Ta có: \(x+2\sqrt{2}.x^2+2x^3=0\)

\(\Leftrightarrow x\left(1+2\sqrt{2}.x+2x^2\right)=0\)

\(\Leftrightarrow x\left[1^2+2.x\sqrt{2}.1+\left(x\sqrt{2}\right)^2\right]=0\)

\(\Leftrightarrow x\left(1+x\sqrt{2}\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+x\sqrt{2}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{\sqrt{2}}\end{cases}}\)

Vậy\(x\in\left\{0;\frac{-1}{\sqrt{2}}\right\}\)

\(x+2\sqrt{2}x^2+2x^3=0\)

\(x\left(1+2\sqrt{2}x+2x^2\right)=0\)

\(x\left(2\sqrt{2}x+1\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2\sqrt{2}x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2x\sqrt{2}}\end{cases}}\)

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b. (x-2)(x+15)=0

x-2=0 hoặc x+15=0

x=2 hoặc x=-15

a. (x-2)(x+4)=0

x-2=0 hoặc x+4=0

x=2 hoặc x=-4

g. (x-3)(x-5)<0

\(\begin{cases}x-3>0\\x-5< 0\end{cases}\)=>\(\begin{cases}x>3\\x< 5\end{cases}\)=> 3<x<5 Vậy x= 4

a.

\(\left(x-2\right)\times\left(x+4\right)=0\)

               \(x=2\)

               \(x=-4\)

Vậy x = 2 hoặc x = - 4.

b.

\(\left(x-2\right)\times\left(x+15\right)=0\)

               \(x=2\)

                 \(x=-15\)

Vậy x = 2 hoặc x = - 15.

c.

\(\left(7-x\right)\times\left(x+19\right)=0\)

               \(x=7\)

                 \(x=-19\)

Vậy x = 7 hoặc x = -19.

d.

\(-5< x< 1\)

\(x\in\left\{-4;-3;-2;-1;0\right\}\)

e.

\(\left|x\right|< 3\)

\(\left|x\right|\in\left\{0;1;2\right\}\)

\(x\in\left\{-2;-1;0;1;2\right\}\)

Chúc bạn học tốt

e. /x/<3 => -3<x<3

Vậy x=-2;-1;0;1;2

d. -5<x<1

=> x= -4;-3;-2;-1;0

c. (7-x)(x+19)=0

7-x=0 hoặc x+19=0

x=7 hoặc x=-19

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

TH1: \(x=0\)

TH2: \(x^2-1=0\Rightarrow x^2=1\Rightarrow x=\sqrt{1}\)hoặc \(x=-\sqrt{1}\)

<=>\(\left(x^3-4x^2\right)+\left(x^2-4x\right)+\left(5x-20\right)=0\)

<=>\(x^2\left(x-4\right)+x\left(x-4\right)+5\left(x-4\right)=0\)

<=>\(\left(x^2+x+5\right)\left(x-4\right)=0\)

Vì \(x^2+x+5>0\)=>x-4=0

<=>x=4

\(a,A=\left\{0;1;2;3;4\right\}\\ b,B=\left\{-16;-13;-10;-7;-4;-1;2;5;8\right\}\\ c,C=\left\{-9;-8;-7;...;7;8;9\right\}\\ d,x^2-3x+1=0\\ \Delta=9-4=5\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{5}}{2}\\x=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\\ \Leftrightarrow D=\left\{\dfrac{3-\sqrt{5}}{2};\dfrac{3+\sqrt{5}}{2}\right\}\)

\(e,2x^3-5x^2+2x=0\\ \Leftrightarrow x\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow E=\left\{0;2\right\}\\ f,F=\left\{0;3;6;9;12;15;18\right\}\)

a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)

\(\Rightarrow x=0\)

c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)

\(\Rightarrow P_{max}=4\) khi \(x=0\)

 x³ - x² - x = 1/3 
<=> x³ = x² + x + 1/3 
<=> 3x³ = 3(x² + x + 1/3) 
<=> 3x³ = 3x² + 3x + 1 
<=> 3x³ + x³ = x³ + 3x² + 3x + 1 
<=> 4x³ = (x + 1)³ 
<=> ³√(4x³) = ³√(x + 1)³ 
<=> ³√4.x = x + 1 
<=> ³√4.x - x = 1 
<=> x(³√4 - 1) = 1 
<=> x = 1/(³√4 - 1) 

Ta có \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)

\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)

\(\Rightarrow x^2+2x-3=x^2-4\)

\(\Rightarrow x^2-x^2+2x=-4+3\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\frac{1}{2}\)

Vậy \(x=-\frac{1}{2}\)

=>20x+190=950

=>20x=760

hay x=38

`20x+190=950`

`20x=760`

`x= 760: 20`

`x= 38`

20 x X = 950 - (1 + 2 + 3 + 4 + 5 + ... + 19)

20 x X = 950 - 190

20 x X = 760

X = 760 : 20

X = 38