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Đột phá môn toán bằng casio

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Hướng dẫn tất cả toàn bộ cách giải toán 12 trắc nghiệm bằng máy tính casio, giúp học sinh đạt điểm tối đa trong kì thi THPT quốc gia sắp đến, gồm 5 chuyên đề: Hàm số, Mũ Loogarit, Nguyên hàm tích phân, Số phức, Hình học Không gian.

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 7tQK0{ÿXQqĺeĺ1KұSzĺ=ĺ.ӃWTXҧ

 7tQKVӕSKӭFOLrQKӧS z : qĺ2ĺ2ĺ1KұSzĺ=ĺ.ӃWTXҧ

 1KұSLOjb

1

L

L

+

7tQK0{ÿXQwYӟL Z ] L= + +1

*L̫L

  2(1 2 )

1 2

L L

L z

L

+ + −

+

=

+ aĺ

2(1 2 )

 

1 2

L L

L L

+ + −

+ + ĺ=ĺ.ӃWTXҧ (3 2 )+ L

%j,3+›¡1*75†1+6’3+&

Trang 15

,/êWKX\ÄWY¬YÉGÜ

7tFKFyKѭӟQJFӫDKDLYHFWѫ

*LҧVӱYHFWѫJJJGAB x  \  1 1 z1 CD xJJJG  \  2 2 z2

7tFKFyKѭӟQJFӫDKDLYHFWѫOjF{QJWKӭFzG= JJJG JJJGAB CD

%˱ͣF wĺ8PjQKuQKFKRYHFWѫA,B,C

%˱ͣF*LҧVӱJiQJJJGAB x  \  1 1 z1

OjYHFWѫAWKuEҩP1

%˱ͣF&KӑQ1

%˱ͣF1KұSx1= y1= z1 NKLÿyWDÿmJiQ[RQJYHFWѫ ABJJJGYjRYHFWѫ AJG

%˱ͣFҨQCÿӇWKRiW

%˱ͣFҨQqĺ5

%˱ͣFҨQ1 'LP ÿӇJiQJLiWUӏFKRYHFWѫ

%˱ͣFҨQ2ÿӇJiQYHFWѫ CD xJJJG  \  2 2 z2

YjRYHFWѫ BJG

%˱ͣF&KӑQ1

%˱ͣF1KұSx2= y2= z2 NKLÿyWDÿmJiQ[RQJYHFWѫCDJJJGYjRYHFWѫ BJG 1KѭYұ\FK~QJWDÿmQKұS[RQJYHFWѫ ABJJJGYj CDJJJGYjRPi\WtQK

%˱ͣFҨQCÿӇWKRiW

%˱ͣF7tFKFyKѭӟQJFӫD ABJJJGYj CDJJJGOj⎡⎣JJJG JJJGAB CD ⎤⎦

ҨQqĺ5ĺ3 JӑLYHFWѫ ABJJJG) ĺO GҩXQKkQ qĺ5ĺ4

JӑLYHFWѫ CDJJJG)ĺ=ĺNӃWTXҧ[= y = z= ĺFKӑQÿiSiQÿ~QJ

6DXNKLQKұS[RQJEѭӟFPi\FKRNӃWTXҧ>9FWA9FWB@

%j,%j,72k19HFWã

Trang 16

ĈӇJLҧLKӋSKѭѫQJWUuQKWDWKӵFKLӋQ

wĺ5ĺ16DXÿyQKұSSKѭѫQJWUuQKYjR

%˱ͣF3KѭѫQJWUuQKÿѭӡQJWKҷQJ G Oj



− = − = − ⇒ ÿiSiQ

+m\YLӃW G OjJLDRWX\ӃQFӫD 3 Yj 4 

 $ %

*L̫L

0һWSKҷQJ 3 Fy9737 uG =   −

0һWSKҷQJ 4 Fy9737 YG =   −

⇒ G Fy97&3 QG = ⎣ ⎦⎡X YG G ⎤

%˱ͣF1KұSYHFWѫ uG Yj YGYjRPi\

;RQJEѭӟFQKұSGӳOLӋX

%˱ͣF7tQKWtFKFyKѭӟQJ QG = ⎣ ⎦⎡X YG G ⎤

qĺ5ĺ3ĺOĺqĺ5ĺ4ĺ= ⇒ NӃWTXҧnG =    − −

%˱ͣF7uPÿLӇPWKXӝF G YjWKXӝF 3  4

*LҧVӱA x y (    ) ∈ d

( ) ( )

