Toán Lớp 9: Giải tìm ra OH hộ mik nha
$\frac{1}{OH^2}$ = $\frac{1}{(-3m + 2)^2}$ + $\frac{1}{(\frac{3m – 2}{m} )^2}$
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Toán Lớp 9: Giải tìm ra OH hộ mik nha
$\frac{1}{OH^2}$ = $\frac{1}{(-3m + 2)^2}$ + $\frac{1}{(\frac{3m – 2}{m} )^2}$
Comments ( 2 )
1 / [ OH² ] = 1 / [ ( -3m + 2 )² ] + 1 / [ ( [ 3m – 2 ] / [ m ] )² ] ( ĐK : m $\ne$ 2/3 )
⇔ 1 / [ OH² ] = 1 / [ ( 2 – 3m )² ] + ( m / [ 3m – 2 ] )²
⇔ 1 / [ OH² ] = 1 / [ ( 2 – 3m )² ] + [ m² ] / [ ( 3m – 2 )² ]
⇔ 1 / [ OH² ] = 1 / [ ( 3m – 2 )² ] + [ m² ] / [ ( 3m – 2 )² ]
⇔ 1 / [ OH² ] = [ m² + 1 ] / [ ( 3m – 2 )² ]
⇔ OH² = [ ( 3m – 2 )² ] / [ m² + 1 ]
⇔ OH = $\dfrac{ | 3m – 2 | }{ \sqrt{ m² + 1 } }$
Giải đáp:
1/(OH^2) = 1/(-3m +2)^2 + 1/((3m-2)/m)^2
<=> 1/(OH^2) = 1/(3m -2)^2 + (m /(3m -2))^2$\\$ <=> 1/(OH^2) = 1/(3m -2)^2 + m^2/(3m-2)^2$\\$ <=> 1/(OH^2) = (1 + m^2)/(3m-2)^2$\\$ <=> OH^2 = (3m -2)^2/(1+m^2)$\\$ <=> OH = (\sqrt{(3m-2)^2})/(\sqrt{1 + m^2})$\\$ <=> OH = (|3m -2|)/(\sqrt{1+m^2})$\\$