Điền các hạng tử thích hợp vào chỗ có dấu * để có hằng đẳng thức:
a) x ²+4x+* = (* + *) ²
b) 9x ² – * +4 = (* + *)
c) x ²+x+* = (* + *)
d) * – 2a+4=(* + *)
e) 4y ² – * = (* – 3x)(* + *)
f) * – 1/4 = (3y-*)(*+*)
g) 8x ³+ * = (* + 2a)(4x ²-*+*)
h) * – 27y ³ = (4x – *)(9y ²+ * + *)
Đáp án+Giải thích các bước giải:
`a) x^2+2.x.2+2^2=(x+2)^2`
`→ x^2+4x+4=(x+2)^2`
`b) (3x)^2-2.3x.2+2^2=(3x+2)^2`
`→9x^2-12x+4=(3x+2)^2`
`c) x^2+2.x. 1/2+(1/2)^2=(x+1/2)^2`
`→x^2+x+1/4=(x+1/2)^2`
`d) (1/2a)^2-2. 1/2a.2+2^2=(1/2a-2)^2`
`→1/4a^2 -2a+4= (1/2a-2)^2`
`e) (2y)^2-(3x)^2=(2y-3x).(2y+3x)`
`→ 4y^2-9x^2=(2y-3x)(2y+3x)`
`f) (3y)^2-(1/2)^2=(3y-1/2).(3y+1/2)`
`→9y^2-1/4=(3y-1/2)(3y+1/2)`
`g) (2x)^3+(2a)^3=(2x+2a).[(2x)^2-2x.2a+(2a)^2]`
`→ 8x^3+8a^3=(2x+2a)(4x^2-4xa+4a^2)`
`h) (4x)^3-(3y)^3=(4x-3y).[(3y)^2+3y.4x+(4x)^2]`
`→ 64x^3-27y^3=(4x-3y)(9y^2+12xy+16x^2)`
Đáp án:
\(\begin{array}{l}
a){\left( {x + 2} \right)^2}\\
b){\left( {3x + 1} \right)^2}\\
c){\left( {x + \dfrac{1}{2}} \right)^2}\\
d){\left( {\dfrac{a}{2} – 2} \right)^2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a){x^2} + 2.2x + {\left( 2 \right)^2}\\
= {x^2} + 4x + 4\\
= {\left( {x + 2} \right)^2}\\
b)9{x^2} – 2.3x.1 + 1 = 9{x^2} + 6x + 1\\
= {\left( {3x + 1} \right)^2}\\
c){x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} = {x^2} + x + \dfrac{1}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2}\\
d){\left( {\dfrac{a}{2}} \right)^2} – 2.\dfrac{a}{2}.2 + 4\\
= \dfrac{{{a^2}}}{4} – 2a + 4\\
= {\left( {\dfrac{a}{2} – 2} \right)^2}\\
e)4{y^2} – {\left( {3x} \right)^2}\\
= \left( {2y – 3x} \right)\left( {2y + 3x} \right)\\
f){\left( {3y} \right)^2} – \dfrac{1}{4} = \left( {3y – \dfrac{1}{2}} \right)\left( {3y + \dfrac{1}{2}} \right)\\
g)8{x^3} + {\left( {2a} \right)^3}\\
= \left( {2x + 2a} \right)\left( {4{x^2} – 2x.2a + 4{a^2}} \right)\\
= \left( {2x + 2a} \right)\left( {4{x^2} – 4xa + 4{a^2}} \right)\\
h){\left( {4x} \right)^3} – {\left( {3y} \right)^3}\\
= \left( {4x – 3y} \right)\left( {9{y^2} + 12xy + 16{x^2}} \right)
\end{array}\)