Toán Lớp 8: các bạn làm giúp mình với nha
a.$x^{2}$ -1-$4y^{2}$ +$4y^{}$
b.$2x^{2}$- $y^{2}$ +xy
c.$x^{2}$- $20x^{}$ +84=0
d.Tìm x,y là số nguyên biết:2xy+4x+2y+1>$5x^{2}$+ $2y^{2}$
e. ($x^{2}$ +4x)$^{2}$ -2($x^{2}$ +4x)-15=0
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Toán Lớp 8: các bạn làm giúp mình với nha
a.$x^{2}$ -1-$4y^{2}$ +$4y^{}$
b.$2x^{2}$- $y^{2}$ +xy
c.$x^{2}$- $20x^{}$ +84=0
d.Tìm x,y là số nguyên biết:2xy+4x+2y+1>$5x^{2}$+ $2y^{2}$
e. ($x^{2}$ +4x)$^{2}$ -2($x^{2}$ +4x)-15=0
Comments ( 1 )
a){x^2} – 1 – 4{y^2} + 4y\\
= {x^2} – \left( {4{y^2} – 4y + 1} \right)\\
= {x^2} – {\left( {2y – 1} \right)^2}\\
= \left( {x – 2y + 1} \right)\left( {x + 2y – 1} \right)\\
b)2{x^2} – {y^2} + xy\\
= 2{x^2} + 2xy – xy – {y^2}\\
= \left( {x + y} \right)\left( {2x – y} \right)\\
c){x^2} – 20x + 84 = 0\\
\Leftrightarrow {x^2} – 6x – 14x + 84 = 0\\
\Leftrightarrow \left( {x – 6} \right)\left( {x – 14} \right) = 0\\
\Leftrightarrow x = 6;x = 14\\
Vậy\,x = 6;x = 14\\
d)2xy + 4x + 2y + 1 > 5{x^2} + 2{y^2}\\
\Leftrightarrow 5{x^2} + 2{y^2} – 2xy – 4x – 2y – 1 < 0\\
\Leftrightarrow \left( {4{x^2} – 4x + 1} \right) + \left( {{x^2} – 2xy + {y^2}} \right)\\
+ {y^2} – 2y + 1 – 3 < 0\\
\Leftrightarrow {\left( {2x – 1} \right)^2} + {\left( {x – y} \right)^2} + {\left( {y – 1} \right)^2} < 3\\
Do:x;y \in Z\\
{\left( {2x – 1} \right)^2} + {\left( {x – y} \right)^2} + {\left( {y – 1} \right)^2} \ge 0\\
+ TH1:\left\{ \begin{array}{l}
{\left( {2x – 1} \right)^2} = 0\\
{\left( {x – y} \right)^2} = 0\\
{\left( {y – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
x = y\\
y = 1
\end{array} \right.\left( {ktm} \right)\\
+ TH2::\left\{ \begin{array}{l}
{\left( {2x – 1} \right)^2} = 1\\
{\left( {x – y} \right)^2} = 0\\
{\left( {y – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
x = y\\
y = 1
\end{array} \right. \Leftrightarrow x = y = 1\\
+ TH3:\left\{ \begin{array}{l}
{\left( {2x – 1} \right)^2} = 0\\
{\left( {x – y} \right)^2} = 1\\
{\left( {y – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
x = y + 1/x = y – 1\\
y = 1
\end{array} \right.\left( {ktm} \right)\\
+ TH4::\left\{ \begin{array}{l}
{\left( {2x – 1} \right)^2} = 0\\
{\left( {x – y} \right)^2} = 0\\
{\left( {y – 1} \right)^2} = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
x = y\\
y = 0/y = 2
\end{array} \right.\left( {ktm} \right)\\
+ TH5:\left\{ \begin{array}{l}
{\left( {2x – 1} \right)^2} = 1\\
{\left( {x – y} \right)^2} = 1\\
{\left( {y – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0/x = 1\\
x = y + 1/x = y – 1\\
y = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = 1
\end{array} \right.\\
+ TH6::\left\{ \begin{array}{l}
{\left( {2x – 1} \right)^2} = 1\\
{\left( {x – y} \right)^2} = 0\\
{\left( {y – 1} \right)^2} = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1/x = 0\\
x = y\\
y = 1/y = 2
\end{array} \right. \Leftrightarrow x = y = 1\\
+ TH7::\left\{ \begin{array}{l}
{\left( {2x – 1} \right)^2} = 0\\
{\left( {x – y} \right)^2} = 1\\
{\left( {y – 1} \right)^2} = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
x = y + 1/x = y – 1\\
y = 1/y = 2
\end{array} \right.\left( {ktm} \right)\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {1;1} \right);\left( {0;1} \right)} \right\}\\
e){\left( {{x^2} + 4x} \right)^2} – 2\left( {{x^2} + 4x} \right) – 15 = 0\\
Dat:{x^2} + 4x = a\\
\Leftrightarrow {a^2} – 2a – 15 = 0\\
\Leftrightarrow {a^2} – 5a + 3a – 15 = 0\\
\Leftrightarrow \left( {a – 5} \right)\left( {a + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 4x – 5 = 0\\
{x^2} + 4x + 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {x – 1} \right)\left( {x + 5} \right) = 0\\
\left( {x + 1} \right)\left( {x + 3} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = – 5\\
x = – 1\\
x = – 3
\end{array} \right.\\
Vậy\,x \in \left\{ { – 5; – 3; – 1;1} \right\}
\end{array}$