S=1.2.3+2.3.(4+1)+3.4.(5+2)+...+n(n+1)[(n+2).(n-1)=

=1.2.3+1.2.3+2.3.4+2.3.4+3.4.5+...+(n-1)n(n+1)+n(n+1)(n+2)=

=2[1.2.3+2.3.4+3.4.5+...+(n-1)n(n+1)]+n(n+1)(n+2)

Đặt 

A=1.2.3+2.3.4+3.4.5+...+(n-1)n(n+1)

4A=1.2.3.4+2.3.4.4+3.4.5.4+...+(n-1)n(n+1).4=

=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+...+(n – 1).n.(n + 1).[(n + 2) – (n – 2)]

=1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)=

= (n – 1).n(n + 1).(n + 2)

2A=\(\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{2}\)

S=2A+n(n+1)(n+2)

Phân tích đa thức thành nhân tử:

\(6a^2b-18ab\)

\(=6ab\left(a-3\right)\)

6a²b-18ab = 6ab ( a-3)

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

\(\left(b\right):\left(x-y-z\right)^2+\left(x+y+z\right)^2\\ =\left[\left(x-y\right)-z\right]^2+\left[\left(x+y\right)+z\right]^2\\ =\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x+y\right)^2+2z\left(x+y\right)+z^2\\ =x^2-2xy+y^2-2xz+2yz+z^2+x^2+2xy+y^2+2xz+2yz+z^2\\ =2x^2+2y^2+2z^2+4yz\)

\(\left(c\right):\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =\left(2y\right)^2=4y^2\)

\(a,64x^2-\left(8a+b\right)^2\)

\(=\left(8x\right)^2-\left(8a+b\right)^2\)

\(=\left[8x-\left(8a+b\right)\right]\left(8x+8a+b\right)\)

\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)

\(b,\dfrac{12}{5}x^2y^2-9x^2-\dfrac{4}{25}y^2\)

\(=-\left(9x^2-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^2\right)\)

\(=-\left[\left(3x\right)^2-2.3.\dfrac{2}{5}x^2y^2+\left(\dfrac{2}{5}y\right)^2\right]\)

\(=-\left(3x-\dfrac{2}{5}y\right)^2\)

1,24² - 0,24²

= (1,24 - 0,24)(1,24 + 0,24)

\(7,\) \(a,\left(2x-3y\right)^2-\left(2x+3y\right)^2=\left(3x-2y\right)^2-\left(3x+2y\right)^2\)

\(\Leftrightarrow4x^2-12xy+9y^2-4x^2-12xy-9y^2=9x^2-12xy+4y^2-9x^2-12xy-4y^2\)

\(\Leftrightarrow-24xy=-24xy\) ( luôn đúng )

Vậy 2 đẳng thức ở 2 vế bằng nhau.

\(b,\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(\Leftrightarrow\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2=\left(ac\right)^2+2acbd+\left(bd\right)^2+\left(ad\right)^2-2adbc+\left(bc\right)^2\)

\(\Leftrightarrow\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2=\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2\) ( luôn đúng )

Vậy 2 đẳng thức ở 2 vế bằng nhau.

 *Ở câu \(b,\) dòng thứ 3, vế phải triệt tiêu \(2acbd-2adbc\) \(=0\) nên mất rồi nha.

cảm ơn mn nhiều =^

`@` `\text {Ans}`

`\downarrow`

`8,`

`a,`

Thay \(x=18;y=41\) vào bt

\(18^2-4\cdot41^2\)

`= 18^2 - (2*41)^2`

`= 18^2 - 82^2`

`= -6400`

`b,`

\(87^2+13^2+26\cdot87\)

`= 87*(87+26) + 169`

`= 87*113 + 169`

`= 9831 + 169`

`= 10000`

\(9,\) \(a,\left(2x+1\right)^2-4\left(x-1\right)\left(x+1\right)=2x-4\)

\(\Leftrightarrow4x^2+4x+1-4\left(x^2-1\right)-2x+4=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+4-2x+4=0\)

\(\Leftrightarrow\left(4x^2-4x^2\right)+\left(4x-2x\right)+\left(1+4+4\right)=0\)

\(\Leftrightarrow2x=-9\)

\(\Leftrightarrow x=-\dfrac{9}{2}\)

Vậy \(S=\left\{-\dfrac{9}{2}\right\}\)

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

\(\Leftrightarrow9-6x+x^2-2\left(2x+4-x^2-2x\right)-3\left(x^2+2x+1\right)-4=0\)

\(\Leftrightarrow9-6x+x^2-4x-8+2x^2+4x-3x^2-6x-3-4=0\)

\(\Leftrightarrow-12x=6\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)

\(c,3x^2+\left(-1-x\right)^2=\left(2x+5\right)\left(2x-5\right)\)

\(\Leftrightarrow3x^2+1+2x+x^2=4x^2-25\)

\(\Leftrightarrow2x=-26\)

\(\Leftrightarrow x=-13\)

Vậy \(S=\left\{-13\right\}\)

Ta có: \(C=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)

\(C=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)

\(C=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\)

Vì: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{ab}\left(-\dfrac{1}{c}\right)=0\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+-\dfrac{3}{abc}=0\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\left(1\right)\)

Thay (1) vào C ta được:

\(C=abc\left(\dfrac{3}{abc}\right)\)

\(\Rightarrow C=3\)

a) \(\left(-x-4\right)^2\)

\(=\left(-x\right)^2-2\cdot\left(-x\right)\cdot4+4^2\)

\(=x^2+8x+16\)

b) \(\left(-5+3x\right)^2\)

\(=\left(-5\right)^2+2\cdot\left(-5\right)\cdot3x+\left(3x\right)^2\)

\(=25-30x+9x^2\)

c) \(\left(-x-3\right)\left(x-3\right)\)

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

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

thank bạn :^