⇒ ⎨ ∈



4

x

x y

x y

y

⎧ =

⎪ + =

+ =

⎪⎩



x+ y+ z

x+ = y− = z



x+ yz



z

Trang 17

7tQKNKRҧQJFiFKJLӳD d1 Yj d2 

 & '

*L̫L

%˱ͣF&KӑQ 01   − ∈ G 01 2   ∈ G2 ⇒0 0JJJJJJG1 2 

%˱ͣF

 



X X 0 0

d d d

u u

=

G G JJJJJJG

G G

*iQuJG JJG1   u2

0 0

JJJJJJG

YjR

7tQKNKRҧQJFiFK9WFA9WFB9WFC

qĺeĺqĺ5ĺ3ĺOĺqĺ5ĺ4ĺ)ĺ

qĺ5ĺ7ĺqĺ5ĺ5ĺ)ĺPĺqĺeĺqĺ 5ĺ3ĺOĺqĺ5ĺ4ĺ)ĺ.ӃWTXҧ

⇒ ĈiSiQ&Ojÿ~QJ

,/êWKX\ÄWY¬YÉGÜ

7URQJNK{QJJLDQYӟLKӋWӑDÿӝ2[\]FKRKDLPһWSKҷQJ 31 Yj 32 YӟL 9737OҫQOѭӧWOjnG1

YjnG2

7tQKJyFJLӳDKDLPһWSKҷQJÿy

7DFyF{QJWKӭF

42 3

46 3

%j,7‡1+*Œ&*,Ÿ$+$,0z73+x1*+$,ô›¢1*

7+x1*ô›¢1*7+x1*9j0z73+x1*

 FRV 



n n

&RV Q Q

n n

JG JJG

JG JJG

JG JJG FRV

Trang 18

&iFKJL̫L

%˱ͣF;iFÿӏQK nJG1

YjnJJG2

%˱ͣF1KұS nJG1

YjnJJG2

YjRPi\

%˱ͣF6ӱGөQJqĺeYjWtFKY{KѭӟQJFӫDKDLYHFWѫ

(31) x+y+z − = (3  2) x+ −y z+ =  

7tQKFRVLQFӫDJyFJLӳDPһWSKҷQJÿy

 & 6

14



*L̫L

%˱ͣF7DFy nJG1       nJJG2 − Oj9737FӫD 31 Yj 32)

%˱ͣF1KұS nJG1

YjnJJG2

YjRPi\

%˱ͣF7tQK

qĺeĺ(ĺqĺ5ĺ3ĺqĺ5ĺ7ĺqĺ5ĺ

4ĺ)ĺPĺ(ĺqĺeĺqĺ5ĺ3ĺ)ĺOĺq ĺeĺ5ĺ4ĺ)ĺ)ĺ)ĺ= ӃWTXҧOjKD\

⇒ 9ұ\ÿiSiQÿ~QJOj&

(d1 Yj d2)

7tQKJyFJLӳDKDLÿѭӡQJWKҷQJ d1 Yj d2)

$ƒ %ƒ

&ƒ 'ƒ

*L̫L

%˱ͣF7DFy uJG1     − uJJG2 OjFiF97&3FӫD d1 Yj d2)

2



16



6



x− = y+ = z

2 1 2

x

=

⎪ = +

⎪ = − +

 FRV 



n n

&RV Q Q

n n

JG JJG

JG JJG

JG JJG FRV

Trang 19

ô

3x −3−x = Oj2

 $ % & '

3x −3−x = Oj2

 $ % & '

 $P< 2 % 2− < < P 2

 &P= 2 'P> KRһF2 P< − 2

 $ @ %>  &>@ '  

 $± %ORJ  2 & 'ORJ 2

 $ % & '

y= − − −x x − Oj

 $ −∞ −   % − +∞  &   − −  '[− −  ]

 $ < <D < < E  %< <D  E>

ORJ   ORJ ORJ   ORJ

+ = +

⎪⎩

2

2

1

   OQ

1

x

2

3

ORJ x + ORJ x + x− = 

a >a ORJ 3 ORJ 4

( )

3

2

1

2

x

x x

⎜ ⎟

2

x

x

y

y

+

+ + =

⎪⎩

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