a)

\(5+\left(-7\right)+9+\left(-11\right)+13+\left(-15\right)\)

\(=\left[5+\left(-7\right)\right]+\left[9+\left(-11\right)\right]+\left[13+\left(-15\right)\right]\)

\(=\left(-2\right)+\left(-2\right)+\left(-2\right)=-6\)

b)

\(\left(-6\right)+8+\left(-10\right)+12+\left(-14\right)+16\)

\(=\left[\left(-6\right)+8\right]+\left[\left(-10\right)+12\right]+\left[\left(-14\right)+16\right]\)

\(=2+2+2=6\)

trả lời giúp em

ĐKXĐ: \(x\ge-2;y\ge-11\)

\(x\left(x+21\right)+y\left(x-33\right)=2\left(y^2+50\right)\)

\(\Leftrightarrow x^2+\left(y+21\right)x-2y^2-33y-100=0\)

\(\Delta=\left(y+21\right)^2+4\left(2y^2+33y+100\right)=\left(3y+29\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-y-21+3y+29}{2}=y+4\\x=\dfrac{-y-21-3y-29}{2}=-2y-25\end{matrix}\right.\)

TH1: \(x=-2y-25\Rightarrow x+2y=-25\)

Mà \(x+2y\ge-2+2.\left(-11\right)=-23>-25\)

\(\Rightarrow\) Pt vô nghiệm

TH2:  \(x=y+4\) thay vào pt dưới:

\(\sqrt{y+6}+2\sqrt{y+11}=\sqrt{\left(3y+10\right)^3}\)

\(\Leftrightarrow\sqrt{y+6}-2+2\sqrt{y+11}-6=\sqrt{\left(3y+10\right)^3}-8\)

\(\Leftrightarrow\dfrac{y+2}{\sqrt{y+6}+2}+\dfrac{2\left(y+2\right)}{\sqrt{y+11}+3}=\dfrac{3\left(y+2\right)\left(3y+14+2\sqrt{3y+10}+4\right)}{\sqrt{3y+10}+2}\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-2\Rightarrow x=2\\\dfrac{1}{\sqrt{y+6}+2}+\dfrac{2}{\sqrt{y+11}+3}=\dfrac{3\left(3y+14+2\sqrt{3y+10}+4\right)}{\sqrt{3y+10}+2}\left(1\right)\end{matrix}\right.\)

Xét (1), ta có:

\(\dfrac{1}{\sqrt{y+6}+2}+\dfrac{2}{\sqrt{y+11}+3}< \dfrac{1}{2}+\dfrac{2}{3}< 2\)

\(\dfrac{3\left(3y+14+2\sqrt{3y+10}+4\right)}{\sqrt{3y+10}+2}=\dfrac{3\left(3y+14\right)}{\sqrt{3y+10}+2}+6>2\)

\(\Rightarrow\left(1\right)\) vô nghiệm

Vậy hệ có nghiệm duy nhất \(\left(x;y\right)=\left(2;-2\right)\)

Thừa số tổng quát: \(n^4+4=n^4+4n^2+4-4n^2=\left(n^2+2\right)-4n^2=\left(n^2+2n+2\right)\left(n^2-2n+2\right)\)

\(=\left[\left(n+1\right)^2+1\right]\left[\left(n-1\right)^2+1\right]\)

Thay vào r làm thôi bạn

thanks

co vay thoi cung thanks . De ot

a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)

b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)

c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)

i: =-12*căn 3/2căn 3=-6

h: =72căn 2/12căn 2=6

g: =25căn 12/5căn 6=5căn 2

f: =(15:5)*căn 6:3=3căn 2

d: =-1/2*6*căn 10=-3căn 10

\(a = \left( { - 2} \right).\left( { - 3} \right) = 2.3 = 6\)

\(b = \left( { - 15} \right).\left( { - 6} \right) = 15.6 = 90\)

\(c = \left( { + 3} \right).\left( { + 2} \right) = 3.2 = 6\)

\(d = \left( { - 10} \right).\left( { - 20} \right) = 10.20 = 200\).