1, xy(x+y)+yz(y+z)+xz(x+z)+2xyz

= x²y+xy²+y²z+yz²+x²z+xz²+2xyz

=(x²y+x²z+xz²+xyz) + ( xy²+y²z+yz²+xyz)

=x(xy+xz+z²+yz)+y(xy+yz+z²+xz)

=(xy+xz+yz+z²).(x+y)

=(x(y+z)+z(y+z)).(x+y)

=((y+z).(x+z)).(x+y)= (x+y)(x+z)(y+z)

2. 3(x-3)(x-7)+(x-4)²+48

=3(x²+4x-21)+x²-8x+16+48

=4x²-4x+1 = (2x-1)²

Thay x=0,5 vào bt trên, ta có : (2.0,5 -1)²=0

3, x²-6x+10

= x²-2.3.x+9+1

=(x-3)²+1 \(\ge\)1 >0 ( do (x-3)² >=0 với mọi x)

=> x²6x+10 >0 với mọi x

4x-x²-5

=-(x²-4x+5)

=- (x²-2.2x+4+1)

= - ((x-2)²+1) = -(x-2)²-1\(\le\)-1 < 0 ( do (x-2)²\(\ge\)0 với mọi x => - (x-2)²\(\le\)0 với mọi x)

vậy, 4x-x²-5<0 với mọi x

Ta có : x² - 6x + 10 

= x² - 6x + 9 + 1 

= (x - 3)² + 1

Mà (x - 3)² \(\ge0\forall x\)

Nên : (x - 3)² + 1 \(\ge1\forall x\)

=> (x - 3)² + 1 \(>0\)(đpcm)

\(x^2-4x+4-y^2\)

\(=\left(x-2\right)^2-y^2\)

\(=\left(x-2-y\right)\left(x-2+y\right)\)

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

\(x^2-4x-y^2+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)

\(x^2-4x+4-y^2\)

\(=\left(x-2\right)^2-y^2\)

\(=\left(x-2-y\right)\left(x-2+y\right)\)

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

\(=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)

Giải:

a) \(3x^2y-6xy^2\)

\(=3xy\left(x-2y\right)\)

Vậy ...

b) \(\left(2x-a\right)x^2-\left(2x-a\right)y\)

\(=\left(2x-a\right)\left(x^2-y\right)\)

\(=\left(2x-a\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)

Vậy ...

c) \(25a^2-c^2\)

\(=\left(5a-c\right)\left(5a+c\right)\)

Vậy ...

d) \(4-36x+81x^2\)

\(=2^2-2.2.9x+\left(9x\right)^2\)

\(=\left(2-9x\right)^2\)

Vậy ...

e) \(\left(x+7\right)2-\left(2x-9\right)2\)

\(=2\left[\left(x+7\right)-\left(2x-9\right)\right]\)

\(=2\left(x+7-2x+9\right)\)

\(=2\left(16-x\right)\)

Vậy ...

f) \(x^2-6x+8\)

\(=x^2-6x+9-1\)

\(=\left(x-3\right)^2-1\)

\(=\left(x-4\right)\left(x-2\right)\)

Vậy ...

a) \(-y^2+\dfrac{1}{9}\)

\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)

\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)

b) \(4^4-256\)

\(=4^4-4^4\)

\(=0\)

c) \(9\left(x-3\right)^2-4\left(x+1\right)^2\)

\(=\left(3x-9\right)^2-\left(2x+2\right)^2\)

\(=\left(3x-9+2x+2\right)\left(3x-9-2x-2\right)\)

\(=\left(5x-7\right)\left(x-11\right)\)

\(a,=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\\ b,=\left(x^2-16\right)\left(x^2+16\right)\\ =\left(x-4\right)\left(x+4\right)\left(x^2+16\right)\\ c,=\left[3\left(x-3\right)-2\left(x+1\right)\right]\left[3\left(x-3\right)+2\left(x+1\right)\right]\\ =\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)\\ =\left(x-11\right)\left(5x-7\right)\\ d,=\left(5x-\dfrac{1}{9}xy\right)\left(5x+\dfrac{1}{9}xy\right)=x^2\left(5-\dfrac{1}{9}y\right)\left(5+\dfrac{1}{9}y\right)\)

\(=\left(x-y\right)^2-2\left(x-y\right)=\left(x-y\right)\left(x-y-2\right)\)

\(a,x^2+6x+9\)

\(=\left(x+3\right)^2\)

\(b,10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x-5\right)^2\)

\(c,8x^3-\frac{1}{8}\)

\(=8x^3-\left(\frac{1}{2}\right)^3\)

\(=\left(8x-\frac{1}{2}\right)\left(64x^2+4x+\frac{1}{4}\right)\)

\(d,8x^3+12x^2+6xy^2+y^3\)

\(=2\left(4x^3+6x^2+3xy^2+\frac{1}{2}y^3\right)\)

hok tốt!

ai tra loi dung minh cho

Điệp viên 007 sai c

c, \(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

Ta có: \(\Delta'=32>0\)

\(\Rightarrow\) Phương trình có 2 nghiệm phân biệt

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=12\\x_1x_2=4\end{matrix}\right.\)

Mặt khác: \(T=\dfrac{x_1^2+x^2_2}{\sqrt{x_1}+\sqrt{x_2}}\)

\(\Rightarrow T^2=\dfrac{x_1^4+x^4_2+2x_1^2x_2^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(x_1^2+x_1^2\right)^2}{x_1+x_2+2\sqrt{x_1x_2}}\) \(=\dfrac{\left[\left(x_1+x_2\right)^2-2x_1x_2\right]^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(12^2-2\cdot4\right)^2}{12+2\sqrt{4}}=1156\)

Mà ta thấy \(T>0\) \(\Rightarrow T=\sqrt{1156}=34\)