Đáp án + Giải thích các bước giải:
Bài 1 :
`a) (9/25 -2,18) : (3 4/5 + 0,2)` $\\$ `= (9/25 - 218/100) : (19/5 + 2/10)` $\\$ `= (9/25 - 109/50) : (38/10 + 2/10)` $\\$ `= (18/50 - 109/50) : 40/10` $\\$ `= -91/50 : 4 = -91/200`
`b) 3/8 * 19 1/3 - 3/8 * 33 1/3 = 3/8 *(19 1/3 - 33 1/3) ` $\\$ `= 3/8 * (19 - 33) = 3/8 * (-14) = [3*(-14)]/8 = [3*(-7)]/4 = -21/4`
`c) 1 4/23 + 5/21 - 4/23 + 0,5 + 16/21 = (1 4/23 - 4/23) + (5/21 + 16/21) + 0,5 ` $\\$ `= 1 + 1 + 0,5 = 2,5 = 5/2`
`d) (2^12 * 3^5 - 4^6 * 81)/((2^2*3)^6 + 8^4 * 3^5) = (2^12 * 3^5 - (2^2)^6 * 3^4)/(2^12 * 3^6 + (2^3)^4 * 3^5)` $\\$ `= (2^12 * 3^5 - 2^12 * 3^4)/(2^12 * 3^6 + 2^12 * 3^5)` $\\$ `= [2^12(3^5 - 3^4)]/[2^12(3^6 + 3^5)] = [2^12*3^4(3 - 1)]/[2^12 * 3^5(3 + 1)]` $\\$ `= [2^12 * 3^4 * 2]/[2^12 * 3^5 * 4] =[2^12 * 3^4 * 2]/[2^12 * 3^4 * 3 * 2 * 2] = 1/(3*2) = 1/6`
`e) 4 * (-1/2)^3 - 2(-1/2)^2 + 3(-1/2) + 1 = 4(-1/8) - 2*1/4 + (-3/2) + 1` $\\$ `= -1/2 - 1/2 - 3/2 + 1 = -3/2`
`g) sqrt(4/81) : sqrt(25/81) - 1 2/5 = 2/9 : 5/9 - 1 2/5 = 2/9 . 9/5 - 1 2/5` $\\$ `= 2/5 - 1 2/5 = (0 - 1) - (2/5 - 2/5) = -1`
Bài 2 :
a) 5x - 7 = 3x + 9
=> 5x - 3x = 9 + 7
=> 2x = 16
=> x = 8
b) `1 3/4*x + 1 1/2 = -4/5 => 7/4 * x + 3/2 = -4/5 ` $\\$ `=> 7/4 * x = -4/5 - 3/2` $\\$ `=> 7/4 * x = -23/10` $\\$ `=> x = (-23/10) : 7/4 = (-23/10)*4/7 = -46/35 `
`c) x + 1/2 = 2^5 : 2^3 => x + 1/2 = 2^2 => x + 1/2 = 4 => x = 4 - 1/2 =7/2`
`d) (x + 1/2)^2 = 4/25 => (x + 1/2)^2 = (pm2/5)^2` $\\$ `=>` \(\left[ \begin{array}{l}x + \frac{1}{2} = \frac{2}{5}\\x + \frac{1}{2} = \frac{-2}{5}\end{array} \right.\) $\\$ `=> `\(\left[ \begin{array}{l}x=\frac{-1}{10}\\x=\frac{-9}{10}\end{array} \right.\)
e) 5x - |9 - 7x| = 3
=> |9 - 7x| = 5x - 3
TH1: 9 - 7x = 5x - 3
=> 9 - 7x - 5x + 3 = 0
=> 12 - 12x = 0
=> 12x = 12 => x = 1
TH2 : -(9 - 7x)= 5x - 3
=> -9 + 7x = 5x - 3
=> -9 + 7x - 5x + 3 = 0
=> -6 + 2x = 0
=> 2x = 6
=> x = 3
g) -5 + |3x - 1| + 6 = |-4|
=> -5 + |3x - 1| + 6 = 4
=> -5 + |3x - 1| = 4 - 6
=> -5 + |3x - 1| = -2
=> |3x - 1| = -2 - (-5) = 3
TH1 : 3x - 1 = 3 => 3x = 4 => `x = 4/3`
TH2 : 3x - 1 = -3 => 3x = -3 + 1 => 3x = -2 => `x = -2/3`
h) `(x - 1)^2 = (x - 1)^4 => (x - 1)^2 - (x - 1)^4 = 0` $\\$ `=> (x - 1)^2[1 - (x- 1)^2] = 0`
TH1 : `(x - 1)^2 = 0 => x = 1`
TH2 : `1 - (x - 1)^2 = 0 => (x - 1)^2 = 1 => `\(\left[ \begin{array}{l}x - 1 = 1\\x-1=-1\end{array} \right.\) $\\$ `=> `\(\left[ \begin{array}{l}x=2\\x=0\end{array} \right.\)
Vậy x = 1,x = 2,x = 0
i) `5^-1 * 25^x = 125 => 1/5 * 25^x = 125 => 25^x = 125 : 1/5 = 125*5 = 625` $\\$ `=> 25^x = 625 => 25^x = 25^2 => x = 2`
k) |x + 1| + |x + 2| + |x + 3| = 4x
VT tổng các GTTĐ nên là số không âm , do đó `4x >= 0 => x >= 0`
`=> x + 1>0,x + 2>0,x + 3 >0`
Ta có : |x + 1| + |x + 2| + |x + 3| = 4x
`<=> x + 1 + x + 2 + x + 3 = 4x` $\\$ `<=> 3x + 6 = 4x => x = 6(tm)`
Vậy x = 6
