Ta có:

\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=\left(a+b+c\right)\left(ab+c\left(a+b\right)\right)-abc\)

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

\(=\left(a+b\right)\left(ab+c^2+c\left(a+b\right)\right)\)

\(=\left(a+b\right)\left(ab+ac+c^2+bc\right)\)

\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

Đồng thời:

\(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)

Tương tự:

\(b^2+1=\left(a+b\right)\left(b+c\right)\)

\(c^2+1=\left(a+c\right)\left(b+c\right)\)

Từ đó:

\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)

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

Do vế trái luôn dương nên vế phải dường

\(\Rightarrow2021x>0\)

\(\Rightarrow x>0\)

\(\Rightarrow\) Tất cả các số hạng trong trị tuyệt đối đều dương.

Do đó pt trở thành:

\(\left|x+\dfrac{1}{1.2}\right|+\left|x+\dfrac{1}{2.3}\right|+...+\left|x+\dfrac{1}{2020.2021}\right|=2021x\)

\(\Rightarrow x+\dfrac{1}{1.2}+x+\dfrac{1}{2.3}+x+\dfrac{1}{3.4}+...+x+\dfrac{1}{2020.2021}=2021x\)

\(\Leftrightarrow2020x+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2020.20210}=2021x\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}=2021x-2020x\)

\(\Leftrightarrow x=1-\dfrac{1}{2021}\)

\(\Leftrightarrow x=\dfrac{2020}{2021}\)

\(12x=-15y=10z\)

\(\Rightarrow\dfrac{12x}{60}=\dfrac{-15y}{60}=\dfrac{10z}{60}\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{-4}=\dfrac{z}{6}\)

Đặt: \(\dfrac{x}{5}=\dfrac{y}{-4}=\dfrac{z}{6}=k\)

\(\Rightarrow x=5k;y=-4k;z=6k\)

Ta có: \(xyz=120\)

\(\Rightarrow5k\cdot-4k\cdot6k=120\)

\(\Rightarrow-120k^3=120\)

\(\Rightarrow k^3=-1\)

\(\Rightarrow k=-1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-1\\\dfrac{y}{-4}=-1\\\dfrac{z}{6}=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-1\cdot5=-5\\y=-4\cdot-1=4\\z=-1\cdot6=-6\end{matrix}\right.\)

Vậy: ... 

a) Đặt: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\) 

Ta có: \(x^2+3y^2-2z^2=-16\)

\(\Rightarrow\left(2k\right)^2+3\cdot\left(3k\right)^2-2\cdot\left(4k\right)^2=-16\)

\(\Rightarrow4k^2+3\cdot9k^2-2\cdot16k^2=-16\)

\(\Rightarrow4k^2+27k^2-32k^2=-16\)

\(\Rightarrow-k^2=-16\)

\(\Rightarrow k^2=16\)

\(\Rightarrow k=\pm4\)

Với k = 4

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{y}{3}=4\\\dfrac{z}{4}=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot4=8\\y=3\cdot4=12\\z=4\cdot4=16\end{matrix}\right.\)

Với k = -4 

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-4\\\dfrac{y}{3}=-4\\\dfrac{z}{4}=-4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot-4=-8\\y=3\cdot-4=-12\\z=4\cdot-4=-16\end{matrix}\right.\) 

Vậy: ...

b) Đặt: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\) 

Ta có: \(2x^2+2y^2-3z^2=-100\)

\(\Rightarrow2\cdot\left(3k\right)^2+2\cdot\left(4k\right)^2-3\cdot\left(5k\right)^2=-100\)

\(\Rightarrow2\cdot9k^2+2\cdot16k^2-3\cdot25k^2=-100\)

\(\Rightarrow18k^2+32k^2-75k^2=-100\)

\(\Rightarrow-25k^2=-100\)

\(\Rightarrow k^2=-\dfrac{100}{-25}=4\)

\(\Rightarrow k=\pm2\)

Với k = 2

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=2\\\dfrac{y}{4}=2\\\dfrac{z}{5}=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=2\cdot4=8\\z=2\cdot5=10\end{matrix}\right.\)

Với k = -2

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-2\\\dfrac{y}{4}=-2\\\dfrac{z}{5}=-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot-3=-6\\y=2\cdot-4=-8\\z=2\cdot-5=-10\end{matrix}\right.\)

Vậy: ... 

\(\dfrac{1-\left|x\right|}{3}=\dfrac{\left|x\right|+3}{-5}\)

\(\Rightarrow-5\cdot\left(1-\left|x\right|\right)=3\cdot\left(\left|x\right|+3\right)\)

\(\Rightarrow-5+5\left|x\right|=3\left|x\right|+9\)

\(\Rightarrow5\left|x\right|-3\left|x\right|=9+5\)

\(\Rightarrow2\left|x\right|=14\)

\(\Rightarrow\left|x\right|=14:2\)

\(\Rightarrow\left|x\right|=7\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

Vậy \(x\in\left\{7;-7\right\}\).

a)Nối K với M .

Xét △BMK và △IMK có:

-MK:cạnh chung.

-^BKM=^IMK( 2 góc so le trong của IM // BC)

-^BMK=^MKI( 2 góc so le trong của AB // IK)

⇒ △BMK = △IMK (g.c.g)

⇒ BM=IK(cctư)

mà AM=BM(M là trung điểm của AB)

⇒AM=IK(ĐPCM).

b) Có ^AMI=^MIK( 2 góc so le trong của AB // IK).

Mà ^MIK=^IKC(2 góc so le trong của MI // BC).

⇒ ^AMI = ^IKC (1).

Xét △AMI và △IKC có:

-^AMI = ^IKC (chứng minh (1)).

-AM=IK(chứng minh câu a)).

-^MAI=^KIC( 2 góc đồng vị của AB // IK).

⇒△AMI=△IKC(g.c.g)(ĐPCM).

c)Từ câu b) , △AMI=△IKC.Suy ra: AI=IC (cctư).

`#3107.101107`

`A(x) = 3x - 9x^2 + 4x + 5x^3 + 7x^2 + 1`

`= (3x + 4x) - (9x^2 - 7x^2) + 5x^3 + 1`

`= 7x - 2x^2 + 5x^3 + 1`

`B(x) = 5x^3 - 3x^2 + 7x + 10`

`A(x) - B(x) = 7x - 2x^2 + 5x^3 + 1 - (5x^3 - 3x^2 + 7x + 10)`

`= 7x - 2x^2 + 5x^3 + 1 - 5x^3 + 3x^2 - 7x - 10`

`= (7x - 7x) + (3x^2 - 2x^2) + (5x^3 - 5x^3) - (10 - 1)`

`= x^2 - 9`

`=> C(x) = x^2 - 9`

`C(x) = 0`

`=> x^2 - 9 = 0`

`=> x^2 = 9 => x^2 = (+-3)^2 => x = +-3`

Vậy, nghiệm của đa thức `C(x)` là `x \in {3; -3}.`

\(H=\left(3x-6\right)^2-3\left|2x-4\right|+2023\)

\(=\left(3x-6\right)^2-2\left|3x-6\right|+2023\)

\(=\left(3x-6\right)^2-2\left|3x-6\right|+1+2022\)

\(=\left(\left|3x-6\right|-1\right)^2+2022\)

Do \(\left(\left|3x-6\right|-1\right)^2\ge0;\forall x\)

\(\Rightarrow H\ge2022\)

\(\Rightarrow H_{min}=2022\) khi \(\left|3x-6\right|-1=0\Rightarrow x=\left\{\dfrac{7}{3};\dfrac{5}{3}\right\}\)