250 và 40

250>40

=> 2.5^3> 5.2^3

a) 12x³ : 4x = (12:4) . (x³ : x) = 3.x²

b) (-2x⁴ ) : x⁴ = [(-2) : 1] . (x⁴ : x⁴) = -2

c) 2x⁵ : 5x² = (2:5) . (x⁵ : x²) = \(\frac{2}{5}\)x³

a) ta thay : \(2^{10}=2^{10}\)

\(3^{20}=\left(3^2\right)^{10}=9^{10}\)

vi \(2< 9\)nen \(2^{10}< 9^{10}\)

vay \(2^{10}< 3^{20}\)

b) \(2.5^3=250\)

\(5.2^3=40\)

vi  \(250>40\) nen \(2.5^3>5.2^3\)

vay \(2.5^3>5.2^3\)

câu A đúng ko vậy bạn

À câu A là 3^30 và 3^30 nha

a. \(A=x^5-3x^3+x^2-x^3-3+2x=x^5-4x^3+x^2+2x-3\)

\(B=x^4-3x-2+5x^2-3x^4+2x^5=2x^5-2x^4+5x^2-3x-2\)

b. \(A+B=x^5-4x^3+x^2+2x-3+2x^5-2x^4+5x^2-3x-2\)

\(=3x^5-2x^4-4x^3+6x^2-x-5\)

lũy thừa giảm dần hay tăng dần v ạ

a) 2³⁰ = (2³)¹⁰ = 8¹⁰

3²⁰ = (3²)¹⁰ = 9¹⁰

Vì 8¹⁰ < 9¹⁰ => 2³⁰ < 3²⁰

b) 10²⁰ = 2²⁰.5²⁰ = 2²⁰.5¹⁰.5¹⁰

5.2³⁰ = 5.2¹⁰.2²⁰

Vì 5¹⁰.5¹⁰ > 5.2¹⁰ => 10²⁰ > 5.2³⁰

c) 2.5³ > 2.4³ = 2.(2²)³ = 2.2⁶ = 2⁷ = 2³.2⁴ = 2³.16 > 2³.5

=> 2.5³ > 5.2³

d) 27¹¹ = (3³)¹¹ = 3³³

81⁸ = (3⁴)⁸ = 3³²

Vì 3³³ > 3³² => 27¹¹ > 81⁸

e) 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰

2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰

Vì 9¹⁰⁰ > 8¹⁰⁰ => 3²⁰⁰ > 2³⁰⁰

Câu e hình như sai đề nên mk sửa lại, nếu ko sai đề thì lm thế này

Vì 3 > 2; 2000 > 300 => 3²⁰⁰⁰ > 2³⁰⁰

\(\Leftrightarrow\left(x-3\right)\left(2x^2+2\right)+5x\left(3-x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\)

=>(x-3)(x-2)(2x-1)=0

=>x=3 hoặc x=2 hoặc x=1/2

\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2+2\right)-5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left[\left(2x^2-4x\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left[2x\left(x-2\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\)

\(\Leftrightarrow2x^3+2x-6x^2-6+15x-5x^2=0\)

\(\Leftrightarrow2x^3-11x^2+17x-6=0\)

\(\Leftrightarrow2x^3-4x^2-7x^2+14x+3x-6=0\)

\(\Leftrightarrow2x^2\left(x-2\right)-7x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^2-7x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-6x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-1\right)-3\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{2;\dfrac{1}{2};3\right\}\)

\(\Leftrightarrow\left(x^4+x^3-3x^2\right)-\left(x^3+x^2-3x\right)-\left(x^2+x-3\right)< 0\)

\(\Leftrightarrow x^2\left(x^2+x-3\right)-x\left(x^2+x-3\right)-1\left(x^2+x-3\right)< 0\)

\(\Leftrightarrow\left(x^2-x-1\right)\left(x^2+x-3\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1-\sqrt{13}}{2}< x< \dfrac{1-\sqrt{5}}{2}\\\dfrac{-1+\sqrt{13}}{2}< x< \dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)

\(y+5x2-\left(32+16x3:6-15\right)=0\)

\(y+10-\left(32+8-15\right)=0\)

\(y+10-\left(40-15\right)=0\)

\(y+10-25=0\)

\(y=25-10\)

\(y=15\)

Lời giải:

a) $0,2x^2+0,4x-7=0$

$\Leftrightarrow 2x^2+4x-70=0$

$\Leftrightarrow x^2+2x-35=0$

$\Leftrightarrow (x-5)(x+7)=0$

$\Rightarrow x=5$ hoặc $x=-7$

b) 

$\frac{1}{2}x^2+11x+60,5=0$

$\Leftrightarrow x^2+22x+121=0$

$\Leftrightarrow (x+11)^2=0\Leftrightarrow x=-11$

c) 

$5x^2+\sqrt{3}-1=0$

$\Leftrightarrow 5x^2=1-\sqrt{3}< 0$ (vô lý)

Vậy  PT vô nghiệm.