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2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
a) 4x2-8x=0
(2x)2-2.2.2x+4-4=0
(2x-2)2 =4
2x-2=2
2x =4
x=2
Nhớ k cho mk nha
Ta có : A = x(x + 1)(x + 2)(x + 3)
=> A = [x(x + 3)].[(x + 1)(x + 2)]
=> A = (x2 + 3x) . (x2 + 3x + 2)
Đặt a = x2 + 3x + 1
Khi đó A = (a - 1)(a + 1)
=> A = a2 - 1
=> A = x2 + 3x + 1 - 1
=> A = x2 + 3x
=> A = x2 + 3x + \(\frac{4}{9}-\frac{4}{9}\)
\(\Rightarrow A=\left(x+\frac{2}{3}\right)^2-\frac{4}{9}\)
Mà \(\left(x+\frac{2}{3}\right)^2\ge0\forall x\)
Nên : \(A=\left(x+\frac{2}{3}\right)^2-\frac{4}{9}\ge-\frac{4}{9}\forall x\)
Vậy Amin = \(\frac{-4}{9}\) , dầu "=" xảy ra khi và chỉ khi x = \(-\frac{2}{3}\)
Bài 1:
a) (x-3)\(^2\)-(x+1)(x-4)=0
<=>x\(^2\)-6x+9-x\(^2\)+4x-x+4=0
<=>-3x+13=0
<=>3x=13
<=> x=\(\dfrac{13}{3}\)
b)x\(^2\)-25=3x+15
<=>(x+5)(x-5)=3(x+5)
<=>(x+5)(x-5)-3(x+5)=0
<=>(x+5)[(x-5)-3]=0
<=>(x+5)(x-8)=0
<=> x+5=0 hoặc x-8=0
*x+5=0 *x-8=0
<=>x=-5 <=>x=8
c)x\(^2\)-10x+25=2(x-5)
<=>(x-5)\(^2\)=2(x-5)
<=>(x-5)\(^2\)-2(x-5)=0
<=>(x-5)[(x-5)-2]=0
<=>(x-5)(x-7)=0
<=>x-5=0 hoặc x-7=0
* x-5=0 *x-7=0
<=>x=5 <=>x=7
d)4x\(^2\)-12x+9=(1-x)\(^2\)
<=>4x\(^2\)-12x+9=1-2x+x\(^2\)
<=>4x\(^2\)-12x+9-1+2x-x\(^2\)=0
<=>3x\(^2\)-10x+9=0
Câu d đến đây mik chịu...
d)
\(4x^2-12x+9=\left(1-x\right)^2\)
\(\Leftrightarrow4x^2-12x+9-1+2x-x^2=0\)
\(\Leftrightarrow3x^2-10x+8=0\)
\(\Leftrightarrow3x^2-6x-4x+8=0\)
\(\Leftrightarrow3x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)