Bạn xem lại câu `4) ,7)` nhé.

`1)`

`x^4 +2x^2+1`

`=(x^2)^2 +2.x^2 .1+1^2`

`=(x^2 +1)^2` 

`2)`

`4x^2 -12xy+9y^2`

`=(2x)^2 - 2.2x.3y + (3y)^2`

`=(2x-3y)^2`

`3)`

`-x^2 -2xy-y^2`

`=-(x^2 +2xy+y^2)`

`=-(x+y)^2`

`4)`

`(x+y)^2 -2(x+y)+1`

`=(x+y)^2 -2(x+y).1+1^2`

`=(x+y-1)^2`

`5)`

`x^3 -3x^2 +3x-1`

`=x^3 -3.x^2 .1+3.x.1^2 -1^3`

`=(x-1)^3`

`6)`

`x^3 +6x^2 +12x+8`

`=x^3 +3.x^2 .2 +3.x.2^2 +2^3`

`=(x+2)^3`

`7)`

`x^3 +1-x^2 -x`

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

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

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

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

`8)`

`(x+y)^3 -x^3 -y^3`

`=x^3 +3x^2 y+3xy^2 +y^3 -x^3 -y^3`

`=3x^2 y+3xy^2`

`=3xy(x+y)`.

\(B=\dfrac{x+1}{x-2}+\dfrac{2}{x+3}-\dfrac{9x-3}{x^2+x-6}\)

\(=\dfrac{x+1}{x-2}+\dfrac{2}{x+3}+\dfrac{-9x+3}{\left(x-2\right)\left(x+3\right)}\) (1)

`a,`

`ĐKXĐ:`\(\left\{{}\begin{matrix}x-2\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-3\end{matrix}\right.\)

`b,` 

(Tiếp tục từ (1))

\(=\dfrac{\left(x+1\right)\left(x+3\right)+2\left(x-2\right)-9x+3}{\left(x-2\right)\left(x+3\right)}\)

\(=\dfrac{x^2+4x+3+2x-4-9x+3}{\left(x-2\right)\left(x+3\right)}\)

`=(x^2 -3x+2)/((x-2)(x+3))`

`=((x-2)(x-1))/((x-2)(x+3))`

`=(x-1)/(x+3)`.

`1)`

Để `D(x)` có nghiệm `=>x(x-3)=0`

`<=>x=0` hoặc `x-3=0`

`<=>x=0` hoặc `x=3`

Vậy `x\in{0;3}`.

`2)`

Để `F(x)` có nghiệm `=> x^2 +x=0`

`<=>x(x+1)=0`

`<=>x=0` hoặc `x+1=0`

`<=>x=0` hoặc `x=-1`

Vậy `x\in{-1;0}`.

`3)`

Để `D(x)` có nghiệm `=> (x-3)(3+x)=0`

`<=>x-3=0` hoặc 3+x=0`

`<=>x=3` hoặc `x=-3`

Vậy `x\in{-3;3}`.

`4)`

Để `F(x)` có nghiệm `=>3x-6x^2 =0`

`<=>3x(1-2x)=0`

`<=>3x=0` hoặc `1-2x=0`

`<=>x=0` hoặc `x=1/2`

Vậy `x\in{0 ; 1/2}`.

`5)`

Để `D(x)` có nghiệm `=>(x-1)(3x+2)=0`

`<=>x-1=0` hoặc `3x+2=0`

`<=>x=1` hoặc `x=-2/3`

Vậy `x\in{-2/3 ;1}`.

`6)`

Để `F(x)` có nghiệm `=>5x^3 +10x=0`

`<=>5x(x^2 +2)=0`

`<=>5x=0` hoặc `x^2 +2=0`

`<=>x=0` hoặc `x^2 =-2` (vô lí)

Vậy `x\in{0}`.

`M-N=(3xyz-3x^2 +5xy-1) - (5x^2 +xyz -5xy+3-y)`

`=(3xyz - xyz) + (-3x^2 - 5x^2) + (5xy+5xy) + (-1-3)+y`

`=2xyz - 8x^2 + 10xy - 4+y`

`N- M = (5x^2 +xyz -5xy+3-y) - (3xyz-3x^2 +5xy-1)`

`=(5x^2 + 3x^2) + (xyz - 3xyz) + (-5xy-5xy) + (3+1)-y`

`=8x^2 - 2xyz - 10xy +4-y`.

`{(3x-y/2 =2 (1)),(x/4 - y/3 = -2/3 (2)):}`

`(1)=>3x = 2 + y/2 <=> x = 2/3 + 1/6 y`

Thay `x=2/3 + 1/6 y` vào `(2)`, ta được:

`(2/3 + 1/6 y)/4 - y/3 = -2/3`

`<=> 1/6 + 1/24 y - y/3 = -2/3`

`<=> 1/6 - 7/24 y = -2/3`

`<=> 7/24 y = 5/6`

`<=> y = 20/7`

Thay `y=20/7` vào `(1)`, ta được:

`3x - (20/7)/2 = 2`

`<=> 3x - 10/7 =2`

`<=>3x=24/7`

`<=>x=8/7`

Vậy `(x;y)=(8/7 ; 20/7)`.

`a)`

`A=(3+x)/(3-x) - (4x^2)/(x^2 -9) - (3-x)/(3+x)`

`=(-(3+x)(x+3)-4x^2 -(3-x)(x-3))/((x+3)(x-3))`

`=(-x^2 -6x-9-4x^2 +x^2 -6x+9)/((x+3)(x-3))`

`=(-4x^2 -12x)/((x+3)(x-3))`

`=(-4x(x+3))/((x+3)(x-3))`

`=(-4x)/(x-3)`

`b)`

`A=-3`

`=>(-4x)/(x-3)=-3`

`<=> -4x=(x-3).(-3)`

`<=>-4x=-3x+9`

`<=>-4x+3x=9`

`<=>-x=9`

`<=>x=-9`

Vậy `x=-9`.

`c)`

`x=3 => A=(-4.3)/(3-3) = (-12)/0 =0`.

Tọa độ giao điểm của `(d)` và `(d')` là:

`(m+1)x+3=2x+3`

`<=>mx+x+3-2x-3=0`

`<=>mx-x=0`

`<=>x(m-1)=0`

`<=>[(x=0),(m=1 (loại)):}`

`=>y=2.0+3=0+3=3`

`=>` Tọa độ giao điểm của `(d)` và `(d')` là `(0;3)`.

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^2-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)^3-3ab\left(a+b\right)-3\left(a+b\right).c\left(a+b+c\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)^3-3ab\left(a+b+c\right)-3\left(a+b\right).c\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b+c\right)^2-3ab-3ab-3bc\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

Ta có:

\(a;b;c>0\)

\(\Rightarrow a+b+c>0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow a=b=c\)

\(A=2020\left(1-\dfrac{a}{b}\right)\left(1-\dfrac{b}{c}\right)\left(1-\dfrac{c}{a}\right)-2021\left(\dfrac{a}{b}-\dfrac{b}{c}+\dfrac{c}{a}\right)^3\)

\(\Rightarrow A=2020.\left(1-1\right)\left(1-1\right)\left(1-1\right)-2021\left(1-1+1\right)^3\)

\(\Rightarrow A=-2021\).

`a(b+1)+b(a+1)=(a+1)(b+1)`

`VT=ab+a+ab+b`

`=2ab+a+b`

`=2.1+a+b`

`=2+a+b(1)`

`VP=(a+1)(b+1)`

`=a(b+1)+1(b+1)`

`=ab+a+b+1`

`=1+a+b+1`

`=2+a+b(2)`

`(1)(2)=>VT=VP`.