\(\left(3x^2-2x+1\right)\left(x^2+2x+3\right)-4x\left(x^2-1\right)-3x^2\left(x^2+2\right)=3x^4+6x^3+9x^2-3x^3-4x^2-6x-4x^3+4x-3x^4-6x^2=0\)

Lời giải:

Từ ĐKĐB suy ra:

$-x^2+5xy+2y^2=3(x^2+y^2)$

$\Leftrightarrow 4x^2-5xy+y^2=0$
$\Leftrightarrow 4x(x-y)-y(x-y)=0$

$\Leftrightarrow (4x-y)(x-y)=0$

$\Rightarrow 4x=y$ hoặc $x=y$.

Nếu $4x=y$. Thay vô PT $(1)$ thì:

$x^2+(4x)^2=1\Rightarrow x=\pm \frac{1}{\sqrt{17}}$

$\Rightarrow x=\pm \frac{4}{\sqrt{17}}$ (tương ứng)

Trường hợp $x=y$ tương tự, ta tìm được $(x,y)=(\pm \frac{1}{\sqrt{2}}; \pm \frac{1}{\sqrt{2}})$

\(\dfrac{2x}{3}-\dfrac{1}{3}=3x-2x+2\Leftrightarrow\dfrac{2x-1}{3}=x+2\)

\(\Rightarrow2x-1=3x+6\Leftrightarrow x=-7\)

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

\(\Rightarrow\dfrac{2}{3}x-\dfrac{1}{3}=3x-2x+2\)

\(\Rightarrow3x-2x-\dfrac{2}{3}x=-\dfrac{1}{3}-2\)

\(\Rightarrow\dfrac{1}{3}x=-\dfrac{7}{3}\Rightarrow x=-7\)

\(\dfrac{2}{3}x-\dfrac{1}{3}=3x-2x+2\)

\(\dfrac{2x}{3}-\dfrac{1}{3}=3x-2x+2\)

\(\dfrac{2x}{3}-\dfrac{1}{3}=x-2\)

\(2x-1=3x+6\)

\(2x-1-3x=6\)

\(2x-3x=6+1\)

\(-x=7\Leftrightarrow x=-7\)

1. \(S=1+3+3^2+3^3+........+3^{2019}+3^{2020}\)

\(\Rightarrow3S=3+3^2+3^3+3^4+........+3^{2020}+3^{2021}\)

\(\Rightarrow3S-S=3^{2021}-1\)

\(\Rightarrow2S=3^{2021}-1\)

\(\Rightarrow S=\frac{3^{2021}-1}{2}\)

2. \(\left(3x-2\right)^3=64\)

\(\Leftrightarrow\left(3x-2\right)^3=4^3\)

\(\Leftrightarrow3x-2=4\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

[3x-2]^3=64

Ta có:64=4^3    

Suy ra:3x-2=4

           3x   =4+2

          3x=6

         x=6:3

         x=2

Đề là gì vậy ???

\(B=x\left(x-2\right)\left(x+2\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(B=x\left(x^2-4\right)-\left(x^3-3x^2+9x+3x^2-9x+27\right)\)

\(B=x^3-4x-\left(x^3+27\right)\)

\(B=-4x-27\)

đề là rút gọn ạ

(x+x+....+x)+(1+2+.......+50)=1875

50x+[(50-1):1+1].(50+1):2=1875

50x+50.51:2=1875

50x+1275

50x=1875-1275=600

x=600:50

x=12

( x + 1 ) + ( x + 2 ) + ( x + 3 ) + (x + 4) +...+ (x + 50) = 1875

(x + x + x + x + ... + x) + (1 + 2 + 3 + 4 + ... + 50) = 1875

x . 50 + (1 + 50) . 50 : 2 = 1875

x . 50 + 51 . 50 : 2 = 1875

x . 50 + 2550 : 2 = 1875

x . 50 + 1275 = 1875

x . 50 = 1875 - 1275 = 600

x = 600 : 50

x = 12

(x + 1)+(x + 2)+ (x + 3 ) +(x +4) +...+ ( x + 50) = 1875

<=> (x+ x + x + x+...+ x ) + (1+2+3+4+...+50 ) = 1875

<=> 50x + 1275 = 1875

<=>        50x       = 1875 - 1275

<=>        50x       = 600

<=>           x         = 600 : 50

<=>            x         = 12

với giải thích hộ mik số trên có chia hết cho 13 ko và có là số chính phương không ạ

Đặt biểu thức trên là A , ta có :

\(A=1+3+3^2+3^3+...+3^{98}\)

\(\Rightarrow3A=3+3^2+3^3+3^4+...+3^{99}\)

\(\Leftrightarrow3A-A=\left(3+3^2+3^3+3^4+...+3^{99}\right)-\left(1+3+3^2+3^3+...+3^{98}\right)\)

\(\Leftrightarrow2A=3^{99}-1\)

\(\Leftrightarrow A=\frac{3^{99}-1}{2}\)

Đặt \(A=1+3+3^2+...+3^{98}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{99}\)

\(\Rightarrow3A-A=.......................=3^{99}-1\)

\(\Rightarrow2A=3^{99}-1\).\(\Rightarrow A=\frac{3^{99}-1}{2}\)

a/ => (6x - 3)(3x - 1) - (2x - 3)(9x - 1) = 0

=> 18x² - 15x + 3 -  18x² + 29x - 3 = 0

=> 14x = 0 => x = 0

b/ ghi dấu rõ vào tớ mới giải đc