Lời giải:

a. Xét tam giác $ABC$ và $HBA$ có:

$\widehat{B}$ chung

$\widehat{BAC}=\widehat{BHA}=90^0$

$\Rightarrow \triangle ABC\sim \triangle HBA$ (g.g)

b.

$BC=\sqrt{AB^2+AC^2}=\sqrt{15^2+20^2}=25$ (cm) - định lý Pitago

$AH=2S_{ABC}:BC=AB.AC:BC=15.20:25=12$ (cm)

$BH=\sqrt{AB^2-AH^2}=\sqrt{15^2-12^2}=9$ (cm) - định lý Pitago

c.

Theo tính chất đường phân giác:

$\frac{DA}{DC}=\frac{AB}{BC}=\frac{15}{25}=\frac{3}{5}$

$DA+DC=AC=20$

$\Rightarrow DA=20:(3+5).3=7,5$ (cm)

$DC=AC-DA=20-7,5=12,5$ (cm)

Hình vẽ:

Đk: \(-1< x< 1\)

Ta có \(2\sqrt{2022\left(1-x^2\right)}\le2023-x^2\) 

Nếu \(0\le x< 1\) thì \(x\left(x+2021\right)\ge0\) 

\(\Leftrightarrow x^2+2021x\ge0\)

\(\Leftrightarrow2023-x^2\le2021x+2023\)

\(\Rightarrow\) \(2\sqrt{2022\left(1-x^2\right)}\le2023-x^2\le2021x+2023\)

\(\Leftrightarrow2\sqrt{2022}\le\dfrac{2021x+2023}{\sqrt{1-x^2}}\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2022=1-x^2\\x=0\end{matrix}\right.\), vô lý.

Vậy nếu \(0\le x< 1\) thì BĐT đúng.

Xét \(-1< x< 0\) thì đặt \(x=-t\left(0< t< 1\right)\). 

BĐT cần chứng minh \(\Leftrightarrow\dfrac{2023-2021t}{\sqrt{1-t^2}}\ge2\sqrt{2022}\)

Ta có \(2023-2021t\) 

\(=2022-2022t+1+t\)

\(=2022\left(1-t\right)+\left(1+t\right)\)

\(\ge2\sqrt{2022\left(1-t\right)\left(1+t\right)}\)

\(=2\sqrt{2022\left(1-t^2\right)}\)

\(\Leftrightarrow\dfrac{2023-2021t}{\sqrt{1-t^2}}\ge2\sqrt{2022}\)

Dấu "=" xảy ra \(\Leftrightarrow2022-2022t=1+t\) \(\Leftrightarrow t=\dfrac{2021}{2023}\) \(\Leftrightarrow x=-\dfrac{2021}{2023}\)

Vậy ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{2021}{2023}\)

Trường hợp \(x\) = - \(\dfrac{2020}{2021}\) thì sao em nhỉ?

a: Ta có: \(AM=MB=\frac{AB}{2}\)

\(AN=NC=\frac{AC}{2}\)

mà AB=AC

nên AM=MB=AN=NC

Xét ΔNCK và ΔMAC có

NC=AM

\(\hat{CNK}=\hat{AMC}\left(=90^0-\hat{NCH}\right)\)

NK=CM

Do đó: ΔNCK=ΔMAC

=>\(\hat{NCK}=\hat{MAC}=90^0\)

=>CK⊥CA

=>CK//AB

Xét tứ giác ABKC có AB//KC và CK⊥CA

nên ABKC là hình thang vuông

pt đã cho \(\Leftrightarrow\left(x-y\right)\left(x+3y\right)=2x+6y-x+y\)

\(\Leftrightarrow\left(x-y\right)\left(x+3y\right)=2\left(x+3y\right)-\left(x-y\right)\)

Đặt \(\left\{{}\begin{matrix}x-y=a\\x+3y=b\end{matrix}\right.\) với \(a,b\inℤ\) và \(b\ge4\)

pt thành \(ab=2a-b\)

\(\Leftrightarrow ab-2a+b-2=-2\)

\(\Leftrightarrow\left(a+1\right)\left(b-2\right)=-2\) (*)

 Vì \(b\ge4\Leftrightarrow b-2\ge2\). Do đó (*) \(\Rightarrow\) \(b-2=2\) hay \(b=4\), nghĩa là dấu "=" phải xảy ra \(\Leftrightarrow x=y=1\). Thử lại, ta thấy không thỏa mãn.

