Mọi ng giúp mình với mình đang cần gấp
\(\)Bài 1 : \(y=\sqrt{x^4-3x^2+5}\Rightarrow y'=\dfrac{1}{2\sqrt{x^4-3x^2+5}}\left(x^4-3x^2+5\right)'=\dfrac{4x^3-6x}{2\sqrt{x^4-3x^2+5}}\)
\(y=cos\left(\dfrac{3x+2}{4x-1}\right)\Rightarrow y'=-sin\left(\dfrac{3x+2}{4x-1}\right).\left(\dfrac{3x+2}{4x-1}\right)'\)
Thấy : \(\left(\dfrac{3x+2}{4x-1}\right)'=\dfrac{\left(3x+2\right)'\left(4x-1\right)-\left(3x+2\right)\left(4x-1\right)'}{\left(4x-1\right)^2}\)
\(=\dfrac{\left(4x-1\right)3-4\left(3x+2\right)}{\left(4x-1\right)^2}=-\dfrac{11}{\left(4x-1\right)^2}\)
Suy ra : \(y'=sin\dfrac{3x+2}{4x-1}.\dfrac{11}{\left(4x-1\right)^2}\)
Bài 2 : a . \(\Delta SAC\) cân tại S ; có O là TĐ AC \(\Rightarrow SO\perp AC\) . CMTT : \(SO\perp BD\)
Suy ra : \(SO\perp\left(ABCD\right)\)
b . Ta có : \(SO\perp AC;AC\perp BD\Rightarrow AC\perp\left(SBD\right)\) . CMTT : \(BD\perp\left(SAC\right)\)
Bài 3 : a . SA \(\perp\left(ABCD\right)\Rightarrow SA\perp BC\)
Có : \(BC\perp AB\) . Suy ra : \(BC\perp\left(SAB\right)\Rightarrow BC\perp SB\)
Suy ra : \(BC\perp\left(SAB\right)\Rightarrow\left(SAB\right)\perp\left(SBC\right)\) ( đpcm )
b . Ta có : \(BD\perp SA;BD\perp AC\Rightarrow BD\perp\left(SAC\right)\Rightarrow SC\perp BD\)
Mà BD // HK \(\Rightarrow SC\perp HK\) (1)
* \(SA=AB\Rightarrow\Delta SAB\) cân tại A có : H là TĐ SB \(\Rightarrow AH\perp SB\)
Có : \(BC\perp\left(SAB\right)\Rightarrow BC\perp AH\) . Suy ra : \(AH\perp\left(SBC\right)\Rightarrow AH\perp SC\) (2)
Từ (1) và (2) suy ra : \(SC\perp\left(AHK\right)\) ( đpcm )
Bài 1 : y=cos(3x+24x−1)⇒y′=−sin(3x+24x−1).(3x+24x−1)′y=cos(3x+24x−1)⇒y′=−sin(3x+24x−1).(3x+24x−1)′
Thấy : =(4x−1)3−4(3x+2)(4x−1)2=−11(4x−1)2=(4x−1)3−4(3x+2)(4x−1)2=−11(4x−1)2
Suy ra :
