Question

 

Tìm x

a) x(2x-1)(x+5)-(2x²+1)(x+4,5)=3,5

b) (2x-5)²+(y-3)²=0

Answer 1

a) \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow2x^3-x^2+10x^2-5x-2x^3-x-9x^2-4,5=3,5\)

\(\Leftrightarrow-6x=8\Leftrightarrow x=-\frac{4}{3}\)

b) \(\left(2x-5\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}2x-5=0\\y-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=3\end{cases}}}\)

Answer 2

a) x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5x(2x−1)(x+5)−(2x2+1)(x+4,5)=3,5

\Leftrightarrow2x^3-x^2+10x^2-5x-2x^3-x-9x^2-4,5=3,5⇔2x3−x2+10x2−5x−2x3−x−9x2−4,5=3,5

\Leftrightarrow-6x=8\Leftrightarrow x=-\frac{4}{3}⇔−6x=8⇔x=−34​

b) \left(2x-5\right)^2+\left(y-3\right)^2=0(2x−5)2+(y−3)2=0

\(\Leftrightarrow\hept{\begin{cases}2x-5=0\\y-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=3\end{cases}}}\)

Answer 3

a) x (2x - 1) (x + 5) - (2x² + 1) (x + 4,5) = 3,5

=> 2x³ - x² + 10x² - 5x - 2x³ - x - 9x² - 4,5 = 3,5

 (2x³ - 2x³) + (-x² - 9x²) + 10x² - (5x - x) - 4,5 = 3,5

 -10x² + 10x² - 4x - 4,5 = 3,5

 (-10x² + 10x²) - 4x - 4,5 = 3,5

  -4x = 3,5 + 4,5

 -4x = 8

 x = 8 : (-4) = -2

Vậy x = 2 hoẵ x = -2

b) (2x - 5)² + (y - 3)² = 0

=> (2x - 5)² = 0 hoặc (y - 3)² = 0

* (2x - 5)² = 0 => x = \(\frac{5}{2}\)

* (y - 3)² = 0 => y = 3

(Mình không chắc câu a nha)