Toán Lớp 8: Câu 1:x^2+y^2+z^2=x(y+z)
Câu 2:Tìm gtnn của B=4x^2+4y^2-4xy-3x
Câu 3:cho 2a^2+9b^2+6ab-14a-30b+29=0.Tính S=(a-1)^2021-(1-b)^2021
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Toán Lớp 8: Câu 1:x^2+y^2+z^2=x(y+z)
Câu 2:Tìm gtnn của B=4x^2+4y^2-4xy-3x
Câu 3:cho 2a^2+9b^2+6ab-14a-30b+29=0.Tính S=(a-1)^2021-(1-b)^2021
Comments ( 1 )
1,\\
x = y = z = 0\\
2,\\
{B_{\min }} = – \dfrac{3}{4} \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = \dfrac{1}{4}
\end{array} \right.\\
3,\\
S = 1
\end{array}\)
1,\\
{x^2} + {y^2} + {z^2} = x\left( {y + z} \right)\\
\Leftrightarrow 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2x.\left( {y + z} \right)\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} = 2xy + 2xz\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} – 2xy – 2xz = 0\\
\Leftrightarrow \left( {{x^2} – 2xy + {y^2}} \right) + \left( {{x^2} – 2xz + {z^2}} \right) + {y^2} + {z^2} = 0\\
\Leftrightarrow {\left( {x – y} \right)^2} + {\left( {x – z} \right)^2} + {y^2} + {z^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – y} \right)^2} = 0\\
{\left( {x – z} \right)^2} = 0\\
{y^2} = 0\\
{z^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x – y = 0\\
x – z = 0\\
y = 0\\
z = 0
\end{array} \right. \Leftrightarrow x = y = z = 0\\
2,\\
B = 4{x^2} + 4{y^2} – 4xy – 3x\\
= \left( {{x^2} – 4xy + 4{y^2}} \right) + \left( {3{x^2} – 3x + \dfrac{3}{4}} \right) – \dfrac{3}{4}\\
= \left[ {{x^2} – 2.x.2y + {{\left( {2y} \right)}^2}} \right] + 3.\left( {{x^2} – x + \dfrac{1}{4}} \right) – \dfrac{3}{4}\\
= {\left( {x – 2y} \right)^2} + 3.\left[ {{x^2} – 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right] – \dfrac{3}{4}\\
= {\left( {x – 2y} \right)^2} + 3.{\left( {x – \dfrac{1}{2}} \right)^2} – \dfrac{3}{4}\\
{\left( {x – 2y} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {x – \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow B = {\left( {x – 2y} \right)^2} + 3{\left( {x – \dfrac{1}{2}} \right)^2} – \dfrac{3}{4} \ge – \dfrac{3}{4},\,\,\,\forall x,y\\
\Rightarrow {B_{\min }} = – \dfrac{3}{4} \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – 2y} \right)^2} = 0\\
{\left( {x – \dfrac{1}{2}} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x – 2y = 0\\
x – \dfrac{1}{2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow {B_{\min }} = – \dfrac{3}{4} \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = \dfrac{1}{4}
\end{array} \right.\\
3,\\
2{a^2} + 9{b^2} + 6ab – 14a – 30b + 29 = 0\\
\Leftrightarrow \left( {{a^2} + 6ab + 9{b^2}} \right) + \left( { – 10a – 30b} \right) + 25 + \left( {{a^2} – 4a + 4} \right) = 0\\
\Leftrightarrow \left[ {{a^2} + 2.a.3b + {{\left( {3b} \right)}^2}} \right] – 10.\left( {a + 3b} \right) + 25 + \left( {{a^2} – 2.a.2 + {2^2}} \right) = 0\\
\Leftrightarrow {\left( {a + 3b} \right)^2} – 2.\left( {a + 3b} \right).5 + {5^2} + {\left( {a – 2} \right)^2} = 0\\
\Leftrightarrow {\left[ {\left( {a + 3b} \right) – 5} \right]^2} + {\left( {a – 2} \right)^2} = 0\\
\Leftrightarrow {\left( {a + 3b – 5} \right)^2} + {\left( {a – 2} \right)^2} = 0\\
{\left( {a + 3b – 5} \right)^2} \ge 0,\,\,\,\forall a,b\\
{\left( {a – 2} \right)^2} \ge 0,\,\,\forall a\\
\Rightarrow {\left( {a + 3b – 5} \right)^2} + {\left( {a – 2} \right)^2} \ge 0,\,\,\,\forall a,b\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {a + 3b – 5} \right)^2} = 0\\
{\left( {a – 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a + 3b – 5 = 0\\
a – 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 2\\
b = 1
\end{array} \right.\\
S = {\left( {a – 1} \right)^{2021}} – {\left( {1 – b} \right)^{2021}}\\
= {\left( {2 – 1} \right)^{2021}} – {\left( {1 – 1} \right)^{2021}}\\
= {1^{2021}} – {0^{2021}}\\
= 1 – 0\\
= 1
\end{array}\)