given that x^3-x=7. evaluate A=x^6-2x^4+x^3+x^2-x
given that a^2-b^2 =1 evaluate A=2(a^6-a^6)-3(a^4+a^4)
find the remainder in the division ò x^30+x^4-x^1975+1 by x-1
Đa thức chia x-1 có ngiệm là 1 nên:
Thay x=1 vào đa thức chia ta có:
130+14-11975+1
=1+1-1+1
=2
Vậy số dư khi chia khi chia x30+x4-x1975+1 cho x-1 là 2
Given that\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x+a}{x+b}\) .Evaluate \(a+b\)
\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}=\frac{\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)+5\left(x+2\right)}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\Rightarrow a=-2;b=5\)
\(\Rightarrow\)\(a+b=-2+5=3\)
Evaluate the expression at
Fill in the blank: ............
Fill in the blank: Fill in the blank:Evaluate the expression at
x3 + 12x + 48x + 64
= (x + 4)2
= (- 4 + 4)2
= 02
= 0
Fill in the blank: ............
x3 - a = (x - 2)(x2 + 2x + 4)
x3 - a = x3 - 8
a = 8
Fill in the blank: (x - 1)3 = x3 - 3x2 + 3x - 1 Fill in the blank: (x + 1)3 = x3 + 3x2 + 3x + 1a + b = 8
(a + b)2 = 82
a2 + b2 + 2ab = 64
a2 + b2 + 2 . 10 = 64
a2 + b2 + 20 = 64
a2 + b2 = 64 - 20
a2 + b2 = 44
(a - b)2
= a2 - 2ab + b2
= 44 - 2 . 10
= 44 - 20
= 24
Given .
Evaluate A at .
Answer: A
A = (x - 5)(x2 + 5x + 25) - x2(x + 3) + 3x2
= x3 - 125 - x3 - 3x2 + 3x2
= - 125
Given .
Evaluate A at .
Answer: A
Given that x/y+z+y/z+x+z/x+y=1
Evaluate A=X^2/y+z+y^2/Z+x+z^2/x+y
\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\\ =>\dfrac{x}{y+z}=1-\dfrac{y}{z+x}-\dfrac{z}{x+y}\\ =>\dfrac{x}{y+z}=\dfrac{(z+x)(x+y)-y(x+y)-z(z+x)}{(z+x)(x+y)}\\ =>\dfrac{x}{y+z}=\dfrac{xz+yz+x^{2}+xy-xy-y^{2}-z^{2}-xz}{(z+x)(x+y)}\\ =>\dfrac{x}{y+z}=\dfrac{x^{2}-y^{2}-z^{2}+yz}{(z+x)(x+y)}\\ =>\dfrac{x^{2}}{y+z}=\dfrac{x^{3}-xy^{2}-xz^{2}+xyz}{(z+x)(x+y)} \ \ \ \ (1)\\ =>\dfrac{y^{2}}{z+x}=\dfrac{y^{3}-yz^{2}-yx^{2}+xyz}{(x+y)(y+z)} \ \ \ \ (2)\\ =>\dfrac{z^{2}}{x+y}=\dfrac{z^{3}-zx^{2}-zy^{2}+xyz}{(y+z)(z+x)} \ \ \ \ (3)\)
Cộng vế vs vế của (1),(2) và (3) ta đc \(\dfrac{x^{2}}{y+z}+\dfrac{y^{2}}{z+x}+\dfrac{z^{2}}{x+y}=0\)
Câu 1 The function mm is defined on the real numbers by m(k) = \dfrac{k+2}{k+8}m(k)= k+8 k+2 . What is the value of 10\times m(2)10×m(2)? Answer: Câu 2 The function ff is defined on the real numbers by f(x)= ax-3f(x)=ax−3. What is the value of a if f(3)=9f(3)=9? Answer: Câu 3 The function ff is defined on the real numbers by f(x)= 2x+a-3f(x)=2x+a−3. What is the value of a if f(-5)=11f(−5)=11? Answer: Câu 4 The function ff is defined