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Làm ơn viết cái đề rõ hơn dc ko vậy?

-_- Viết ra đi cậu. Khó nhìn chết được.

\(2014a^2+b^2+c^2\) / \(a^2\) = \(a^2+2014b^2+c^2\) /b\(^2\) = \(a^2+b^2+2014c^2\) /c\(^2\)

P = \(2015a^2+b^2\) /c\(^2\) + \(2015b^2\) +\(c^2\) / a\(^2\) + 2015\(c^2+a^2\)/b\(^2\)

1. \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

\(VP=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2\)

\(\Rightarrow VT=VP\)

2. \(a^4-b^4=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\)

\(VP=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)=\left(a^2-b^2\right)\left(a^2+b^2\right)=a^4+a^2b^2-b^2a^2-b^4=a^4-b^4\)

\(\Rightarrow VT=VP\)

3. \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(bx+ay\right)^2\)

\(VT=\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(VP=\left(ax-by\right)^2+\left(bx+ay\right)^2=a^2x^2-2axby+b^2y^2+b^2x^2+2bxay+a^2y^2=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(\Rightarrow VT=VP\)

1; 7³.5².5⁴.7⁶:(5⁵.7⁸)

= (7³.7⁶).(5².5⁴) : (5⁵.7⁸)

= 7⁹.5⁶: (5⁵.7⁸)

= (7⁹:7⁸).(5⁶:5⁵)

= 7.5

= 35

2; 3³.a⁷.3.a²:(3⁴.a⁶)

= (3³.3).(a⁷.a²): (3⁴.a⁶)

= 3⁴.a⁹: (3⁴.a⁶)

= (3⁴:3⁴).(a⁹:a⁶)

= a³

3; 7³.11⁴.a⁸.b⁷: 7².11².a⁵.b⁶

=  (7³:7²).(11⁴.11²).(a⁸.a⁵).(b⁷.b⁶)

= 7.11⁶.a¹³.b¹³

Ta có công thức tổng quát như sau:

\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)

Áp dụng ta có:

\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\) 

\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)

______

\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)

\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)

_____

\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)

\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)

b) Đặt \(\frac{a}{5}=\frac{b}{2}=\frac{c}{3}=k\Rightarrow\hept{\begin{cases}a=5k\\b=2k\\c=3k\end{cases}}\)

Khi đó a² + b² + c² = 38

<=> (5k)² + (2k)² + (3k)² = 38

=> 25k² + 4k² + 9k² = 38

=> 38k² = 38

=> k² = 1

=> k = \(\pm\)1

Khi k = 1

=> a = 5 ; b = 2 ; c = 3

Khi k = -1

=> a = -5 ; b = -2 ; c = -3

Vậy các cặp số (a;b;c) thỏa mãn bài toán là : (5;2;3) ; (-5;-2;-3)

\(\frac{a}{2}=\frac{b}{3}\)và \(a^2+b^2=52\)

\(\Rightarrow\frac{a^2}{2^2}=\frac{b^2}{3^2}\)và \(a^2+b^2=52\)

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

\(\frac{a^2}{2^2}=\frac{b^2}{3^2}=\frac{a^2+b^2}{2^2+3^2}=\frac{52}{13}=4\)

\(\frac{a^2}{2^2}=4\Rightarrow a^2=16\Rightarrow a=\pm4\)

\(\frac{b^2}{3^2}=4\Rightarrow b^2=36\Rightarrow b=\pm6\)

Còn bài kia tí mình làm cho :>

a)  đặt \(\frac{a}{2}=\frac{b}{3}=k\)

\(\Rightarrow a=2k;b=3k\)

ta có \(a^2+b^2=52\)

\(\left(2k\right)^2+\left(3k\right)^2=52\)

\(2k.2k+3k.3k=52\)

\(4k^2+9k^2=52\)

\(k^2\left(4+9\right)=52\)

\(k^213=52\)

\(k^2=4\)

\(\Rightarrow k=\pm2\)

do đó 

\(\frac{a}{2}=k\Leftrightarrow\frac{a}{2}=\pm2\Leftrightarrow\hept{\begin{cases}\frac{a}{2}=2\\\frac{a}{2}=-2\end{cases}\Leftrightarrow\hept{\begin{cases}a=2.2=4\\a=2.\left(-2\right)=-4\end{cases}}}\)

\(\frac{b}{3}=k\Leftrightarrow\frac{b}{3}=\pm2\Leftrightarrow\hept{\begin{cases}\frac{b}{3}=2\\\frac{b}{3}=-2\end{cases}\Leftrightarrow\hept{\begin{cases}b=3.2=6\\b=3.\left(-2\right)=-6\end{cases}}}\)

vậy cặp a,b thỏa mãn là \(\left(a=4;b=6\right)\)\(\left(a=-4;b=-6\right)\)