2c) Ta có :

Bài 1 ( ảnh 2 )
a) x^4 - 3x^2 - 4= 0
= ( x^2 - 4 ).( x^2 + 1 ) = 0
= ( x - 2 ).( x + 2 ).( x^2 + 1 ) = 0
=> ( x - 2 ).( x + 2 ) = 0 ( vì x^2 + 1 > 0 )
=> x = 2 hoặc x = -2
b) x^4 - 5x^2 + 4 = 0
<=> ( x^2 - 1 ).( x^2 - 4 ) = 0
<=> ( x - 1 ).( x + 1 ).( x - 4 ).( x + 4 ) = 0
<=> x = 1 hoặc x = -1 hoặc x = 4 hoặc x = -4
c) x^4 + 5x^2 + 6 = 0
<=> ( x^2 + 2 ).( x^2 +3 ) = 0
Do x^2 + 2 > 0. x^2 = 3 > 0
=> Phương trình trên vô nghiệm
 

3a) Áp dụng bất đẳng thức Cosi cho 2 số ( x+ 3)^ 4 và (x +5) ^4: ( x+ 3)^ 4 + (x +5) ^4 >= 2* {(x+3)(x+5)}^2
= 2* (x^2 + 8x +15)^2 = 2* {(x+4)^2 -1}^2
Do (x+4)^2 >=0
=> (x+4)^2 -1>= -1
=> {(x+4)^2 -1}^2 >=1
=> VT pt >= 2 ( VP ptr)
Dấu " =" xảy ra khi ( x+ 3)^ 4 = (x +5) ^4 <=> x + 3 = x + 5 hoặc x + 3 = - (x + 5)
<=> x = - 4

3b)
(x + 1)^4 + (x + 3)^4 = 16
đặt t = (x + 1)(x + 3)
<<nhớ hai hằng đẳng thức:
a² + b² = (a - b)² + 2ab và a² + b² = (a + b)² - 2ab
Có:
*(x + 1)² + (x + 3)² = [(x + 1) - (x + 3)]² + 2(x + 1)(x + 3) = 4 + 2t
*(x + 1)^4 + (x + 3)^4 = [(x + 1)² + (x + 3)²]² - 2(x + 1)²(x + 3)² =
= [4 + 2t]² - 2t²
Ta có pt:
(4 + 2t)² - 2t² = 16 <=> 16 + 16t + 4t² - 2t² = 16
<=> t = 0 hoặc t = -8
*(x + 1)(x + 3) = 0 <=> x=-1 or x = -3
* (x + 1)(x + 3) = - 8 <=> x² + 4x + 11= 0 vn
Vậy tập nghiệm của pt là: {-1;-3}

4a

4d

3a

3b) (x+2)(x+3)(x+1)(x+6)=-36
<=>(x^2-x-6)(x+1)(x+6)=-36
<=>(x^3+x^2-x^2-x-6x-6)(x+6)=-36
<=>(x^3-7x-6)(x+6)=-36
<=>x^4+6x^3-7x^2-42x-6x-36=-36
<=>x^4+6x^3-7x^2-42x-6x=0
<=>x^4+6x^3-7x^2-48x=0
<=>x(x^3+6x^2-7x-48)=0
<=>x(x^3+3x^2+3x^2+9x-16x-48)=0
<=>x[x^2(x+3)+3x(x+3)-16(x+3)]=0
<=>x(x+3)(x^2+3x-16)=0
<=> x = 0 or x +3=0 or x^2+3x-16=0
<=>x=0 or x= -3 or x = - 3 + căn 73/2 or x = - 3 - căn 73/2

3c

3e

3f) 6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0
<=> 6x^4 - 5x^3 - 30x^3 + x^2 + 25x^2 + 36x^2 - 5x - 30x + 6 = 0
<=> 6x^4 - 5x^3 + x^2 - 30x^3 + 25x^2 - 5x + 36x^2 - 30x + 6 = 0
<=> x^2.( 6x^2 - 5x + 1 ) - 5x.( 6x^2 - 5x + 1 ) + 6.( 6x^2 - 5x + 1 ) = 0
<=> ( 6x^2 - 5x + 1 )( x^2 - 5x + 6 ) = 0
<=> 6x^2 - 5x +1 = 0 or x^2 - 5x + 6 = 0
<=> x =1/3 or x = 1/2 or x = 2 or x =3
3g) trên ảnh

