HÖ thèng c¸c bµi tËp «n tËp To¸n THCS
Ph¬ng tr×nh qui vÒ ph¬ng tr×nh bËc hai
Bµi 1 Gi¶i c¸c ph¬ng tr×nh sau:
a) x− 6 x+ 5 = 0 b) − 2x+ 5 x+ 7 = 0 c) − x+ 8x− 9 = 0
1
20
=
+
x e) (2x+1)(x−1)=−2x f) x4 −13x2 +36=0
g) 9x4 +6x2+1=0 h) 2x4 −5x2 +3=0 i) − x4 +5x2+6= 0
5
100
5
100
=
−
+
2
2 2
2
=
+ +
x x
x
l)
1
1 2 2
1 2
1
+
+
=
−
− + +
+
x
x x
x x
x
Bµi 2 Gi¶i c¸c ph¬ng tr×nh sau:
a)x5 −x3 −x2 +1= 0 b) 6x4+7x3−36x2 −7x+6=0 c)2x3+7x2+7x+2=0
d) x3 −8x2 −8x+1=0e) x3+ x2 +4=0 g)x3 −5x2 +8x−4= 0
Bµi 3 Gi¶i c¸c ph¬ng tr×nh sau:
a) x(x+1)(x+2)(x+3)+1=0 b) ( x + 2 ) ( x + 3 ) ( x − 7 ) ( x − = 8 ) 144
c) (x−1)(x−3)(x+5)(x+7) =297 d) (x−1)(x+2)(x+3)(x+6)=108
e) (x−1)(x−3)(x−5)(x−7)=20 f) (4x+1)(12x−1)(3x+2)(x+1)=4
g) (6 5) (2 3 2)( 1) 35
= + +
x h) (12 7) (2 3 2)(2 1) 3
= + +
x
i) ( 1) (2 2 1)(2 3) 18
= + +
x j) ( x − 4 ) ( x − 5 ) ( x − 8 ) ( x − 10 ) = 72 x2
k) ( x + 10 ) ( x + 12 ) ( x + 15 ) ( x + 18 ) = 2 x2
Bµi 4 Gi¶i c¸c ph¬ng tr×nh sau:
a) ( 3) (4 5)4 2
= + +
x b) ( 1) (4 3)4 82
=
− +
x
c) (x− 2) (4 + 6 −x)4 = 82 d) (x− 2) (4 + x− 4)4 = 64
Bµi 5 Gi¶i c¸c ph¬ng tr×nh sau:
a) x4 −10x3 +26x2 −10x+1=0 b) x4−4x3−6x2 −4x+1=0
c) x4+2x3 −x2−2x+1=0 d) x4+3x3−14x2 −6x+4=0
e) x4 − 4 x3 − 9 x2 + 8 x + = 4 0 f) x4 + 5 x3 + 10 x2 + 15 x + = 9 0
Bµi 6 Gi¶i c¸c ph¬ng tr×nh sau:
5
3 5
2
2
= +
− + +
−
+
x x
x x
x
5
7
2
− +
− +
x x x x
10
4
+
2 2
4
= + +
x
Bµi 7 Gi¶i c¸c hÖ ph¬ng tr×nh sau:
=
−
−
= +
+
3
1
2 2
xy y x
y xy
x
a
= +
=
+
10
58
2 2
y x
y
x b
−=
+
= +
−
2
13
2 2
y x
y xy
x c
= + +
= +
+
2
4
2 2
xy y x
y xy
x d
Bµi 8 Gi¶i c¸c hÖ ph¬ng tr×nh sau:
.
a
− = −
=
−
=
−
y
x x y
x
y y
x b
4 3
4
3
+
=
−
+
=
−
x y x y
y x y
x c
2 2
2
2 2 2
2 2
= +
=
+
y xy y
x xy
x d
3 2
3
2 2
2
- Biªn so¹n néi dung: ThÇy NguyÔn Cao C êng – 0904.15.16.50