Toán Lớp 8: viet cac da thuc duoi dang hang dang thuc
a,x^2+5x+25/4
b,16^2-8x+1
c,4x^2+12xy+9y^2
d,(x+3)(x+4)(x+5)(x+6)+1
e,x^3+3x^2+3x+1
g,27y^2-9y^2+y-1/27
h,8x^6+12x^4y+6x^2y^2+y^3
i,(x+y)^3(x-y)^3
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Toán Lớp 8: viet cac da thuc duoi dang hang dang thuc
a,x^2+5x+25/4
b,16^2-8x+1
c,4x^2+12xy+9y^2
d,(x+3)(x+4)(x+5)(x+6)+1
e,x^3+3x^2+3x+1
g,27y^2-9y^2+y-1/27
h,8x^6+12x^4y+6x^2y^2+y^3
i,(x+y)^3(x-y)^3
Comments ( 1 )
a){x^2} + 5x + \dfrac{{25}}{4} = {\left( {x + \dfrac{5}{2}} \right)^2}\\
b)16{x^2} – 8x + 1 = {\left( {4x} \right)^2} – 2.4x + 1 = {\left( {4x – 1} \right)^2}\\
c)4{x^2} + 12xy + 9{y^2} = {\left( {2x + 3y} \right)^2}\\
d)\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right)\left( {x + 6} \right) + 1\\
= \left( {x + 3} \right)\left( {x + 6} \right).\left( {x + 4} \right)\left( {x + 5} \right) + 1\\
= \left( {{x^2} + 9x + 18} \right).\left( {{x^2} + 9x + 20} \right) + 1\\
Dat:\left( {{x^2} + 9x + 18} \right) = a\\
\Leftrightarrow a.\left( {a + 2} \right) + 1\\
= {a^2} + 2a + 1\\
= {\left( {a + 1} \right)^2}\\
= {\left( {{x^2} + 9x + 18 + 1} \right)^2}\\
= {\left( {{x^2} + 9x + 19} \right)^2}\\
e){x^3} + 3{x^2} + 3x + 1 = {\left( {x + 1} \right)^3}\\
g)27{y^3} – 9{y^2} + y – \dfrac{1}{{27}} = {\left( {3y – \dfrac{1}{3}} \right)^2}\\
h)8{x^6} + 12{x^4}y + 6{x^2}{y^2} + {y^3}\\
= {\left( {2{x^2} + y} \right)^3}\\
i){\left( {x + y} \right)^3}{\left( {x – y} \right)^3}\\
= {\left( {{x^2} – {y^2}} \right)^3}
\end{array}$