Tài liệu Phương trình chứa căn

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ngocyenduong

Tham gia: 01/08/2016

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Giaûi caùc phöông trình sau : 1/ x +2x 3  2x  x  3 =0 ( nhaåm nghieäm x =1; x=3 ) 3 Giaûi : Ñk : 3 ≤ x ≤ 2 Nhaùp : 3  2x =ax+b thay x=1 vaø x=3 => heä 2 1  a  1  a  b  2 : <=>  3 3   3a  b  b   2 Töông töï : x  3 =mx+n thay x=1 vaø x=3 1  m  2 2  m  n => heä :  <=>  0  3m  n n  3  2 1 3 1 3 x2 +2x3  x+  3  2x + x+  x  3 =0 2 2 2 2 2 2 3 3  1 1   x    (3  2x)  x    (3  x) 2 2 2 2 2 <=> x +2x3 +  + =0 1 3 1 3  x   3  2x x   3 x 2 2 2 2 1 2 3 9 1 2 3 9 x  x   (3  2x) x  x   (3  x) 2 4 2 4 <=> x2 +2x3 + 4 +4 =0 1 3 1 3  x   3  2x x   3 x 2 2 2 2 1 2 1 3 1 2 1 3 x  x x  x 2 4 + 4 2 4 =0 <=> x2 +2x3 + 4 1 3 1 3  x   3  2x x   3 x 2 2 2 2 1 1 4 4 <=> (x2+2x3)(1+ + ) =0 1 3 1 3  x   3  2x x   3 x 2 2 2 2 2 <=> x +2x3=0 <=> x=1  x=3 2/ x2 +3x3  14  x  11  x = 0 3/ 16x2 54x +57 4 2x  1 4 7  4x =0 4/ 2x2 5x +10  12  6x  26x  12 =0 5/ x2 2x+3  4x  1  4  2x =0 9 19 6/ 4x2  x + 2 2  x  11x  5 =0 5 5 2 7/ 2x x+18  12  6x 3 26x  12 =0 8/ x2 +5 3 4x  1  4  2x =0 Giaûi caùc phöông trình sau : 1/ x +2x 3  2x  x  3 =0 3 Giaûi : Ñk : 3 ≤ x ≤ 2 1 3 1 3 x2 +2x3  x+  3  2x + x+  x  3 =0 2 2 2 2 2 2 3 3  1 1   x    (3  2x)  x    (3  x) 2 2 2 2 2 <=> x +2x3 +  + =0 1 3 1 3  x   3  2x x   3 x 2 2 2 2 1 2 3 9 1 2 3 9 x  x   (3  2x) x  x   (3  x) 2 4 2 4 <=> x2 +2x3 + 4 +4 =0 1 3 1 3  x   3  2x x   3 x 2 2 2 2 1 2 1 3 1 2 1 3 x  x x  x 2 4 2 4 4 2 4 =0 <=> x +2x3 + + 1 3 1 3  x   3  2x x   3 x 2 2 2 2 1 1 4 4 <=> (x2+2x3)(1+ + ) =0 1 3 1 3  x   3  2x x   3 x 2 2 2 2 2 <=> x +2x3=0 <=> x=1  x=3 2/ x2 +3x3  14  x  11  x = 0 Giaûi : ñk : 14≤ x ≤ 11 1 26 1 23 Pt : x2 +3x10 + x +  14  x  x +  11  x =0 7 7 7 7 2 2 26  23  1  1  x    (14  x)   x    (11  x) 7 7  7 7  2 <=> x +3x10 +  + =0 1 26 1 23 x  14  x  x   11  x 7 7 7 7 2 1 2 3 10 1 2 3 10 x  x x  x 49 49 + 49 49 49 =0 <=> x2 +3x10 + 49 1 26 1 23 x  14  x  x   11  x 7 7 7 7 1 1 49 49 <=> (x2 +3x10)(1+ + ) =0 1 26 1 23 x  14  x  x   11  x 7 7 7 7 2 <=> x +3x10=0 <=> x=2  x=5 3/ 16x2 54x +57 4 2x  1 4 7  4x =0 1 7 