(4x-5)/3>(7-x)/5
Tìm x biết
a) (7^x-2+7^x-1+7)/57=(5^x+5^2x+1+5^2x+3)/131
b) (4x-3)^4=(4x-3)^2
7.1 cho f (x) = X^5 + 3x^2-5x^3-x^7+x^3+2x^2+X^5-4x^2+2x^7
cho g(x)=x^4+4x^3-5x^8-x^7+x^3+x^2-2x^7+x^4- 4x^2-x^8
tham khảo
f(x) = x5 + 3x2 − 5x3 − x7 + x3 + 2x2 + x5 − 4x2 + x7
= (x5 + x5) + (3x2 + 2x2 – 4x2) + (-5x3 + x3) + (-x7 + x7)
= 2x5 + x2 – 4x3.
= 2x5 - 4x3 + x2
Đa thức có bậc là 5
g(x) = x4 + 4x3 – 5x8 – x7 + x3 + x2 – 2x7 + x4 – 4x2 – x8
= (x4 + x4) + (4x3 + x3) – (5x8 + x8) – (x7 + 2x7) + (x2 – 4x2)
= 2x4 + 5x3 – 6x8 – 3x7 – 3x2
= -6x8 - 3x7 + 2x4 + 5x3 - 3x2.
Đa thức có bậc là 8.
1)x^6+3x^5+4x^4+4x^3+4x^2+3x+1
2)(x+y+z)^2+(x-2)^2
3)(a-b)^3+(b-c)^3+(c-a)^3
4)10(x^7+y^7+z^7)=7(x^2+y^2+z^2)(x^5+y^5+z^5)
5) D=(x-7)*(x+5)-(x-4)*(x+3)
6) E= 4x*(x2-x-1)-(x+3)*(x2-2)
7) F= 5x*(x-3)*(x-1)-4x*(x2-2x)
tìm x:
a)7(x-3)-5(3-x)=11x-5
b)-4x+15+7x=25+8
c)4x-5(-3+x)=7
d)4(x-1)-3(x-2)=-|-5|
may bai nay de ma
sao ban khong tu lam di
nhân các vế ta được
a) 7x-21-15+5x=11x-5
12x-36=11x-5
12x-11x=-5+36
x=31
b) -4x+15+7x=25+8
3x+15=33
3x= 33-15=18
x=18/3=6
c)4x-5(-3+x)=7
4x- 5x+ 15=7
-x+15=7
15-7=x
8=x
d) vì vế phải có chứa dấu giá trị tuyệt đối nên có 2 trường hợp xảy ra
TH1 4(x-1) -3(x-2)=-(5)
4x-3x -10=-5
x=-5+10=5
TH2 4(x-1) -3(x-2)=-(-5)
4x-3x-10=5
x=15
1) (\(\dfrac{1}{2}\)x + 3)*(x2- 4x- 6)
2) (6x2 -9x +15)*(\(\dfrac{2}{3}\)x+1)
3) (3x2 -x+5)*(x3+5x-1)
4) (x-1)*(x+1)*(x-2)
5) D=(x-7)*(x+5)-(x-4)*(x+3)
6) E= 4x*(x2-x-1)-(x+3)*(x2-2)
7) F= 5x*(x-3)*(x-1)-4x*(x2-2x)
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)