Tìm x:
\(8x^2-\left(2x+5\right)\left(4x-2\right)-9=0\)
\(\Leftrightarrow8x^2-\left(8x^2-4x+20x-10\right)-9=0\)
\(\Leftrightarrow8x^2-8x^2+4x-20x+10-9=0\)
\(\Leftrightarrow-16x+1=0\)
\(\Leftrightarrow-16x=-1\)
\(\Leftrightarrow x=\dfrac{-1}{-16}=\dfrac{1}{16}\)
Vậy \(x=\dfrac{1}{16}\)
Bài 1:
\(a,8x^2-\left(2x+5\right)\left(4x-2\right)-9=0\)
\(\Rightarrow8x^2-\left(8x^2+16x-10\right)-9=0\)
\(\Rightarrow8x^2-8x^2-16x+10-9=0\)
\(\Rightarrow-16x+1=0\)
\(\Rightarrow x=\dfrac{1}{16}\)
Tìm x:
a) 8x2 - (2x + 5)(4x - 2) - 9 =0
8x2 - (8x2 - 4x + 20x - 10) - 9= 0
8x2 - (8x2 + 16x - 10) - 9 =0
8x2 - 8x2 - 16x + 10 - 9 = 0
- 16x + 1 =0
- 16x = - 1
x = \(\frac{1}{16}\)
Vậy \(x = \frac{1}{16}\)
PTĐTTNT:
3/ x2 - 3x + 2
= x2 - x - 2x + 2
= x(x - 1) - 2(x - 1)
= (x - 1)(x - 2)
pn coi lại câu 1 và 2 nhé
C1 :
\(a,8x^2-\left(2x+5\right)\left(4x-2\right)-9=0\)
\(\Rightarrow8x^2-\left(8x^2+16x-10\right)-9=0\)
\(\Rightarrow8x^2-8x^2-16x+10-9=0\)
\(\Rightarrow-16x+1=0\)
\(\Rightarrow x=\dfrac{1}{16}\)
1/ 5x2-10x-x+2
=5x(x-2)-(x-2)
=(x-2)(5x-1)
2/ 3x2-6xy+3y2-12z2
=3(x2-2xy+y2-4z2 )
3{(x-y)2-4z2 }
3(x-y-2z)(x-y+2z)
x2-3x+2
=x2-2x-x+2
=x(x-2)-(x-2)
=(x-1)(x-2)
Tìm x:
8x2 - (2x + 5)(4x - 2) - 9 = 0
8x2 - (8x2 - 4x + 20x - 10) - 9 = 0
8x2 - 8x2 +4x - 20x + 10 - 9 = 0
-16x + 1 = 0
=> -16x = -1
=> x = \(\dfrac{1}{16}\)
Phân tích Đa thức thành nhân tử:
con a bn xem lại đi nhé
mạn phép chữa đề:
b) 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2 - 4z2)
= 3[(x2 - 2xy + y2) - 4z2]
= 3[(x - y)2 - 4z2]
= 3(x - y + z)(x - y - z)
c) x2 - 3x + 2
C1:
x2 - 3x + 2
= x2 - 3x - 1 + 3
= (x2 - 1) - (3x - 3)
= (x - 1)(x + 1) - 3(x - 1)
= (x - 1)(x + 1 - 3)
=(x - 1)(x - 2)
C2: x2 - 3x + 2
= 3x2 - 2x2 - 3x + 2
= (3x2 - 3x) - (2x2 - 2)
= 3x(x - 1) - 2(x2 - 1)
= 3x(x - 1) - 2(x - 1)(x + 1)
= (x - 1)[3x - 2(x + 1)]
= (x - 1)(3x - 2x - 2)
= (x - 1)(x - 2)
C3: x2 - 3x + 2
= x2 - x - 2x + 2
= (x2 - x) - (2x - 2)
= x(x - 1) - 2(x - 1)
= (x - 2)(x - 1)
Chúc Bn Hok Tốt
\(\)
1/ 5x2 - 10 - x2 + 2 ( sửa đề )
= 5( x2 - 2) - (x2 - 2)
= 4(x2 - 2)
2/ 3x2 - 6xy + 3y2 - 12z2 ( sửa đề)
= 3(x2 - 2xy + y2 - 4z2)
= 3[ ( x - y)2 - ( 2z)2]
= 3( x - y - 2z)( x - y + 2z)
3/ x2 - 3x + 2
= x2 - x - 2x + 2
= x( x - 1) - 2(x - 1)
= ( x - 1)( x - 2)