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Đáp án:
a)⎧⎪ ⎪⎨⎪ ⎪⎩x6=y4⇔x24=y16x8=z−3⇔x24=z−9⇔x24=y16=z−9=x+y+z24+16−9=231⇔⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x=24.231=4831y=16.231=3231z=−9.231=−1831Vậyx=4831;y=3231;z=−1831b)⎧⎪ ⎪⎨⎪ ⎪⎩x3=y10⇔x9=y30y15=z4⇔y30=z8⇔x9=y30=z8=3x27=3x−y27−30=6−3=−2⇔⎧⎪⎨⎪⎩x=−18y=−60z=−16c)x:y:z=3:2:5⇔x3=y2=z5=2x6=4y8=z−2x+4y5−6+8=147=2⇔⎧⎪⎨⎪⎩x=6y=4z=10Vậyx=6;y=4;z=10d)3x=4y=2z⇔3x12=4y12=2z12⇔x4=y3=z6=y−z3−6=6−3=−2⇔⎧⎪⎨⎪⎩x=−8y=−6z=−12Vậyx=−8;y=−6;z=−12
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Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
\dfrac{x}{6} = \dfrac{y}{4} \Leftrightarrow \dfrac{x}{{24}} = \dfrac{y}{{16}}\\
\dfrac{x}{8} = \dfrac{z}{{ - 3}} \Leftrightarrow \dfrac{x}{{24}} = \dfrac{z}{{ - 9}}
\end{array} \right.\\
\Leftrightarrow \dfrac{x}{{24}} = \dfrac{y}{{16}} = \dfrac{z}{{ - 9}} = \dfrac{{x + y + z}}{{24 + 16 - 9}} = \dfrac{2}{{31}}\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 24.\dfrac{2}{{31}} = \dfrac{{48}}{{31}}\\
y = 16.\dfrac{2}{{31}} = \dfrac{{32}}{{31}}\\
z = - 9.\dfrac{2}{{31}} = \dfrac{{ - 18}}{{31}}
\end{array} \right.\\
Vậy\,x = \dfrac{{48}}{{31}};y = \dfrac{{32}}{{31}};z = - \dfrac{{18}}{{31}}\\
b)\left\{ \begin{array}{l}
\dfrac{x}{3} = \dfrac{y}{{10}} \Leftrightarrow \dfrac{x}{9} = \dfrac{y}{{30}}\\
\dfrac{y}{{15}} = \dfrac{z}{4} \Leftrightarrow \dfrac{y}{{30}} = \dfrac{z}{8}
\end{array} \right.\\
\Leftrightarrow \dfrac{x}{9} = \dfrac{y}{{30}} = \dfrac{z}{8} = \dfrac{{3x}}{{27}} = \dfrac{{3x - y}}{{27 - 30}} = \dfrac{6}{{ - 3}} = - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 18\\
y = - 60\\
z = - 16
\end{array} \right.\\
c)x:y:z = 3:2:5\\
\Leftrightarrow \dfrac{x}{3} = \dfrac{y}{2} = \dfrac{z}{5} = \dfrac{{2x}}{6} = \dfrac{{4y}}{8}\\
= \dfrac{{z - 2x + 4y}}{{5 - 6 + 8}} = \dfrac{{14}}{7} = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 6\\
y = 4\\
z = 10
\end{array} \right.\\
Vậy\,x = 6;y = 4;z = 10\\
d)3x = 4y = 2z\\
\Leftrightarrow \dfrac{{3x}}{{12}} = \dfrac{{4y}}{{12}} = \dfrac{{2z}}{{12}}\\
\Leftrightarrow \dfrac{x}{4} = \dfrac{y}{3} = \dfrac{z}{6} = \dfrac{{y - z}}{{3 - 6}} = \dfrac{6}{{ - 3}} = - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 8\\
y = - 6\\
z = - 12
\end{array} \right.\\
Vậy\,x = - 8;y = - 6;z = - 12
\end{array}$
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