\(\dfrac{9}{4}\)

Ta có

\(\frac{3}{x-3}\)\(-\)\(\frac{5}{x-5}=\frac{4}{x-4}-\)\(\frac{6}{x-6}\) ( ĐKXĐ x khác 3;4;5;6)

<=> \(\left(\frac{3}{x-3}+1\right)-\left(\frac{5}{x-5}+1\right)=\left(\frac{4}{x-4}+1\right)-\left(\frac{6}{x-6}+1\right)\)

<=>\(\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)

<=> \(\frac{x}{x-3}-\frac{x}{x-5}-\frac{x}{x-4}+\frac{x}{x-6}=0\)

<=>\(x\left(\frac{1}{x-3}-\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}\right)\)=0

<=> x =0 vì \(\frac{1}{x-3}-\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}\)=0

<=> x=0 hoặc x=4,5

<=> Trung bình cộng là (4,5+0):2=2,25=\(\frac{9}{4}\)

`a, = 5^3 xx 3^3 = 15^3`

`b, = 12^3 xx 4^3 = 48^3`

`c, = 625^3 xx 8^3 = 5000^3`

`d, = 625 ^3 xx 125^3 = 78125^3`

`e, = 5^20 : 5^14 = 5^6`

1,\(125\cdot27=5^3\cdot3^3=\left(5\cdot3\right)^3=15^3\)

2, \(12^3\cdot64=12^3\cdot4^3=\left(12\cdot4\right)^3=48^3\)

3, \(25^6\cdot8^3=\left(5^2\right)^6\cdot\left(2^3\right)^3=5^8\cdot2^9\)

4, \(25^6\cdot125^3=\left(5^2\right)^3\cdot\left(5^3\right)^3=5^6\cdot5^9=5^{15}\)

5,\(625^5:25^7=\left(5^4\right)^5:\left(5^2\right)^7=5^{20}:5^{14}=5^6\)

\(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)(2)

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

\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)(1)

Đặt \(x^2-7x+10=t\)

\(\Rightarrow\left(1\right)=t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2\)

Mà \(x^2-7x+10=t\)nên \(\left(2\right)=\left(x^2-7x+11\right)^2\)

Vậy \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)\(=\left(x^2-7x+11\right)^2\)

\(\left(x+1\right)^3-x\left(x-3\right)\left(x+3\right)-6\left(x-1\right)\left(x+2\right)=13\)

\(\Leftrightarrow x^3+3x^2+3x+1-x\left(x^2-9\right)-6\left(x^2+x-2\right)=13\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+9x-6x^2-6x+12=13\)

\(\Leftrightarrow-3x^2+6x=0\)

\(\Leftrightarrow-3\left(x^2-2\right)=0\)

\(\Leftrightarrow x^2-2=0\Leftrightarrow x^2=2\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

\(\left(\frac{3}{7}\right)^5.\left(\frac{7}{3}\right)^{-1}.\left(\frac{5}{6}\right)^6.\left(\frac{343}{625}\right)^{-2}\)

\(=\left(\frac{3}{7}\right)^5.\left(\frac{3}{7}\right).\left(\frac{5}{2.3}\right)^6.\left(\frac{5^4}{7^3}\right)^2\)

\(=\frac{3^5}{7^5}.\frac{3}{7}.\frac{5^6}{2^6.3^6}.\frac{5^8}{7^6}=\frac{1}{2^6}.\frac{3^6}{3^6}.\frac{5^{14}}{1}.\frac{1}{7^{12}}=\frac{5^{14}}{2^6.7^{12}}\)

Nó đâu có phải là số nguyên đâu

3/7 x 5/6 +5/6 x3/7

= (3/7x5/6) + (5/6x3/7)

= 15/42 + 15/42  = 30/42 = 5/7

\(=\text{[}\left(x+3\right)\left(x+6\right)\text{]}\text{[}\left(x+4\right)\left(x+5\right)\text{]}+1=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1.\text{Đ}\text{ạt}:x^2+9x+18=a\Rightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1=a\left(a+2\right)+1=a^2+2a+1=\left(a+1\right)^2=\left(x^2+9x+19\right)^2\)

a: \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)\)

\(=\left[a^2+\left(2a+3\right)\right]\left[a^2-\left(2a+3\right)\right]\)

\(=\left(a^2\right)^2-\left(2a+3\right)^2\)

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

b: \(\left(-a^2-2a+3\right)^2\)

\(=\left(a^2+2a-3\right)^2\)

\(=a^4+4a^2+9+4a^3-18a-6a^2\)

\(=a^4+4a^3-2a^2-18a+9\)

c: \(\left(x-y-z\right)^2\)

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

\(=x^2-2xy-2xz+y^2+2yz+z^2\)

d: \(\left(x+y+z\right)\left(x-y-z\right)\)

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

\(=x^2-y^2-2yz-z^2\)

1a/ z² - 6z + 5 - t² - 4t = z² - 2 . 3z + 3² - 4 - t² - 4t = (z² - 2 . 3z + 3²) - (2² + 2 . 2t + t²) = (z - 3)² - (2 + t)²

b/ x² - 2xy + 2y² + 2y² + 1 = x² - 2xy + y² + y² + 2y + 1 = (x² - 2xy + y²) + (y² + 2y + 1) = (x - y)² + (y + 1)²

c/ 4x² - 12x - y² + 2y + 8 = (2x)² - 12x - y² + 2y + 3² - 1 = [ (2x)² - 2 . 3 . 2x + 3² ] - (y² - 2y + 1) = (2x - 3)² - (y - 1)²

2a/ (x + y + 4)(x + y - 4) = x² + xy - 4x + xy + y² - 4y + 4x + 4y + 16 = x² + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y² + 16

= x² + 2xy + y² + 4² = (x + y)² + 4²

b/ (x - y + 6)(x + y - 6) = x² + xy - 6x - xy - y² + 6y + 6x + 6y - 36 = x² + (xy - xy) + (-6x + 6x) + (6y + 6y) - y² - 36

= x² - y² + 12y - 6² = x² - (y² - 12y + 6²) = x² - (y² - 2 . 6y + 6²) = x² - (y - 6)²

c/ (y + 2z - 3)(y - 2z - 3) = y² -2yz - 3y + 2yz - 4z² - 6z - 3y + 6z + 9 = y² + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z² + 9

= y² - 6y - 4z² + 9 = (y² - 6y + 9) - 4z² = (y - 3)² - (2z)²

d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x² + 4y² + 6yz - 2xy + 6yz + 9z² - 3xz = (2xy - 2xy) + (3xz - 3xz) - x² + (6yz + 6yz) + 9z² + 4y²

= -x² + 4y² + 12yz + 9z² = (4y² + 12yz + 9z²) - x² = [ (2y)² + 2 . 2 . 3yz + (3z)² ] - x² = (2y + 3z)² - x²

:v dễ mà có trong nâng cao mới hc qua :3

a, x2+10x+26+y2+2y

=(x2+2.x.5+52)+(12+2.1.y+y2)

=(x+5)2+(y+1)2

b, x2−2xy+2y2+2y+1

=x2−2xy+y2+y2+2y+1

=(x2−2.x.y+y2)+(y2+2.y.1+12)

=(x−y)2+(y+1)2

c,z2−6z+5−t2−4t

=−(t2+4t−z2+6z−5)

=−(t2+2.t.2+22−z2+2.z.3−32)

=−((t2+2.t.2+22)−(z2−2.z.3+32))

=−((t+2)2−(z−3)2)

=(z−3)2−(t+2)2

(9+x)(x-3)