\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

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= (5x-25) + (5x - x²)

= 5(x-5) + x(5-x)

= 5(x-5) - x(x-5)

= (5 - x)(x - 5)

\(10x-25-x^2=-\left(x^2-2\cdot x\cdot5+5^2\right)=-\left(x-5\right)^2\)

= (3x + 1 - x - 1)(3x + 1 + x + 1)

= 2x(4x + 2)

Em áp dụng hđt số 3 trong sgk nhé.

(3x+1+x+1) (3x+1-x-1) = (4x+2) 2x

                                      = 4x(2x+1)

\(a,x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right).\)

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

\(b,9x^2+y^2+6xy=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

\(c,6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x^2-2.x.3+3^2\right)=-\left(x-3\right)^2\)

a) \(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}x+8y\right)\left(\frac{1}{5}x-8y\right)\)

b) \(x^3+\frac{1}{27}=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)

c) \(-x^3+9x^2-27x+27\)

\(=27-x^3+9x^2-27x\)

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

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

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

\(=\left(3-x\right)\left(9-6x+x^2\right)=\left(3-x\right)\left(9-3x-3x+x^2\right)\)

\(=\left(3-x\right)\left[3\left(3-x\right)-x\left(3-x\right)\right]=\left(3-x\right)\left(3-x\right)\left(3-x\right)=\left(3-x\right)^3\)

(Nhớ k cho mình với nha!, Mình chắc chắn là mình làm đứng luôn đó! Chúc may mắn nhá!)

a/ Ta có: \(\frac{1}{25}x^2-64y^2=\left(\frac{1}{5}x\right)^2-\left(8y\right)^2=\left(\frac{1}{5}x-8y\right)\left(\frac{1}{5}x+8y\right)\)

b/ \(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)

c/ Đề sai

c/ Ta có:  -x³ + 9x² - 27x + 27 = -(x³ - 9x² + 27x - 27) 

 \(=-\left[x^3-3.3.x^2+3.3^2.x-3^3\right]=-\left(x-3\right)^3\)

b: \(\left(x^2+4\right)^2-16x^2\)

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

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

c: \(x^5-x^4+x^3-x^2\)

\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)

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

Lời giải:

a. Bạn xem lại đề

b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

\(=(x-2)^2(x+2)^2\)

c.

\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)

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

a) 7x.(-y)+2(y-x)2

=>-7xy+4y-4x

b)(x^2+4)-16x^2

=>x^2+4-16x^2

=>-15x^2+4

c)x^5-x^4+x^3-x^2

=>x^4(x-1)+x^2(x-1)

=>(x^4+x^2)(x-1)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)

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

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

10x - 25 - x²

= x²- 10x - 25

= - ( x² +10x +25)

= -(x² + 2.x.5+5² )

= - (x+5 )²

\(a,8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3.\left(2x\right)^2.y+3.2x.y^2+y^3=\left(2x+y\right)^3\)

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

\(c,4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)

1, \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2, \(x^2-10x+25=\left(x-5\right)^2\) 

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

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

1) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2) \(x^2-10x+25=\left(x-5\right)^2\)

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

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

1: \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)

2: \(x^2-10x+25=\left(x-5\right)^2\)

3: \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

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