a) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\) (sửa \(\dfrac{x}{2}\rightarrow x^2\))

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

\(=x^6-\dfrac{1}{27}\)

b) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)

\(=\dfrac{1}{27}x^3+8y^3\)

Lưu ý : Áp dụng hằng đẳng thức đáng nhớ \(a^3\pm b^3=...\)

Thank you

a, |x^2 - 3x| = 0

=> x^2 - 3x = 0

=> x(x - 3) = 0

=> x = 0 hoặc x - 3 = 0 

=> x = 0 hoặc x = 3

vậy_

\(\left|a^2-3a\right|=0\)

\(\Rightarrow a^2-3a=0\)

\(\Rightarrow a\left(a-3\right)=0\)

\(\Rightarrow\hept{\begin{cases}a=0\\a=3\end{cases}}\)

Trả lời

Vì lx²-3xl=0

Nên x²-3x=0

=>x(x-3)=0

x=0.           x-3=0

                    x=3

Học tốt

1) |3x-18| = 45

TH1: 3x - 18 = 45

3x = 63

x = 21

TH2: 3x -18 = -45

3x = -27

x = -9

KL:...

phần còn lại b tự làm nha

\(D=\left(2-3\right)^3+2x\left(x-3\right)+\left(3x+1\right)\left(9x^2-3x+1\right)-6x\left(7x-3\right)+3x^3\)

\(=-1+2x^2-6x+27x^3-9x^2+3x+9x^2-3x+1-42x^2+18x+3x^3\)

\(=-40x^2+12x+30x^3\)

=> Biểu thức có phụ thuộc vào biến x 

Vậy đề sai .

Ta có:(3x-7)²⁰⁰⁹=(3x-7)²⁰⁰⁷

\(\Rightarrow\)(3x-7)²⁰⁰⁹-(3x-7)²⁰⁰⁷=0

​\(\Rightarrow\)​(3x-7)²⁰⁰⁷ [(3x-7)²-1]=0​​

\(\Rightarrow\)\(\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)

\(\Rightarrow\)​\(\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)​

\(\Rightarrow\)\(\orbr{\begin{cases}3x=7\\3x-7=1\end{cases}}\)

​\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\3x=8\end{cases}}\)​

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{8}{3}\end{cases}}\)

Vậy......

\(\left(3x-7\right)^{2009}=\left(3x-7\right)^{2007}\)

\(\Rightarrow\left(3x-7\right)^{2009}-\left(3x-7\right)^{2007}=0\)

\(\Rightarrow\left(3x-7\right)^{2007}\left[\left(3x-7\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x-7=0\\3x-7=\pm1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x\in\left\{\frac{8}{3};2\right\}\end{cases}}\)

d: ta có: \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

Tìm min:

$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$

Tìm min

$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$

$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)

Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$

$\Leftrightarrow x=\frac{-1}{4}$

Tìm min

$H=5x^2-x+1=5(x^2-\frac{x}{5})+1$

$=5[x^2-\frac{x}{5}+(\frac{1}{10})^2]+\frac{19}{20}$

$=5(x-\frac{1}{10})^2+\frac{19}{20}\geq \frac{19}{20}$
Vậy $H_{\min}=\frac{19}{20}$. Giá trị này đạt tại $x-\frac{1}{10}=0$

$\Leftrightarrow x=\frac{1}{10}$

a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555

=> 101x +5050 = 5555

=> 101x = 505

=> x = 505 : 101 = 5

Vậy, x = 5

b)1+2+3+4+...+x=820

=> ( x+1) x :2 = 820

=> (x+1)x = 1640

Mà 1640 = 40 . 41

=> x = 40 ( vì {x+1} - x = 1)

Vậy, x = 40

c) 3^x+1 = 9.27=243

=> 3^x+1 = 3⁵

=>x + 1 = 5

=> x = 4

Vậy, x=4

d) x+2x+3x+...+99x+100x=15150

=> [( 100 + 1) x 100 :2 ] x = 15150

=> 5050x = 15150

=> x = 15150:5050 = 3

Vậy, x =3

e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550

=> 100x + 5050 = 205550

=> 100x =  205550 - 5050= 200500

=> x =  200500 : 100 = 2005

Vậy, x = 2005

f)3^x+3^x+1+3^x+2=351

=> 3^x + 3^x . 3 + 3^x x 9 = 351

=> 3^x ( 1+3+9) = 351

=> 3^x  . 13 = 351

=> 3^x  = 351 :13=27 mà 27 = 3³

=> x=3

Vậy, x=3

mình đg cần gấp á

a) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5555\)

\(\Rightarrow x+x+1+x+2+x+3+...+x+100=5555\)

\(\Rightarrow101\cdot x+5050=5555\)

\(\Rightarrow101\cdot x=5555-5050\)

\(\Rightarrow101\cdot x=505\)

\(\Rightarrow x=505:101\)

\(\Rightarrow x=5\)

b) \(1+2+3+4+...+x=820\)

\(\Rightarrow\left(x+1\right)\cdot\left[\left(x-1\right):1+1\right]:2=820\)

\(\Rightarrow\left(x+1\right)\cdot\left(x+1-1\right):2=820\)

\(\Rightarrow\left(x+1\right)\cdot x:2=820\)

\(\Rightarrow x\cdot\left(x+1\right)=820\cdot2\)

\(\Rightarrow x\cdot\left(x+1\right)=1640\)

Ta thấy: \(40\cdot41=1640\)

Vậy: \(x=40\)