Toán Lớp 8: Tìm x, biết: 3x(x – 5) – x2 + 25 = 0
Rút gọn các biểu thức sau:
a. 2x(3x + 2) – 3x(2x + 3)
b. (x + 2)3 + (x – 3)2 – x2(x + 5)
c. (3×3 – 4×2 + 6x) : 3x
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Toán Lớp 8: Tìm x, biết: 3x(x – 5) – x2 + 25 = 0
Rút gọn các biểu thức sau:
a. 2x(3x + 2) – 3x(2x + 3)
b. (x + 2)3 + (x – 3)2 – x2(x + 5)
c. (3×3 – 4×2 + 6x) : 3x
Comments ( 2 )
Tìm x:
3x(x-5)-x^2+25=0
<=>3x(x-5)-(x^2-5^2)=0
<=>3x(x-5)-(x-5)(x+5)=0
<=>(x-5)(3x-x-5)=0
<=>(x-5)(2x-5)=0
<=>\(\left[ \begin{array}{l}x-5=0\\2x-5=0\end{array} \right.\)
<=>\(\left[ \begin{array}{l}x=5\\2x=5\end{array} \right.\)
<=>\(\left[ \begin{array}{l}x=5\\x=\dfrac{5}{2}\end{array} \right.\)
Vậy S={5;5/2}
Rút gọn các biểu thức sau:
a)2x(3x+2)-3x(2x+3)
=6x^2+4x-6x^2-9x
=-5x
b)(x+2)^3+(x-3)^2-x^2(x+5)
=x^3+6x^2+12x+8-x^2+6x-9-x^3-5x^2
=18x-1
c)(3x^3-4x^2+6x):3x
=3x^3:3x-4x^2:3x+6x:3x
=x^2-4/3x+2.
Giải đáp + Lời giải và giải thích chi tiết:
Tìm x, biết: 3x(x – 5) – x2 + 25 = 0
3x (x – 5) – $x^2$ + 25 = 0
3x (x – 5) – ($x^2$ + 25) = 0
3x (x – 5) – (x + 5)(x – 5) = 0
(3x – x – 5)(x – 5) = 0
(2x – 5)(x – 5) = 0
Nếu TH1: 2x – 5 = 0 $\Rightarrow$ x = $\dfrac{5}{2}$
TH2: x – 5 = 0 $\Rightarrow$ x = 5
Vậy $\Rightarrow$ x $\in$ $\begin{cases}\dfrac{5}{2}\end{cases}$;5 $\left.\begin{matrix} \end{matrix}\right\}$.
Rút gọn các biểu thức sau:
a) 2x (3x + 2) – 3x (2x + 3)
2x (3x + 2) – 3x (2x + 3)
= (2x . 3x) + (2x . 2) – (3x . 2x) – (3x . 3)
= 6$x^2$ + 4x – 6$x^2$ – 9x
= -5x
b) $(x + 2)^3$ + $(x – 3)^2$ – x2 (x + 5)
= ($x^2$ + 6$x^2$ + 12x + 8) + ($x^2$ – 6x + 9) – ($x^3$ + 5$x^2$)
= ($x^3$ + 6$x^2$ + 12x + 8) + ($x^2$ – 6x + 9) – ($x^3$ + 5$x^2$)
= $x^3$ + 6$x^2$ + 12x + 8 + $x^2$ – 6x + 9 – $x^3$ – 5$x^2$
= ($x^3$ – $x^3$) + (6$x^2$ + $x^2$ – 5$x^2$) + (12x – 6x) + 9
= 2$x^2$ + 6x + 9
c) (3$x^3$ – 4$x^2$ + 6x) : 3x
= (3$x^3$ : 3x) – (4$x^2$: 3x) + (6x : 3x)
= $x^2$ – $\dfrac{4}{3}$x + 2
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