Toán Lớp 8: Tìm giá trị nhỏ nhất:b) x^2 -2x+y^2-4y+7
C) 4x^2+4x
d)(x-1)(x+2)(x+3)(x+6)
e)A=6x-x^2+5
Chứng minh các đẳng thức sau bằng nhau:
a)(x+y)^2-y^2=x(x+2y)
b)(x^2+y^2)^2-(2xy)^2=(x+y)^2(x-y)^2
c)(x+y)^3-(x^3+y^3)=3xy(x+y)
d)(a+b)^2+2(a+b)(a-b)+(a-b)^2=4a^2
e)(a+b+c)^2+(b+c-a)^2+(c+a-b)^2+(a+b-c)^2=4(a^2+b^2+c^2)
f)(2x+3)(4x^2+9)(2x-3)=16x^4-81
Mong mọi người giúp ạ, cần gấppp
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Comments ( 1 )
b){x^2} – 2x + {y^2} – 4y + 7\\
= {x^2} – 2x + 1 + {y^2} – 4y + 4 + 2\\
= {\left( {x – 1} \right)^2} + {\left( {y – 2} \right)^2} + 2 \ge 2\\
\Leftrightarrow GTNN = 2\,khi:x = 1;y = 2\\
c)4{x^2} + 4x\\
= 4{x^2} + 4x + 1 – 1\\
= {\left( {2x + 1} \right)^2} – 1 \ge – 1\\
\Leftrightarrow GTNN = – 1\,khi:x = – \dfrac{1}{2}\\
d)\left( {x – 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 6} \right)\\
= \left( {{x^2} + 5x – 6} \right)\left( {{x^2} + 5x + 6} \right)\\
= {\left( {{x^2} + 5x} \right)^2} – 36 \ge – 36\\
\Leftrightarrow GTNN = – 36\,khi:\left[ \begin{array}{l}
x = 0\\
x = – 5
\end{array} \right.\\
e)A = 6x – {x^2} + 5\\
= – \left( {{x^2} – 6x + 9} \right) + 9 + 5\\
= – {\left( {x – 3} \right)^2} + 14 \le 14\\
\Leftrightarrow GTLN:A = 14\,khi:x = 3\\
B2)\\
a){\left( {x + y} \right)^2} – {y^2}\\
= \left( {x + y + y} \right)\left( {x + y – y} \right)\\
= x\left( {x + 2y} \right)\\
b){\left( {{x^2} + {y^2}} \right)^2} – {\left( {2xy} \right)^2}\\
= \left( {{x^2} + {y^2} + 2xy} \right)\left( {{x^2} + {y^2} – 2xy} \right)\\
= {\left( {x + y} \right)^2}{\left( {x – y} \right)^2}\\
c){\left( {x + y} \right)^3} – \left( {{x^3} + {y^3}} \right)\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3} – {x^3} – {y^3}\\
= 3xy\left( {x + y} \right)\\
d){\left( {a + b} \right)^2} + 2\left( {a + b} \right)\left( {a – b} \right) + {\left( {a – b} \right)^2}\\
= {\left( {a + b + a – b} \right)^2}\\
= {\left( {2a} \right)^2}\\
= 4{a^2}\\
e){\left( {a + b + c} \right)^2} + {\left( {b + c – a} \right)^2} + {\left( {c + a – b} \right)^2} + {\left( {a + b – c} \right)^2}\\
= {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac\\
+ {b^2} + {c^2} + {a^2} + 2bc – 2ab – 2ac\\
+ {c^2} + {a^2} + {b^2} + 2ac – 2bc – 2ab\\
+ {a^2} + {b^2} + {c^2} + 2ab – 2bc – 2ac\\
= 4\left( {{a^2} + {b^2} + {c^2}} \right)\\
f)\left( {2x + 3} \right)\left( {4{x^2} + 9} \right)\left( {2x – 3} \right)\\
= \left( {2x + 3} \right)\left( {2x – 3} \right)\left( {4{x^2} + 9} \right)\\
= \left( {4{x^2} – 9} \right)\left( {4{x^2} + 9} \right)\\
= 16{x^4} – 81
\end{array}$