8a^2}}+\frac{7}{8.\left(\frac{1}{2}\right)^2}=5\)

Dấu bằng xảy ra khi \(a=\frac{1}{2}\)

ai làm giải thích hộ mình cái

a, P>0

Có \(P^2=x+2\sqrt{x\left(2-x\right)}+2-x=2+2\sqrt{2x-x^2}=\sqrt{1-\left(x^2-2x+1\right)}+2=2+\sqrt{1-\left(x-1\right)^2}\)

Luôn có: \(1-\left(x-1\right)^2\le1\)=> \(0\le\sqrt{1-\left(x-1\right)^2}\le1\)<=> \(0\le2\sqrt{1-\left(x-1\right)^2}\le4\)

<=> \(2\le2+2\sqrt{1-\left(x-1\right)^2}\le2+2\)

<=> \(2\le P^2\le4\)

<=> \(\sqrt{2}\le P\le2\)(do P>0)

minP xảy ra <=> \(\sqrt{1-\left(x-1\right)^2}=0\)

<=> \(\left(x-1\right)^2=1\) <=> \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)(t/m)

maxP xảy ra<=> \(\sqrt{1-\left(x-1\right)^2}=1\)

<=> \(\left(x-1\right)^2=0\) <=> x=1(t/m)

Số đó là 1125. K mk~ mk k lại!!! @gmail.com.vn!!!! ^_^

98 + 2000 = 2098 

nha  Viên Đạn Lạc

y phải là 4 vì 4 chia 5 dư 4 và chia hết cho 2

 ta có: 5+1+4= 10

vậy x=5