Đáp án + Giải thích các bước giải:
Câu 1 :
a) `5x - 1 = 2(x - 11) <=> 5x - 1 = 2x - 22 <=> 5x - 2x = -22 + 1` $\\$ `<=> 3x = -21 <=> x = -7`
Vậy `S = {-7}`
b) `(x + 4)/2 - (x - 1)/3 = 1 <=> [3(x + 4)]/6 - [2(x - 1)]/6 = 6/6` $\\$ `<=> 3(x + 4) - 2(x - 1) = 6 <=> 3x + 12 - 2x + 2 = 6` $\\$ `<=> x + 14 = 6 <=> x = -8`
Vậy `S = {-8}`
`c) (x + 4)^2 + (3 + x)(3 - x) = 1 <=> x^2 + 8x + 16 + 9 - x^2 = 1` $\\$ `<=> 8x + 16 + 9 = 1 <=> 8x + 25 = 1 <=> 8x = -24` $\\$ `<=> x = -3`
Vậy `S = {-3}`
`d) (x - 2)^2 = 9 <=> (x - 2)^2 = (pm3)^2` $\\$ `<=>`\(\left[ \begin{array}{l}x-2=3\\x-2=-3\end{array} \right.\) $\\$ `<=>` \(\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.\)
Vậy `S = {5;-1}`
Câu 2 :
`a) (m + 2)x - 5 = 4`
Thay x = 3 vào phương trình trên ta có :
`(m + 2)*3 - 5 = 4 <=> (m + 2)*3 = 9` $\\$ `<=> m + 2 = 3 <=> m = 1`
Vậy m = 1
`b) (x + 5)/2006 + (x + 6)/2005 + (x + 7)/2004 = -3` $\\$ `<=> ((x + 5)/2006 + 1) + ((x + 6)/2005 + 1) + ((x + 7)/2004 + 1) = -3` $\\$ `<=> (x + 5 + 2006)/2006 + (x + 6 + 2005)/2005 + (x + 7 + 2004)/2004 = -3` $\\$ `<=> (x + 2011)/2006 + (x + 2011)/2005 + (x + 2011)/2004 = -3` $\\$ `<=> (x + 2011)(1/2006 + 1/2005 + 1/2004) = -3`
Vì `1/2006+1/2005+1/2004 ne 0 `
`<=>x + 2011 = 0 <=> x = -2011`
Vậy phương trình có nghiệm duy nhất là -2011
Đề 2 :
Câu 1 :
`a) x + 3 = 5x - 18 <=> 5x - x = 3 + 18 <=> 4x = 21 <=> x = 21/4`
Vậy `S = {21/4}`
`b) (5x + 3)/6 + 1 = (5x + 1)/3 <=> (5x + 3)/6 + 6/6 = [2(5x + 1)]/6 <=> (5x + 3 + 6)/6 = [2(5x + 1)]/6` $\\$ `<=> 5x + 3 + 6 = 2(5x + 1) ` $\\$ `<=> 5x + 9 = 10x + 2 <=> 5x - 10x = 2-9 <=> -5x = -7` $\\$ `<=> x= 7/5`
Vậy `S = {7/5}`
`c) (3x - 2)^2 + (2x + 5)^2 = 13x^2 <=> (3x)^2 - 2*3x*2 + 2^2 + (2x)^2 + 2*2x*5 + 5^2 = 13x^2` $\\$ `<=> 9x^2 - 12x + 4 + 4x^2 + 20x + 25 = 13x^2` $\\$ `<=> 13x^2 + 8x + 29 = 13x^2 <=> 13x^2 - 13x^2 + 8x + 29 = 0` $\\$ `<=> 8x + 29 = 0 <=> x = -29/8`
Vậy `S = {-29/8}`
`d) (x + 3)^2 - 0,16 = 0 <=> (x + 3)^2 - 16/100 = 0` $\\$ `<=> (x + 3)^2 - 4/25 = 0 <=> (x + 3)^2 - (2/5)^2 = 0` $\\$ `<=> (x + 3 - 2/5)(x + 3 + 2/5) = 0` $\\$ `<=> (x + 13/5)(x + 17/5) = 0 <=> `\(\left[ \begin{array}{l}x=\frac{-13}{5}\\x=\frac{-17}{5}\end{array} \right.\)
Vậy `S = {-13/5;-17/5}`