on the real numbers by f(x) = 2 + x-x^2f(x)=2+x−x 2 . What is the value of f(-3)f(−3)? Answer: Câu 5 Given a real number aa and a function ff is defined on the real numbers by f(x)=-6\times|3x|-4f(x)=−6×∣3x∣−4. Compare: f(a)f(a) f(-a)f(−a) Câu 6 There are ordered pairs (x;y)(x;y) where xx and yy are integers such that \dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8} x 5 + 4 y = 8 1 Câu 7 Given a negative number kk and a function ff is defined on the real numbers by f(x)=\dfrac{6}{13}xf(x)= 13 6 x. Compare: f(k)f(k) f(-k)f(−k) Câu 8 Given a positive number kk and a function ff is defined on the real numbers by f(x)=\dfrac{-3}{4}x+4f(x)= 4 −3 x+4. Compare: f(k)f(k) f(-k)f(−k). Câu 9 A=(1+2+3+\ldots+90) \times(12 \times34-6 \times 68):(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})=A=(1+2+3+…+90)×(12×34−6×68):( 3 1 + 4 1 + 5 1 + 6 1 )= Câu 10 Given that \dfrac{2x+y+z+t}{x}=\dfrac{x+2y+z+t}{y}=\dfrac{x+y+2z+t}{z}=\dfrac{x+y+z+2t}{t} x 2x+y+z+t = y x+2y+z+t = z x+y+2z+t = t x+y+z+2t . The negative value of \dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z} z+t x+y + t+x y+z + x+y z+t + y+z t+x is
given that l x-1 l + l y+5 l =0 evaluate x -y = ?
Ta có |x-1| và |y+5| luôn lớn hơn hoặc bằng 0 mà |x-1|+|y+5|=0 suy ra x-1=0 suy ra x=1 và y+5=0 suy ra y=-5 suy ra x-y=1+5=6
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0
M) (2x+3)(-x+7)=0
given three distinct real numbers x;y;z such as x^3+y^3+z^3= 3xyz evaluate P =2016 xyz/(x+y)(y+z)(z+x)
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0
c. x^2-5x +6 = 0
<=> x^2 - 5x = -6
<=> - 4x = -6
<=> x= -6/-4
Mình chỉ phân tích đa thức thành nhân tử thôi , phần còn lại bạn tự tính nha keo dài lắm
A) 2x2(x+3) - x(x+3) = 0 <=> x(x - 3)(2x-1)=0
B) (2x+5)2 - (x+2)2=0 <=> (x+3)(3x+7)=0
C) (x2-2x) - (3x-6)=0 <=> (x-2)(x-3)=0
D) (2x-7)(2x-7-6x+18)=0 <=> (2x-7)(-4x+11)=0
E) (x-2)(x+1) - (x-2)(x+2)=0 <=> (x-2)*(-1)=0 <=> x-2=0
G) (2x-3)(2x+2-5x)=0 <=> (2x-3)(-3x+2)=0
H) (1-x)(5x+3+3x-7)=0 <=> (1-x)(8x-4)=0
F) (x+6)*3x=0
I) (x-3)(4x-1-5x-2)=0 <=> (x-3)(-x-3)=0
K) (x+4)(5x+8)=0
H) (x+3)(4x-9)=0
B> <2X+5>2-<X+2>2=0
<2X+5-X-2><2X+X+2>=0
<X+3><3X+7>=0
X+3=0 HOẶC 3X+7=0
X=-3 HOẶC X=-7/3
C>X2-5X+6=0
X2-4X+4-X+2=0
<X-2>2-<X-2>=0
<X-2.><X-3>=0
X-2=0 HOẶC X-3=0
X=2 HOẶC X=3
D> <2X-7><2X-7-6<X-3>>=0
<2X-7><-4X+11>=0
2X-7=0 HOẶC -4X+11=0
X=7/2 HOẶC X=11/4
E><X-2><X+1>=X2-4
<X-2><X+1>-<X2-4>=0
<X-2><X+1>-<X-2><X+2>=0
-X+2=0
X=2
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