Do \(x+1=\left(x+1\right)^2\)

\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy x=0; -1

Ta có: \(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(\Rightarrow P=\dfrac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(\Rightarrow P=\dfrac{5^{32}-1}{2}\)

Do \(n^2+4⋮n+2\)

\(\Rightarrow\left(n^2+2n\right)-\left(2n+4\right)+8⋮n+2\)

\(\Rightarrow n\left(n+2\right)-2\left(n+2\right)+8⋮n+2\)

\(\Rightarrow\left(n-2\right)\left(n+2\right)+8⋮n+2\)

Lại có \(\left(n-2\right)\left(n+2\right)⋮n+2\)

\(\Rightarrow8⋮n+2\)

Mà \(n\in N\)

\(\Rightarrow n+2\inƯ\left(8\right)=\left\{1;2;4;8\right\}\)

Mặt khác \(n+2\ge2\)

Vậy n=0;2;6

Chỉ cần mình không tưởng tượng nữa thì sẽ thoát khỏi được đàn cá mập vây quanh chiếc thuyền ! 

Ta có: x²+2xy+4x+4y+3y²+3=0

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(4x+4y\right)+2y^2+3=0\)

\(\Leftrightarrow[\left(x+y\right)^2+4\left(x+y\right)+4]+2y^2=1\)

\(\Leftrightarrow\left(x+y+2\right)^2=1-2y^2\)

Do \(y^2\ge0\Rightarrow1-2y^2\le1\)

\(\Rightarrow B^2=\left(x+y+2\right)^2\le1\)

\(\Rightarrow\left\{{}\begin{matrix}B\le1\\B\ge-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}B_{max}=1\\B_{min}=-1\end{matrix}\right.\)

5050 ai tích mình mình tích lại cho

Do \(\dfrac{4^x}{2^{x+y}}=8\)

\(\Rightarrow4^x=8.2^{x+y}\)

\(\Rightarrow\left(2^2\right)^x=2^3.2^{x+y}\)

\(\Rightarrow2^{2x}=2^{x+y+3}\)

\(\Rightarrow2x=x+y+3\)

\(\Rightarrow x=y+3^{\left(1\right)}\)

Mà \(\dfrac{9^{x+y}}{3^{5y}}=243\)

\(\Rightarrow9^{x+y}=243.3^{5y}\)

\(\Rightarrow\left(3^2\right)^{x+y}=3^5.3^{5y}\)

\(\Rightarrow3^{2x+2y}=3^{5+5y}\)

\(\Rightarrow2x+2y=5+5y\)

\(\Rightarrow2x=5+3y^{\left(2\right)}\)

Từ (1) và (2) suy ra: 2(y+3)=5+3y

\(\Rightarrow2y+6=5+3y\)

\(\Rightarrow3y-2y=6-5\)

\(\Rightarrow y=1\)

Thay y=1 vào (1) ta được: x=1+3=4

Vậy x=3;y=1

Do \(\dfrac{a}{c}=\dfrac{c}{b}\)

\(\Rightarrow ab=c^2\)

Ta có: \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b^2-a^2}{a^2+ab}=\dfrac{\left(b^2-ab\right)+\left(ab-a^2\right)}{a\left(a+b\right)}=\dfrac{b\left(b-a\right)+a\left(b-a\right)}{a\left(a+b\right)}=\dfrac{\left(b+a\right)\left(b-a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)

a) x-2=-3

x=-3+2

x=-1

x+4=-2

b) x-2=-6

x=-6+2

x=-4

c) x-(-4)=1

x = 1 + (-4)

x = -3

d) x+8=-5+4

x+8=1

x=1-8

x=-7

a)Ta có:220=2².5.11

240=2⁴.3.5

300=2².3.5²

^{\(\Rightarrow\)}\(ƯCLN\)(220;240;300)=2².5=20

\(BCNN\left(220;240;300\right)=2^4.3.5^2.11=13200\)

b)Ta có:40=2³.5

75=3.5²

106=2.53

\(\RightarrowƯCLN\)(40;75;106)=1

\(BCNN\left(40;75;106\right)=2^3.3.5^2.53=31800\)

c)Ta có:18=2.3²

36=2².3²

72=2³.3²

\(\RightarrowƯCLN\)(18;36;72)=2.3²=\(18\)

\(BCNN\left(18;36;72\right)=2^3.3^2=72\)