x^2(2x + 3) – 2x^3 = 3.
tìm x, trình bày ra luôn
2x^3+4x^2/3.(x+2)
rút gọn
trình bày ra luôn
b) 4x(2 – x) + (2x + 1)^2 = 2.
c) (x – 3)3 – x^2 (x – 9) = 0.
tìm x, trình bày ra hết lun
c: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+9x^2=0\)
hay x=1
b) 4x(2 – x) + (2x + 1)^2 = 2.
c) (x – 3)^3 – x^2 (x – 9) = 0.
tìm x, trình bày ra hết lun
b) 4x(2-x)+(2x+1)^2=2
8x-4x^2+4x^2+4x+1-2=0
(8x+4x)+(-4x^2+4x^2)+(1-2)=0
12x + 0 -1 =0
12x=1
x=1/12
Vậy x= 1/2
c) (x-3)^3-x^2(x-9)=0
x^3-9x^2+27x-x^3+9x^2=0
(x^3-x^3)+(-9x^2+9x^2)+27x=0
0 + 0 + 27x=0
x= 0
Vậy x=0
a) (1,0 điểm) 4x^2 + 8x.
b) (1,0 điểm) x^2 – 9 .
c) (1,0 điểm) 2x^3 – 3x^2 + 2x – 3.
phân tích đa thức thành nhân tử, trình bày ra luôn
\(a,=4x\left(x+2\right)\\ b,=\left(x-3\right)\left(x+3\right)\\ c,=x^2\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2+1\right)\)
a)4x2+8x b)x2-9
=4x(x+2) =x2-32
=(x-3)(x+3)
c)2x3-3x2+2x-3
=2x3+2x-(3x2+3)
=2x(x2+1)-3(x2+1)
=(2x-3)(x2+1)
4x(2 – x) + (2x + 1)^2 = 2.
tìm x nha trình bày ra lun
\(\Leftrightarrow8x-4x^2+4x^2+4x+1=2\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
\(4x\left(2-x\right)+\left(2x+1\right)^2=2\)
\(\Leftrightarrow8x-4x^2+4x^2+4x+1=2\)
\(\Leftrightarrow12x=1\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
4x(2 – x) + (2x + 1)^2 = 2.
tìm x nha mn trình bày ra hết lun
\(\Rightarrow8x-4x^2+4x^2+4x+1=2\\ \Rightarrow12x=1\Rightarrow x=\dfrac{1}{12}\)
4x(2-x)+(2x+1)2=2
4x2-4xx+(2x)2+2.2x1+12=2
8x-4x2+4x2+4x+1=2
12x+1=2
12x=2-1
12x=1
x=1/12
2x^2+7x-9=0
tìm x trình bày ra luôn
\(2x^2+7x-9=0\Leftrightarrow\left(x-1\right)\left(x+\dfrac{9}{2}\right)=0\)
\(x\in\left\{1;\dfrac{-9}{2}\right\}\)
<=> 2x2 -2x +9x -9 =0
<=> 2x(x-1) + 9 (x-1) = 0
<=> (2x+9)(x-1) = 0
<=> 2x + 9 =0 hoặc x-1 = 0
<=> x= -9/2 hoặc x= 1
a.6(x-2)=8(3x+1)
b.2x-(3-7x)=5(x+3)
c.(x-1)^2=(x+3)(x+2)
d.(3x-9)(4x+5)=0
e.x^2-3x+2=0
f.x^2-4x+4=0
giải phương trình
trình bày hết luôn
\(a,6\left(x-2\right)=8\left(3x+1\right)\\ \Leftrightarrow6x-12=24x+8\\ \Leftrightarrow18x+20=0\\ \Leftrightarrow x=-\dfrac{10}{9}\\ b,2x-\left(3-7x\right)=5\left(x+3\right)\\ \Leftrightarrow2x-3+7x=5x+15\\ \Leftrightarrow9x-3-5x-15=0\\ \Leftrightarrow4x-18=0\\ \Leftrightarrow x=\dfrac{9}{2}\\ c,\left(x-1\right)^2=\left(x+3\right)\left(x+2\right)\\ \Leftrightarrow x^2-2x+1=x^2+5x+6\\ \Leftrightarrow7x+5=0\\ \Leftrightarrow x=-\dfrac{5}{7}\\ d,\left(3x-9\right)\left(4x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)
\(e,x^2-3x+2=0\\ \Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\\ \Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\\ \left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ f,x^2-4x+4=0\\ \Leftrightarrow x^2-2.2+2^2=0\\ \Leftrightarrow\left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ x=2\)
a, \(6x-12=24x+8\Leftrightarrow18x=-20\Leftrightarrow x=-\dfrac{20}{18}=-\dfrac{10}{9}\)
b, \(2x-3+7x=5x+15\Leftrightarrow4x=18\Leftrightarrow x=\dfrac{9}{2}\)
c, \(x^2-2x+1=x^2+5x+6\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)
d, \(\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)
e, \(x^2-3x+2=0\Leftrightarrow x^2-2x-x+2=0\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow x=1;x=2\)
f, \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
a) (1,0 điểm) (x – 1)(2x + 3) – 2x 2 + 3x.
b) (1,0 điểm) (x + 3)2 – (x + 2) (x – 2).
rút gọn biểu thức, trình bày ra lun
b: \(=x^2+6x+9-x^2+4=6x+13\)