tìm x y ∈ z biết x.y = 2016 và x+y=-95 tìm x,y biết $\frac{2x}{3}$ = $\frac{5y}{4}$ và x+ y=25 20/11/2021 Bởi Ariana tìm x y ∈ z biết x.y = 2016 và x+y=-95 tìm x,y biết $\frac{2x}{3}$ = $\frac{5y}{4}$ và x+ y=25
Đáp án: 1) \(\left[ \begin{array}{l}y = – 32\\y = – 63\end{array} \right. \to \left[ \begin{array}{l}x = – 63\\x = – 32\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}1)xy = 2016\\ \to x = \dfrac{{2016}}{y}\left( {y \ne 0} \right)\\Thay:x = \dfrac{{2016}}{y}\\Pt\left( 2 \right):\dfrac{{2016}}{y} + y = – 95\\ \to {y^2} + 95y + 2016 = 0\\ \to {y^2} + 32y + 63y + 2016 = 0\\ \to y\left( {y + 32} \right) + 63\left( {y + 32} \right) = 0\\ \to \left( {y + 32} \right)\left( {y + 63} \right) = 0\\ \to \left[ \begin{array}{l}y = – 32\\y = – 63\end{array} \right. \to \left[ \begin{array}{l}x = – 63\\x = – 32\end{array} \right.\\2)\dfrac{2}{3}x = \dfrac{5}{4}y\\ \to 8x = 15y\\ \to x = \dfrac{{15y}}{8}\\Thay:x = \dfrac{{15y}}{8}\\Pt\left( 2 \right):\dfrac{{15y}}{8} + y = 25\\ \to \dfrac{{23}}{8}y = 25\\ \to y = \dfrac{{200}}{{23}}\\ \to x = \dfrac{{375}}{{23}}\end{array}\) Bình luận
Đáp án+ Giải thích các bước giải: $1, x + y = – 95 → y = – 95 – x$ $thay$ $vào$ $xy = 2016$ $ta$ $có:$ $x(-95 – x) = 2016$ $→ – x² – 95x = 2016$ $→ x² + 95x + 2016 = 0$ $→ (x² + 63x) + (32x + 2016) = 0$ $→ x(x + 63) + 32(x + 63) = 0$ $→ (x + 32)(x + 63) = 0$ $→\left[ \begin{array}{l}x + 32 = 0\\x+ 62 = 0\end{array} \right.→\left[ \begin{array}{l}x=-32\\x=-63\end{array} \right.$ $→\left[ \begin{array}{l}y = – 63\\y = -32\end{array} \right.$ $Vậy$ $(x ; y) = (- 32 ; – 63) ; (- 63 ; – 32)$ $2, x + y = 25 → x = 25 – y$ $thay$ $vào$ $\frac{2x}{3} = \frac{5y}{4}$ $ta có:$ $\frac{2(25-y)}{3}=\frac{5y}{4}$ $→ 8(25 – y)= 15y$ $→ 200 – 8y = 15y$ $→ 23y = 200$ $→ y = \frac{200}{23}$ $→ x = 25 – \frac{200}{23} = \frac{375}{23}$ $Vậy$ $(x ;y) = (\frac{375}{23} ; \frac{200}{23})$ Bình luận
Đáp án:
1) \(\left[ \begin{array}{l}
y = – 32\\
y = – 63
\end{array} \right. \to \left[ \begin{array}{l}
x = – 63\\
x = – 32
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)xy = 2016\\
\to x = \dfrac{{2016}}{y}\left( {y \ne 0} \right)\\
Thay:x = \dfrac{{2016}}{y}\\
Pt\left( 2 \right):\dfrac{{2016}}{y} + y = – 95\\
\to {y^2} + 95y + 2016 = 0\\
\to {y^2} + 32y + 63y + 2016 = 0\\
\to y\left( {y + 32} \right) + 63\left( {y + 32} \right) = 0\\
\to \left( {y + 32} \right)\left( {y + 63} \right) = 0\\
\to \left[ \begin{array}{l}
y = – 32\\
y = – 63
\end{array} \right. \to \left[ \begin{array}{l}
x = – 63\\
x = – 32
\end{array} \right.\\
2)\dfrac{2}{3}x = \dfrac{5}{4}y\\
\to 8x = 15y\\
\to x = \dfrac{{15y}}{8}\\
Thay:x = \dfrac{{15y}}{8}\\
Pt\left( 2 \right):\dfrac{{15y}}{8} + y = 25\\
\to \dfrac{{23}}{8}y = 25\\
\to y = \dfrac{{200}}{{23}}\\
\to x = \dfrac{{375}}{{23}}
\end{array}\)
Đáp án+ Giải thích các bước giải:
$1, x + y = – 95 → y = – 95 – x$
$thay$ $vào$ $xy = 2016$ $ta$ $có:$
$x(-95 – x) = 2016$
$→ – x² – 95x = 2016$
$→ x² + 95x + 2016 = 0$
$→ (x² + 63x) + (32x + 2016) = 0$
$→ x(x + 63) + 32(x + 63) = 0$
$→ (x + 32)(x + 63) = 0$
$→\left[ \begin{array}{l}x + 32 = 0\\x+ 62 = 0\end{array} \right.→\left[ \begin{array}{l}x=-32\\x=-63\end{array} \right.$
$→\left[ \begin{array}{l}y = – 63\\y = -32\end{array} \right.$
$Vậy$ $(x ; y) = (- 32 ; – 63) ; (- 63 ; – 32)$
$2, x + y = 25 → x = 25 – y$
$thay$ $vào$ $\frac{2x}{3} = \frac{5y}{4}$ $ta có:$
$\frac{2(25-y)}{3}=\frac{5y}{4}$
$→ 8(25 – y)= 15y$
$→ 200 – 8y = 15y$
$→ 23y = 200$ $→ y = \frac{200}{23}$
$→ x = 25 – \frac{200}{23} = \frac{375}{23}$
$Vậy$ $(x ;y) = (\frac{375}{23} ; \frac{200}{23})$