(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2.................999 x 2 x 2 x 2 ?

A.(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2 > 999 x 2 x 2 x 2 = 9990 > 7992

B.(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2 < 999 x 2 x 2 x 2 = 9990 < 7992

C.(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2 = 999 x 2 x 2 x 2 = 9990 = 7992

Đáp án A nhé

(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2.................999 x 2 x 2 x 2 ?

A.(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2 > 999 x 2 x 2 x 2 = 9990 > 7992

B.(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2 < 999 x 2 x 2 x 2 = 9990 < 7992

C.(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2 = 999 x 2 x 2 x 2 = 9990 = 7992

Trả lời đúng , mình sẽ cho 5 vote

\(\Rightarrow x-\left(-x+x+3\right)-\left(x+3-x+2\right)=0\\ \Rightarrow x-3-5=0\\ \Rightarrow x=8\)

Sai rồi bạn ơi, bằng -2 mới đúng,bn thử lại coi

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

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

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

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

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

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

Thôi nhầm chỗ HĐT rồi , bạn làm giống bạn Lê Tài Bảo Châu  nhé ! NHầm

=>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

 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 \(x=-\frac{1}{2}\)

nhầm rồi

1cong1=

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

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