 Vậy pt đã cho không có nghiệm nguyên dương.

a) ĐK: \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\4-x^2\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm2\)

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

\(A=\left[\dfrac{x+2}{x-2}-\dfrac{1}{x+2}+\dfrac{x-4}{\left(x+2\right)\left(x-2\right)}\right]\cdot\left(x^2-4\right)\)

\(A=\left[\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-4}{\left(x+2\right)\left(x-2\right)}\right]\cdot\left(x^2-4\right)\)

\(A=\dfrac{x^2+4x+4-x+2+x-4}{\left(x+2\right)\left(x-2\right)}\cdot\left(x+2\right)\left(x-2\right)\)

\(A=x^2+4x+2\)

b) \(A=14\)

\(\Leftrightarrow x^2+4x+2=14\)

\(\Leftrightarrow x^2+4x-12=0\)

\(\Leftrightarrow x^2-2x+6x-12=0\)

\(\Leftrightarrow x\left(x-2\right)+6\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

a; (\(x+y\))² - 4.(\(x-y\))²

   = \(x^2+2xy+y^2\) - 4\(x^2+8xy-4y^2\)

  =  (\(x^2-4x^2\))  + (2\(xy+8xy\)) + (y² - 4y²)

= - 3\(x^2\) + 10\(xy\)  - 3y²

b; (\(x+y\))³ - 2\(x^3\) + (\(x-y\))³

  = \(x^3+3x^2y+3xy^2+y^3\) - 2\(x^3\) + \(x^3-3x^2y+3xy^2-y^3\) 

= (\(x^3\) + \(x^3\)- 2\(x^3\)) + (3\(x^2y-3xy^2\)) + (3\(xy^2\) + 3\(xy^2\)) + (y³-y³)

= 0 + 0 + 6\(xy^2\) + 0

= 6\(xy^2\) 

Lời giải:

Gọi đa thức dư khi chia $f(x)$ cho $(x-2))(x^2+1)$ là $ax^2+bx+c$

Ta có:

$f(x)=(x-2)(x^2+1)Q(x)+ax^2+bx+c$

$f(2) = 4a+2b+c=7(1)$

$f(x) = (x-2)(x^2+1)Q(x)+a(x^2+1)+bx+(c-a)$

$=(x^2+1)[(x-2)Q(x)+a]+bx+(c-a)$

$\Rightarrow bx+(c-a)=3x+5$

$\Rightarrow b=3; c-a=5(2)$
Từ $(1); (2)\Rightarrow a=\frac{-4}{5}; b=3; c=\frac{21}{5}$

Vậy đa thức dư là $\frac{-4}{5}x^2+3x+\frac{21}{5}$

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

\(M=\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right):\dfrac{2}{x+2}\\ =\left(\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x+2}{2}\\ =\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}\\ =\dfrac{4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}\\ =\dfrac{2}{x-2}\)

Để `M=1` Thì

\(\dfrac{2}{x-2}=1\\ \Leftrightarrow\dfrac{2}{x-2}-1=0\\ \Leftrightarrow\dfrac{2}{x-2}-\dfrac{x-2}{x-2}=0\\ \Leftrightarrow2-x+2=0\\ \Leftrightarrow4-x=0\\ \Leftrightarrow x=4\)

a) Để \(M\) xác định thì \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\\dfrac{2}{x+2}\ne0\end{matrix}\right.\Rightarrow x\ne\pm2\)

Khi đó: \(M=\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right):\dfrac{2}{x+2}\)

\(=\left[\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\cdot\dfrac{x+2}{2}\)

\(=\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}\)

\(=\dfrac{4}{2\left(x-2\right)}=\dfrac{2}{x-2}\)