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

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

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

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

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

\(x^3+1=0\)

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

\(x=-1\)

c) \(4x^2-25=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

\(a,x\left(x+2\right)-3x-6=0\)

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

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

\(\Leftrightarrow x^3+3x^2+3x+1-3x^2-3x=0\)

\(\Leftrightarrow x^3+1=0\)

\(\Rightarrow x^3=1\Rightarrow x=1\)

\(c,4x^2-25=0\)

\(\Leftrightarrow\left(2x+5\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}2x+5=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

đặt 3^x ra ngoài bạn nhé bàn phím mik hỏng rồi ;-;

$\Rightarrow 3^x(1+3+3^2+3^3)=1080$

$\Rightarrow 3^x.40=1080$

$\Rightarrow 3^x=27=3^3$

$\Rightarrow x=3$

\(3^x+3^{x+1}+3^{x+2}+3^{x+3}=1080\)

\(\Rightarrow3^x+3^x.3^1+3^x.3^2+3^x.3^3=1080\)

\(3^x.\left(1+3+9+27\right)=1080\)

\(3^x.40=1080\)

\(3^x=27\)

\(3^x=3^3\)

Vậy: x = 3

Thu gọn biểu thức:

\(9x\left(x+5\right)-\left(3x+2\right)\left(3x-2\right)\)

\(=9x^2+45x-\left(9x^2-4\right)\)

\(=45x+4\)

Tìm x. biết:

\(\left(3x-2\right)^2=49x^2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=7x\\3x-2=-7x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};\dfrac{1}{5}\right\}\)

a: Ta có: \(9x\cdot\left(x+5\right)-\left(3x+2\right)\left(3x-2\right)\)

\(=9x^2+45x-9x^2+4\)

=45x+4

b: Ta có: \(\left(3x-2\right)^2=49x^2\)

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

\(\Leftrightarrow\left(4x+2\right)\left(10x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{5}\end{matrix}\right.\)

$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`

b: \(3x^2-2x-1=0\)

=>\(3x^2-3x+x-1=0\)

=>\(\left(x-1\right)\left(3x+1\right)=0\)

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

a: Bạn ghi lại đề đi bạn

bn nhân biểu thức cần tính vs 2  sẽ ra kq

Câu hỏi của Vương Trương Quang - Toán lớp 9 - Học toán với OnlineMath

x²+3x²+3x+1-3x²-3x = 0

  => x³+1 = 0

  => x³     = 0-1

  => x³     = -1

  => x       = -1

\(x^3+3x^2+3x+1-3x^2-3x=0\)0

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

\(\Leftrightarrow x^3+1=0\)

\(\Leftrightarrow x^3=1\)

\(\Leftrightarrow x^3=1^3\)

\(\Rightarrow x=1\)

∈x³ + 3x² + 3x + 1 - 3x² - 3x = 0

x³ + 1 = 0

(x + 1)(x² -x + 1) = 0

=> x + 1 = 0 (do x² -x + 1 > 0 ∀x)

=> x = -1

Vậy x ∈ {-1}