=\(\sqrt{2\left(12-6\sqrt{3}\right)}-\sqrt{2\left(28+10\sqrt{3}\right)}\)

=\(\sqrt{2\left(3-\sqrt{3}\right)2}-\sqrt{2\left(5+\sqrt{3}\right)^2}\)

=\(\sqrt{2}\left(3-\sqrt{3}\right)-\sqrt{2}\left(5+\sqrt{3}\right)=\sqrt{2}\left(3-\sqrt{3}-5-\sqrt{3}\right)\)

=\(\sqrt{2}\left(-2-2\sqrt{3}\right)\)=\(-2\sqrt{2}-2\sqrt{6}\)

3:

\(6a+11b-6\left(a+7b\right)\)

\(=6a+11b-6a-42b=-31b⋮31\)

Ta có: \(\left(6a+11b\right)-6\left(a+7b\right)⋮31\)

\(6a+11b⋮31\)

Do đó: \(6\left(a+7b\right)⋮31\)

=>\(a+7b⋮31\)

Ta có: \(\left(6a+11b\right)-6\left(a+7b\right)⋮31\)

\(a+7b⋮31\)

Do đó: \(6a+11b⋮31\)

4:

\(5a+2b⋮17\)

=>\(12\left(5a+2b\right)⋮17\)

=>\(60a+24b⋮17\)

=>\(51a+17b+9a+7b⋮17\)

=>\(17\left(3a+b\right)+\left(9a+7b\right)⋮17\)

mà \(17\left(3a+b\right)⋮17\)

nên \(9a+7b⋮17\)

Ta có: ( x - 2) x ( y + 3) = -13 = (-13) x 1 = (-1) x 13

* Nếu x - 2 = -13 => x = (-13) + 2 = -11

         y + 3 = 1 => y = 1-3 = -2 

* Nếu x-2 = -1 => x = (-1) + 2 = 1

         y + 3 = 13 => y = 13 - 3 = 10

Vậy có 2 cặp x;y x;y(-11;-2)

                          x;y(1;10)

Có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

\(\Leftrightarrow\)\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có:

\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}=\frac{x^2-y^2+2z^2}{4-9+2\cdot16}=\frac{108}{27}=4\)

\(\Rightarrow\begin{cases}x=4;x=-4\\y=6;y=-6\\z=8;z=-8\end{cases}\)

Vậy pt có nghiệm là \(\left[\begin{array}{nghiempt}x=4;y=6;z=8\\x=-4;y=-6;z=-8\end{array}\right.\)

@Nguyễn Đình Dũng

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) = k

Từ \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) => \(\begin{cases}x=2k\\y=3k\\z=4k\end{cases}\) => \(\begin{cases}x^2=4k^2\\y^2=9k^2\\2z^2=32k^2\end{cases}\)

mà: x² - y² + 2z² = 108 => 4k² - 9k² + 32k² = 108

=> 27k² = 108

=> k² = 4

=> k = \(\pm\) 2

Rồi bạn tự tìm x,y,z...

\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\\ \Rightarrow x^3=\left(\sqrt[3]{9+4\sqrt{5}}\right)^3+\left(\sqrt[3]{9-4\sqrt{5}}\right)^3+3\sqrt[3]{9+4\sqrt{5}}\sqrt[3]{9-4\sqrt{5}}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)

\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3.x\\ \Rightarrow x^3=18+3x\)

\(y=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

\(y^3=\left(\sqrt[3]{3+2\sqrt{2}}\right)^3+\left(\sqrt[3]{3-2\sqrt{2}}\right)^3+3\sqrt[3]{3+2\sqrt{2}}\sqrt[3]{3-2\sqrt{2}}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)

\(\Rightarrow y^3=3+2\sqrt{2}+3-2\sqrt{2}+3y\)

\(\Rightarrow y^3=6+3y\)

\(P=x^3+y^3-3\left(x+y\right)+1993\)

\(P=18+3x+6+3y-3\left(x+y\right)+1993\)

\(P=2017+3\left(x+y\right)-3\left(x+y\right)\)

\(P=2017\)

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

\(=\left|2\sqrt{3}+3\right|-\left|2\sqrt{3}-3\right|\)

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

\(=6\)

em ko 

uwu

60,392

HOK TOT!!!

=60,392

thi cua bn de qa ta

de tui dai tan 3 trang lun

Ko! Còn mấy câu khác nữa!!!!! Nhưng mấy câu ấy cô trả lời hết rùi!!!!! Còn mấy câu mà mình hỏi mấy bạn là những câu hỏi tư duy, học sinh phải tự làm!!!!!!!

29C

30B

31B