phân tích thành nhân tử: 4x2-4x+1-(2x-1)b+\(\dfrac{b^2}{4}\)
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29 tháng 1
a: \(x^2+2x+1+4x+4\)
\(=\left(x^2+2x+1\right)+\left(4x+4\right)\)
\(=\left(x+1\right)^2+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x+1+4\right)\)
\(=\left(x+1\right)\left(x+5\right)\)
b: Sửa đề: \(2x^3+6x^2+x^2+3x\)
\(=2x^2\left(x+3\right)+x\left(x+3\right)\)
\(=\left(x+3\right)\left(2x^2+x\right)\)
\(=x\left(x+3\right)\left(2x+1\right)\)
c: \(\dfrac{1}{2}x^2+\dfrac{1}{4}x+\dfrac{1}{4}x+1\)
\(=\dfrac{1}{4}x\left(\dfrac{1}{4}x+1\right)+\left(\dfrac{1}{4}x+1\right)\)
\(=\left(\dfrac{1}{4}x+1\right)\left(\dfrac{1}{4}x+1\right)=\left(\dfrac{1}{4}x+1\right)^2\)
12 tháng 10 2023
a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)
b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)
c: \(2x-1-x^2\)
\(=-\left(x^2-2x+1\right)=-\left(x-1\right)^2\)
d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
e: \(9-x^2=3^2-x^2=\left(3-x\right)\left(3+x\right)\)
g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)
\(=\left(5-x\right)\left(5+3x\right)\)
h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)
\(=\left(x+1+2x-1\right)\left(x+1-2x+1\right)\)
\(=3x\left(-x+2\right)\)
i: \(=x^2y^2-4xy+4-3\)
\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)
k: \(=y^2-\left(x-1\right)^2\)
\(=\left(y-x+1\right)\left(y+x-1\right)\)
l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)
m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)
T
11 tháng 12 2023
a) $4x^2+4x+1$
$=(2x)^2+2\cdot2x\cdot1+1^2$
$=(2x+1)^2$
b) $x^2+6x-y^2+9$
$=(x^2+6x+9)-y^2$
$=(x^2+2\cdot x\cdot3+3^2)-y^2$
$=(x+3)^2-y^2$
$=(x+3-y)(x+3+y)$
$\text{#}Toru$
11 tháng 12 2023
a: \(4x^2+4x+1\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)
\(=\left(2x+1\right)^2\)
b: \(x^2+6x-y^2+9\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3+y\right)\left(x+3-y\right)\)
KV
9 tháng 12 2023
a) x² - 9
= x² - 3²
= (x - 3)(x + 3)
b) 4x² - 1
= (2x)² - 1²
= (2x - 1)(2x + 1)
c) x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x² - 2²)(x² + 4)
= (x - 2)(x + 2)(x + 4)
d) x² - 4x + 4
= x² - 2.x.2 + 2²
= (x - 2)²
e) x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)
f) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
CP
17 tháng 7 2021
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
17 tháng 7 2021
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
\(4x^2-4x+1-\left(2x-1\right)b+\dfrac{b^2}{4}\)
\(=\left(2x-1\right)^2-2.\left(2x-1\right).\left(\dfrac{1}{2}b\right)+\left(\dfrac{b}{2}\right)^2\)
\(=\left(2x-1-\dfrac{1}{2}b\right)^2\)
Ta có: \(\left(4x^2-4x+1\right)-\left(2x-1\right)b+\dfrac{b^2}{4}\)
\(=\left(2x-1\right)^2-2\cdot\left(2x-1\right)\cdot\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2\)
\(=\left(2x-\dfrac{1}{2}b-1\right)^2\)