Toán Lớp 8: Bài 1: Tìm GTNN của biểu thức:
A= x^2 +3x – 2
Bài 2: Phân tích đa thức sau thành nhân tử:
a) 7xy + 7xz
b) x^2 – 4x – 25y^2 +4
Bài 3: Tìm x
a) x^2 – 25 – (x+5) =0
b) x^2 -4x +3 = 0
c) x – (x + 1)^2 = 0
Bài 4:
Tìm GTLN, GTNN của biểu thức sau:
a) x^2 – 6xx – 11
b) -x^2 + 6x +11
Leave a reply
Comments ( 1 )
B1)A = {x^2} + 3x – 2\\
= {x^2} + 2.\dfrac{3}{2}x + \dfrac{9}{4} – \dfrac{9}{4} – 2\\
= {\left( {x – \dfrac{3}{4}} \right)^2} – \dfrac{{17}}{4} \ge \dfrac{{ – 17}}{4}\\
\Leftrightarrow GTNN:A = \dfrac{{ – 17}}{4}\\
B2)a)7xy + 7xz\\
= 7x\left( {y + z} \right)\\
b){x^2} – 4x – 25{y^2} + 4\\
= {\left( {x – 2} \right)^2} – 25{y^2}\\
= \left( {x – 2 + 5y} \right)\left( {x – 2 – 5y} \right)\\
B3)a){x^2} – 25 – \left( {x + 5} \right) = 0\\
\Leftrightarrow \left( {x + 5} \right)\left( {x – 5} \right) – \left( {x – 5} \right) = 0\\
\Leftrightarrow \left( {x – 5} \right)\left( {x + 5 – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = – 4
\end{array} \right.\\
Vậy\,x = 5,x = – 4\\
b){x^2} – 4x + 3 = 0\\
\Leftrightarrow {x^2} – 4x + 4 = 1\\
\Leftrightarrow {\left( {x – 2} \right)^2} = 1\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\\
Vậy\,x = 3,x = 1\\
c)x – {\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow x – {x^2} – 2x – 1 = 0\\
\Leftrightarrow – {x^2} – x – 1 = 0\\
\Leftrightarrow – {x^2} – 2.\dfrac{1}{2}x – \dfrac{1}{4} + \dfrac{1}{4} – 1 = 0\\
\Leftrightarrow – {\left( {x + \dfrac{1}{2}} \right)^2} – \dfrac{3}{4} = 0\\
\Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = – \dfrac{3}{4}\\
\Leftrightarrow x \in \emptyset \\
Vậy\,x \in \emptyset \\
B4)a){x^2} – 6x – 11\\
= {x^2} – 6x + 9 – 20\\
= {\left( {x – 3} \right)^2} – 20 \ge – 20\\
\Leftrightarrow GTNN = – 20\,khi:x = 3\\
b) – {x^2} + 6x + 11\\
= – {x^2} + 6x – 9 + 20\\
= – {\left( {x – 3} \right)^2} + 20 \le 20\\
\Leftrightarrow GTLN = 20\,khi:x = 3
\end{array}$