\(\sqrt{2x+1}< 2=>\sqrt{2x+1}^2< 4\)

2x+1<4 => 2x<3 => x<3/2

=\(\dfrac{2^{19}.27^3-15.\left(-4\right)^9.9^4}{6^9.2^{10}+\left(-12\right)^{10}}\)

=\(\dfrac{2^{19}.3^9-2^{18}.3^9.5}{6^9.2^{10}+6^{10}.2^{10}}\)

=\(\dfrac{2^{18}.3^9\left(2-2.5\right)}{6^9.2^{10}\left(1+6\right)}\)

=\(\dfrac{2^{18}.3^9.\left(-8\right)}{3^9.2^{19}.7}\)

=\(\dfrac{-8}{14}=\dfrac{-4}{7}\)

1. a) 625/5n=5³ => 5n=625/5³=5⁴/5³=5 =>n=1

b) (-2n)/-128=4 =>-2n=4.(-128)=-2.256 =>n=256

c) (3/7)ⁿ=81/2401=(3/7)⁴ => n=4

2. 32<2ⁿ<512

<=> 2⁵<2ⁿ<2⁹

=> n=6;7;8

3. (x-1)⁴=16=2⁴ => x-1=2 =>x=3

a) a+b=S => (a+b)²=S² => a²+2ab+b²=S² mà ab=P => 2ab=2P

=> a²+2P+b²=S² =>A= a²+b²=S²-2P

b) B=a³+b³=(a+b)(a²-ab+b²)=> B=S.(S²-2P-P)=S(S²-3P)=S³-3S.P

2x²-3x+1=0

<=>(2x²-2x)-(x-1)=0

<=>2x(x-1)-(x-1)=0

<=>(2x-1)(x-1)=0

\(< =>\left\{{}\begin{matrix}2x-1=0=>x=\dfrac{1}{2}\\x-1=0=>x=1\end{matrix}\right.\)

2x²-3x-5=0

<=>(2x²+2x)-(5x+5)=0

<=>2x(x+1)-5(x+1)=0

<=>(2x-5)(x+1)=0

\(< =>\left\{{}\begin{matrix}2x-5=0=>x=\dfrac{5}{2}\\x+1=0=>x=-1\end{matrix}\right.\)

(x+1)²⁰⁰⁶+(y-1)²⁰⁰⁸=0

=> (x+1)²⁰⁰⁶=(y-1)²⁰⁰⁸=0

=>x+1=y-1=0

=>x=-1 và y=1

C=5x¹⁰-y¹⁵+2007=5.(-1)¹⁰-1¹⁵+2007=2011

Giải

Đổi:  0,09 tạ = 9 kg;  2,7 tấn = 2700kg

Tổng số sư tử và hổ là: 2700 : 20 : 9 = 15 con.

Giải bài toán "tổng tỉ"

Số sư tử là: 15 : (2 +3) x2 = 6 con

Số hổ là: 15 - 6  = 9 con

ĐS: sư tử: 6 con; Hổ: 9 con

(-2/3+3/7):4/5+(-1/3+4/7):4/5=(-2/3+3/7-1/3+4/7):4/5

=0:4/5=0

\(\dfrac{x-2}{5}=\dfrac{x}{3}=>3\left(x-2\right)=5x\)

\(< =>3x-6=5x=>x=-3\)

\(\dfrac{x+23}{x+40}=\dfrac{3}{4}=>4\left(x+23\right)=3\left(x+40\right)\)

\(4x+92=3x+120=>x=28\)

(5x-1)(2x-1/3)=0

=>\(\left\{{}\begin{matrix}5x-1=0=>x=\dfrac{1}{5}\\2x-\dfrac{1}{3}=0=>x=\dfrac{1}{6}\end{matrix}\right.\)

Vậy x=\(\dfrac{1}{5};\dfrac{1}{6}\)