Neidio i'r prif gynnwys
Datrys ar gyfer I (complex solution)
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Datrys ar gyfer I
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Datrys ar gyfer R (complex solution)
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Datrys ar gyfer R
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Problemau tebyg o chwiliad gwe

Rhannu

IRR\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)
Lluoswch ddwy ochr yr hafaliad â \left(r+1\right)^{2}.
IR^{2}\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)
Lluosi R a R i gael R^{2}.
IR^{2}\left(r^{2}+2r+1\right)=22000+\left(r+1\right)^{2}\left(-18000\right)
Defnyddio'r theorem binomaidd \left(a+b\right)^{2}=a^{2}+2ab+b^{2} i ehangu'r \left(r+1\right)^{2}.
IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)
Defnyddio’r briodwedd ddosbarthu i luosi IR^{2} â r^{2}+2r+1.
IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r^{2}+2r+1\right)\left(-18000\right)
Defnyddio'r theorem binomaidd \left(a+b\right)^{2}=a^{2}+2ab+b^{2} i ehangu'r \left(r+1\right)^{2}.
IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000-18000r^{2}-36000r-18000
Defnyddio’r briodwedd ddosbarthu i luosi r^{2}+2r+1 â -18000.
IR^{2}r^{2}+2IR^{2}r+IR^{2}=4000-18000r^{2}-36000r
Tynnu 18000 o 22000 i gael 4000.
\left(R^{2}r^{2}+2R^{2}r+R^{2}\right)I=4000-18000r^{2}-36000r
Cyfuno pob term sy'n cynnwys I.
\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I=4000-36000r-18000r^{2}
Mae'r hafaliad yn y ffurf safonol.
\frac{\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I}{R^{2}r^{2}+2rR^{2}+R^{2}}=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}
Rhannu’r ddwy ochr â R^{2}r^{2}+2rR^{2}+R^{2}.
I=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}
Mae rhannu â R^{2}r^{2}+2rR^{2}+R^{2} yn dad-wneud lluosi â R^{2}r^{2}+2rR^{2}+R^{2}.
I=\frac{2000\left(2-18r-9r^{2}\right)}{R^{2}\left(r+1\right)^{2}}
Rhannwch 4000-36000r-18000r^{2} â R^{2}r^{2}+2rR^{2}+R^{2}.
IRR\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)
Lluoswch ddwy ochr yr hafaliad â \left(r+1\right)^{2}.
IR^{2}\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)
Lluosi R a R i gael R^{2}.
IR^{2}\left(r^{2}+2r+1\right)=22000+\left(r+1\right)^{2}\left(-18000\right)
Defnyddio'r theorem binomaidd \left(a+b\right)^{2}=a^{2}+2ab+b^{2} i ehangu'r \left(r+1\right)^{2}.
IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)
Defnyddio’r briodwedd ddosbarthu i luosi IR^{2} â r^{2}+2r+1.
IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r^{2}+2r+1\right)\left(-18000\right)
Defnyddio'r theorem binomaidd \left(a+b\right)^{2}=a^{2}+2ab+b^{2} i ehangu'r \left(r+1\right)^{2}.
IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000-18000r^{2}-36000r-18000
Defnyddio’r briodwedd ddosbarthu i luosi r^{2}+2r+1 â -18000.
IR^{2}r^{2}+2IR^{2}r+IR^{2}=4000-18000r^{2}-36000r
Tynnu 18000 o 22000 i gael 4000.
\left(R^{2}r^{2}+2R^{2}r+R^{2}\right)I=4000-18000r^{2}-36000r
Cyfuno pob term sy'n cynnwys I.
\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I=4000-36000r-18000r^{2}
Mae'r hafaliad yn y ffurf safonol.
\frac{\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I}{R^{2}r^{2}+2rR^{2}+R^{2}}=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}
Rhannu’r ddwy ochr â R^{2}r^{2}+2rR^{2}+R^{2}.
I=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}
Mae rhannu â R^{2}r^{2}+2rR^{2}+R^{2} yn dad-wneud lluosi â R^{2}r^{2}+2rR^{2}+R^{2}.
I=\frac{2000\left(2-18r-9r^{2}\right)}{\left(R\left(r+1\right)\right)^{2}}
Rhannwch 4000-18000r^{2}-36000r â R^{2}r^{2}+2rR^{2}+R^{2}.