4a) 2x+1/3x+2=x+1/x-2
ĐK: x#-2/3,x#2
<=> (2x+1)(x-2)=(x+1)(3x+2)
<=>(2x+1)(x-2)-(x+1)(3x+2)=0
<=>2x^2-4x+x-2-3x^2-5x-2=0
<=>-x^2-8x-4=0
<=>x^2+8x+4=0
<=> x = - 4 + 2 căn 3 (TM) or x =-4-2 căn 3 (TM)
b) 1+2/x-2=10/x+3-50/(2-x)(x+3)
ĐK: x#2,x#-3
<=> 2/x-2-10/x+3+50/(2-x)(x+3) = - 1
<=> 2/x-2-10/x+3-50/(x-2)(x+3) = - 1
<=> 2(x+3)-10(x-2)-50/(x-2)(x+3)=-1
<=> -8x-24/(x-2)(x+3) = -1
<=> -8(x+3)/(x-2)(x+3)=-1
<=> -8/x-2=-1
<=> -(x - 2) = -8
<=> -x +2 =-8
<=> -x=-10
<=>x=10(TM)

2c) Ta có :
(x+1)(x+2)(x+3)(x+4) = 3
(x+1)(x+4)(x+2)(x+3) = 3
(x^2+5x+4)(x^2+5x+6) = 3 (1)
Đặt : t = (x^2+5x+4) => (x^2+5x+6) = t+2
(1) => t.(t+2) = 3 => t^2+2t-3 = 0 => t = -3 or t = 1
* Với t = -3 => ko có gt x thích hợp
* Với t = 1 => x = -5 - căn 13/2, x = - 5 + căn 13/2
* Vậy : .....
3b) (x+2)(x+3)(x+1)(x+6)=-36
<=>(x^2-x-6)(x+1)(x+6)=-36
<=>(x^3+x^2-x^2-x-6x-6)(x+6)=-36
<=>(x^3-7x-6)(x+6)=-36
<=>x^4+6x^3-7x^2-42x-6x-36=-36
<=>x^4+6x^3-7x^2-42x-6x=0
<=>x^4+6x^3-7x^2-48x=0
<=>x(x^3+6x^2-7x-48)=0
<=>x(x^3+3x^2+3x^2+9x-16x-48)=0
<=>x[x^2(x+3)+3x(x+3)-16(x+3)]=0
<=>x(x+3)(x^2+3x-16)=0
<=> x = 0 or x +3=0 or x^2+3x-16=0
<=>x=0 or x= -3 or x = - 3 + căn 73/2 or x = - 3 - căn 73/2

4c) x+3/(x+1)^2=4x-2/(2x-1)^2
ĐK: x#-1,x#1/2
<=> x+3/(x+1)^2=2(2x-1)/(2x-1)^2
<=>x+3/(x+1)^2=2/2x-1
<=> (x+3)(2x-1)=2(x+1)^2
<=> 2x^2-x+6x-3=2(x^2+2x+1)
<=> 5x-3=4x+2
<=> x = 5(TM)
d) x+1/x+2+x-1/x-2=2x+1/x+1
ĐK:x#-2,x#2,X#-1
<=> (x-2)(x+1)^2+(x+2)(x+1)(x-1)-(x+2)(x-2)(2x+1)/(x+2)(x-2)(x+1) = 0
<=> x^2+4x/(x+2)(x-2)(x+1)=0
<=>4x+x^2=0
<=>x(4+x)=0
<=> x = 0 or 4+x=0
<=> x=0 or x=-4

5a) 4/2x+1+3/2x+2=2/2x+3+1/2x+4
ĐKXĐ: x#{-1/2;-1;-3/2;-2}
<=> 4(2x+2)+3(2x+1)/(2x+1)(2x+2)=2(2x+4)+2x+3/(2x+3)(2x+4)
<=> 14x+11/(2x+1)(2x+2)-6x+11/(2x+3)(2x+4)=0
<=>(2x+3)(2x+4)(14x+11)-(2x+1)(2x+2)(6x+11)/(2x+1)2(x+1)(2x+3)(2x+4)=0
<=> 32x^3+160x^2+244x+110/(2x+14)2(x+1)(2x+3)(2x+4)=0
<=>16x^3+80x^2+112x+55=0
<=> (2x+5)(8x^2+20x+11)=0
<=>x=-5/2orx= -5 + cawn3/4 or x =-5 - căn 3/4(TM hết)
5b) ĐKXĐ : x # {0 ,1,3,4,7,10}
(1/x^2+5x+4)+ (1/x^2+11x+28)+ (1/x^2+17x+70)-3/4x-2 = 0
<=>1/(x+1)(x+4) + 1/(x+4)(x+7)+1/(x+7)(x+10) - 3/4x - 2=0
<=>1/3(1/(x+1) - 1/(x+4)+1/(x+4)-1/(x+7)+1/(x+7)-1/(x+10)...
9/4x - 3/2) = 0
<=>1/(x+1) - 1/(x+10)-9/4x - 3/2 = 0(quy đồng mẫu thức)
<=>-6x^3 - 75x^2 - 119x - 90 = 0
<=>6x^3 + 75x^2 + 119x + 90 = 0
Tự giải