Bài 1: Giải các phương trình sau:
Câu 1.
a) 3x – 2 = 2x – 3 b) 3 – 4y + 24 + 6y = y + 27 + 3y
c) 7 – 2x = 22 – 3x d) 8x – 3 = 5x + 12
e) x – 12 + 4x = 25 + 2x – 1 f) x + 2x + 3x – 19 = 3x + 5
g) 11 + 8x – 3 = 5x – 3 + x h) 4 – 2x + 15 = 9x + 4 – 2x
2. a) 5 – (x – 6) = 4(3 – 2x) b) 2x(x + 2)2 – 8x2 = 2(x – 2)(x2 + 2x + 4)
c) 7 – (2x + 4) = – (x + 4) d) (x – 2)3 + (3x – 1)(3x + 1) = (x + 1)3
e) (x + 1)(2x – 3) = (2x – 1)(x + 5) f) (x – 1)3 – x(x + 1)2 = 5x(2 – x) – 11(x + 2)
g) (x – 1) – (2x – 1) = 9 – x h) (x – 3)(x + 4) – 2(3x – 2) = (x – 4)2
i) x(x + 3)2 – 3x = (x + 2)3 + 1 j) (x + 1)(x2 – x + 1) – 2x = x(x + 1)(x – 1)
3. a) 1,2 – (x – 0,8) = –2(0,9 + x) b) 3,6 – 0,5(2x + 1) = x – 0,25(2 – 4x)
c) 2,3x – 2(0,7 + 2x) = 3,6 – 1,7x d) 0,1 – 2(0,5t – 0,1) = 2(t – 2,5) – 0,7
e) 3 + 2,25x +2,6 = 2x + 5 + 0,4x f) 5x + 3,48 – 2,35x = 5,38 – 2,9x + 10,42
4.a) (5x-2)/3=(5-3x)/2 b)(10x+3)/12=1+((6+8x)/9)
c)2(x+3/5)=5-(13/5+x) d)7/8x-5(x-9)=(20x+1,5)/6
e)(7x-1)/6+2x=(16-x)/5 f)4(0,5-1,5x)=-(5x-6)/3
g)(3x+2)/2-(3x+1)/6=5/3+2x h)(x+4)/5-(x+4)=x/3-(x-2)/2
i) (4x+3)/5-(6x-2)/7=(5x+4)/3+3 k)(5x+2)/6-(8x-1)/3=(4x+2)/5-5
m)(2x-1)/5-(x-2)/3=(x+7)/15 n)1/4(x+3)=3-1/2(x+1)-1/3(x+2)
Bài 2 Tìm giá trị của k sao cho:
a. Phương trình: 2x + k = x – 1 có nghiệm x = – 2.
b. Phương trình: (2x + 1)(9x + 2k) – 5(x + 2) = 40 có nghiệm x = 2
c. Phương trình: 2(2x + 1) + 18 = 3(x + 2)(2x + k) có nghiệm x = 1
a) \(({x^2} + 2x + 3) + (3{x^2} - 5x + 1) = ({x^2} + 3{x^2}) + (2x - 5x) + (3 + 1) = 4{x^2} - 3x + 4\);
b) \(\begin{array}{l}(4{x^3} - 2{x^2} - 6) - ({x^3} - 7{x^2} + x - 5) = 4{x^3} - 2{x^2} - 6 - {x^3} + 7{x^2} - x + 5\\ = (4{x^3} - {x^3}) + ( - 2{x^2} + 7{x^2}) - x + ( - 6 + 5) = 3{x^3} + 5{x^2} - x - 1\end{array}\);
c) \(\begin{array}{l} - 3{x^2}(6{x^2} - 8x + 1) = - 3{x^2}.6{x^2} - - 3{x^2}.8x + - 3{x^2}.1\\ = - 18{x^{2 + 2}} + 24{x^{2 + 1}} - 3{x^2} = - 18{x^4} + 24{x^3} - 3{x^2}\end{array}\);
d) \(\begin{array}{l}(4{x^2} + 2x + 1)(2x - 1) = (4{x^2} + 2x + 1).2x - (4{x^2} + 2x + 1).1 = 4{x^2}.2x + 2x.2x + 1.2x - 4{x^2} - 2x - 1\\ = 8{x^{2 + 1}} + 4{x^{1 + 1}} + 2x - 4{x^2} - 2x - 1 = 8{x^3} + 4{x^2} + 2x - 4{x^2} - 2x - 1 = 8{x^3} - 1\end{array}\);
e) \(\begin{array}{l}({x^6} - 2{x^4} + {x^2}):( - 2{x^2}) = {x^6}:( - 2{x^2}) - 2{x^4}:( - 2{x^2}) + {x^2}:( - 2{x^2})\\ = - \dfrac{1}{2}{x^{6 - 2}} + {x^{4 - 2}} - \dfrac{1}{2}{x^{2 - 2}} = - \dfrac{1}{2}{x^4} + {x^2} - \dfrac{1}{2}.\end{array}\);
g)
\(({x^5} - {x^4} - 2{x^3}):({x^2} + x)=x^3-2x^2\)