3x x2 + 15 = 33
Phương trình 2 + 3 x + 1 - 2 a . 2 - 3 x - 4 = 0 có 2 nghiệm phân biệt x 1 , x 2 , thỏa mãn x 1 - x 2 = log 2 + 3 3 . Khi đó a thuộc khoảng
c) 3x.2 + 15 = 33
\(\text{c) 3^x.2 + 15 = 33}\\ 3^x.2=33-15\\ 3^x.2=18\\ 3^x=18:2\\ 3^x=9\\ 3^2=9\\ x=2\)
c) 3x.2 + 15 = 33
> 3x.2 = 33 - 15
> 3x.2 = 18
> 3x = 18 : 2
> 3x = 9
> x = 2
Tìm x
a) 3x – 2 = 19
b) [43 - (56 - x)].12 = 384
c) 3x.2 + 15 = 33
a) 3x – 2 = 19
3x = 19 + 2
3x = 21
x = 21:3
x = 7
b) [43 - (56 - x)].12 = 384
43 – (56 – x) = 384:12
43 – (56 – x) = 32
56 – x = 43 – 32
56 – x = 11
x = 56 – 11
x = 45
c) 3x.2 + 15 = 33
3x.2 = 33 - 15
3x.2 = 18
3x = 18 : 2
3x = 9
3x = 33
x = 2.
a) 3x – 2 = 19
3x = 19 + 2
3x = 21
x = 21:3
x = 7
b) [43 - (56 - x)].12 = 384
43 – (56 – x) = 384:12
43 – (56 – x) = 32
56 – x = 43 – 32
56 – x = 11
x = 56 – 11
x = 45
c) 3x.2 + 15 = 33
3x.2 = 33 - 15
3x.2 = 18
3x = 18 : 2
3x = 9
3x = 33
x = 2.
Tìm giá trị của a để phương trình 2 + 3 x + 1 - a 2 - 3 x - 4 = 0 có 2 nghiệm phân biệt thỏa mãn: x 1 - x 2 = log 2 + 3 3 , ta có a thuộc khoảng:
A. - ∞ ; - 3
B. - 3 ; + ∞
C. 3 ; + ∞
D. 0 ; + ∞
Giải các phương trình sau:
a) 2 x + 5 6 − 1 3 2 x + 5 x − 10 = 0 ;
b) 4 x − 1 x + 5 = x 2 − 25 ;
c) 3 x − 3 2 − x − 3 x + 2 4 = 0 ;
d) x x + 3 3 − x 4 x + 3 = 0 .
a) A = -3x(x-5) +3( x2 -4x) -3x-10
b) B = 4x( x2 -7x +2) – 4( x3 -7x2 +2x -5)
c) C = 5x( x2 – x) – x2( 5x-5) -15
d) D = 7( x2 -5x+3)- x( 7x-35) -14
e) E = x2 - 4x - x( x-4) -15
A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10
A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10
A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10
A = 0 + (3\(x-3x\)) - 10
A = 0 - 10
A = - 10
Tìm X
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47) f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100 h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6 k) 17/ (x + 3)3 : 3 – 1 = – 10
e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
Tìm X
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47) f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100 h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6 k) 17/ (x + 3)3 : 3 – 1 = – 10
Tìm X
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47) f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100 h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6 k) 17/ (x + 3)3 : 3 – 1 = – 10
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)