Giaûi : ÑK:  ≤ x ≤ 2 4 2 Pt : 16x 48x +36 + 2x+54 2x  1 +168x4 7  4x =0 (2x  5) 2  16(2x  1) 4[(4  2x) 2  (7  4x)] <=> (4x6)2 + + =0 2x  5  4 2x  1 4  2x  7  4x (2x  3) 2 4(2x  3) 2 2 <=> (4x6) + + =0 2x  5  4 2x  1 4  2x  7  4x 1 4 <=> (2x3)2[ 4+ + ] =0 2x  5  4 2x  1 4  2x  7  4x 3 <=> (2x3)2 =0 <=> x= 2 2 4/ 2x 5x +10  12  6x  26x  12 =0 6 Giaûi : ñk :  ≤ x ≤ 2 13 2 Pt : 2x 5x+ 2+ 42x 12  6x +2x+4 26x  12 =0 (4  2x) 2  (12  6x) (2x  4) 2  (26  12x) <=> 2x2 5x+2 + + =0 4  2x  12  6x 2x  4  26  12x 4x 2  10x  4 4x 2  10x  4 2 <=> 2x 5x+2+ + =0 4  2x  12  6x 2x  4  26  12x 2 2 <=> (2x2 5x+2)(1+ + )=0 4  2x  12  6x 2x  4  26  12x <=> 2x2 5x+2=0 <=> x=2  x= 1 2 5/ x2 2x+3  4x  1  4  2x =0 ( nhaåm nghieäm x =2; x=0 ) 1 Giaûi : ñk :  ≤ x≤ 2 2 2 Pt : x 2x +(x+1) 4x  1 +(2x) 4  2x =0 (x  1) 2  (4x  1) (2  x) 2  (4  2x) <=> x2 2x + + =0 x  1  4x  1 2  x  4  2x x 2  2x x 2  2x 2 <=> x 2x + + =0 x  1  4x  1 2  x  4  2x 1 1 <=> (x2 2x) (1+ + ) =0 x  1  4x  1 2  x  4  2x <=> x2 2x =0 <=> x=0  x=2 9 19 6/ 4x2  x + 2 2  x  11x  5 =0 5 5 5 Giaûi : ñk :  ≤ x≤ 2 11 4 14 Pt : 4x2 3x1 +( x+ )2 2  x +(2x+2) 11x  5 =0 5 5 2 14   4   x    4(2  x)  2x  2  2  (11x  5) 5 5 2 <=> 4x 3x1+  + =0 4 14 2x  2  11x  5  x  2 2x 5 5 16 2 12 4 x  x 2 25 25 + 4x  3x  1 =0 <=> 4x2 3x1+ 25 4 14  x   2 2  x 2x  2  11x  5 5 5 4 1 25 <=> (4x2 3x1)( 1+ + ) =0 4 14 2x  2  11x  5  x  2 2x 5 5 1 4 2 7/ 2x x+18  12  6x 3 26x  12 =0 6 Giaûi : ñk :  ≤ x ≤ 2 13 2 Pt : 2x 5x+ 2+ 42x 12  6x +6x+123 26x  12 =0 2 (4  2x) 2  (12  6x) 3. (2x  4)  (26  12x)  2 <=> 2x 5x+2 + + =0 4  2x  12  6x 2x  4  26  12x 3. 4x 2  10x  4  4x 2  10x  4 2 <=> 2x 5x+2+ + =0 4  2x  12  6x 2x  4  26  12x 2 6 <=> (2x2 5x+2)(1+ + )=0 4  2x  12  6x 2x  4  26  12x <=> 4x2 3x1 =0 <=> x=1  x= <=> 2x2 5x+2=0 <=> x=2  x= 1 2 8/ x2 +5 3 4x  1  4  2x =0 1 Giaûi : ñk :  ≤ x≤ 2 2 2 Pt : x 2x +3(x+1)3 4x  1 +(2x) 4  2x =0 3  (x  1) 2  (4x  1)  (2  x) 2  (4  2x) 2 <=> x 2x + + =0 x  1  4x  1 2  x  4  2x 3(x 2  2x) x 2  2x 2 <=> x 2x + + =0 x  1  4x  1 2  x  4  2x 3 1 <=> (x2 2x) (1+ + ) =0 x  1  4x  1 2  x  4  2x <=> x2 2x =0 <=> x=0  x=2
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