a:

ĐKXĐ: \(x\ne0;y\ne0\)

 \(\dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{\dfrac{1}{x}-\dfrac{1}{y}}=\dfrac{x+y}{xy}:\dfrac{x-y}{xy}=\dfrac{x+y}{xy}\cdot\dfrac{xy}{x-y}=\dfrac{x+y}{x-y}\)

b:

ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)

 \(\dfrac{\dfrac{x}{x+1}-\dfrac{x-1}{x}}{\dfrac{x}{x-1}-\dfrac{x+1}{x}}\)

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

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

\(=\dfrac{x-1}{x+1}\)

c:

ĐKXĐ: \(x\ne-1\)

 \(1-\dfrac{x}{1-\dfrac{x}{x+1}}\)

\(=1-\dfrac{x}{\dfrac{x+1-x}{x+1}}\)

\(=1-\dfrac{x}{\dfrac{1}{x+1}}\)

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

\(B=\dfrac{-\left(2x^2+3\right)+2x^2+x+5}{2x^2+3}=-1+\dfrac{2\left(x+\dfrac{1}{4}\right)^2+\dfrac{39}{8}}{2x^2+3}>-1\)

\(B=\dfrac{2x^2+3-2x^2+x-1}{2x^2+3}=1-\dfrac{2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}}{2x^2+3}< 1\)

\(\Rightarrow-1< B< 1\)

Mà B nguyên \(\Rightarrow B=0\)

\(\Rightarrow x+2=0\Rightarrow x=-2\)

\(P=\dfrac{3\left(x^2+1\right)}{3\left(x^2-x+1\right)}=\dfrac{2\left(x^2-x+1\right)+x^2+2x+1}{3\left(x^2-x+1\right)}=\dfrac{2}{3}+\dfrac{\left(x+1\right)^2}{3\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge\dfrac{2}{3}\)

\(P=\dfrac{2\left(x^2-x+1\right)-x^2+2x-1}{x^2-x+1}=2-\dfrac{\left(x-1\right)^2}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le2\)

\(\Rightarrow\dfrac{2}{3}\le P\le2\)

Mà P nguyên \(\Rightarrow\left[{}\begin{matrix}P=1\\P=2\end{matrix}\right.\)

- Với \(P=1\Rightarrow\dfrac{x^2+1}{x^2-x+1}=1\Rightarrow x^2+1=x^2-x+1\)

\(\Rightarrow x=0\)

- Với \(P=2\Rightarrow\dfrac{x^2+1}{x^2-x+1}=2\Rightarrow x^2+x=2\left(x^2-x+1\right)\)

\(\Rightarrow x^2-2x+1=0\Rightarrow x=1\)

Vậy \(x=\left\{0;1\right\}\)

1: \(\dfrac{DM}{DE}=\dfrac{2}{5}\)

\(\dfrac{DN}{DF}=\dfrac{6}{15}=\dfrac{2}{5}\)

Do đó: \(\dfrac{DM}{DE}=\dfrac{DN}{DF}\)

2: Xét ΔDEF có \(\dfrac{DM}{DE}=\dfrac{DN}{DF}\)

nên MN//EF

Ta thấy: \(2024\equiv1\) (\(mod\) \(2023\))
\(20242024\equiv1909\) (\(mod\) \(2023\))
...
\(2024...2024:2023\) dư một số nào đó là một trong các số từ \(1\) đến \(2022\) (\(2023\) số).
* Xét \(2024\) số: \(2024;20242024;...;20242024...2024\) (Gồm \(2024\) bộ số \(2024\))
 + Lấy \(2024\) số trên chia cho \(2023\), ta có \(2024\) số dư từ \(0\) đến \(2022\).
\(\Rightarrow\) Tồn tại hai số chia cho \(2023\) có cùng số dư.
Giả sử hai số đó là \(a=2024...2024\) (\(i\) bộ số \(2024\)) và \(b=2024...2024\) (\(j\) bộ số \(2024\)) \(\left(1\le i\le j\le2024\right)\)
+ \(a-b=2024...2024\cdot10^{4i}\) (\(j-i\) bộ số \(2024\)) chia hết cho \(2023\)
+ \(ƯCLN\left(10^{4i};2023\right)=1\)
\(\Rightarrow2024...2024\) (\(j-i\) bộ số \(2024\)) chia hết cho \(2023\) \(\left(đpcm\right)\).

\(x^4-3x+2=x\left(x^3+ax^2+bx-2\right)-\left(x^3+ax^2+bx-2\right)\)

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

Đồng nhất hệ số 2 vế ta được:

\(\left\{{}\begin{matrix}a-1=0\\b-a=0\\b+2=3\end{matrix}\right.\) \(\Rightarrow a=b=1\)

\(x^4-3x+2=\left(x-1\right)\left(x^3+ax^2+bx-2\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+x-2\right)=\left(x-1\right)\left(x^3+ax^2+bx-2\right)\)
\(\Rightarrow x^3+x^2+x-2=x^3+ax^2+bx-2\)
\(\Rightarrow1\cdot x^2+1\cdot x=ax^2+bx\)
\(\Rightarrow a=1\) và \(b=1\)

\(B=3x^2+3y^2+z^2+5xy-3yz-3xz-2x-2y+3\\\Rightarrow4A=12x^2+12y^2+4z^2+20xy-12yz-12xz-8x-8y+12\\\\=[(9x^2+18xy+9y^2)-(12xz+12yz)+4z^2]+[(2x^2+4xy+2y^2)-(8x+8y)+8]+(x^2-2xy+y^2)+4\\=[(3x+3y)^2-2\cdot(3x+3y)\cdot2z+(2z)^2]+[2(x^2+2xy+y^2)-8(x+y)+8]+(x-y)^2+4\\=(3x+3y-2z)^2+2[(x+y)^2-4(x+y)+4]+(x-y)^2+4\\=(3x+3y-2z)^2+2(x+y-2)^2+(x-y)^2+4\)

Ta thấy: \(\left\{{}\begin{matrix}\left(3x+3y-2z\right)^2\ge0\forall x,y,z\\2\left(x+y-2\right)^2\ge0\forall x,y\\\left(x-y\right)^2\ge0\forall x,y\end{matrix}\right.\)

\(\Rightarrow\left(3x+3y-2z\right)^2+2\left(x+y-2\right)^2+\left(x-y\right)^2+4\ge4\forall x,y,z\)

\(\Leftrightarrow4B\ge4\Leftrightarrow B\ge1\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}3x+3y-2z=0\\x+y-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x\\2x=2\\2z=6x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=3\end{matrix}\right.\)

Vậy \(Min_B=1\) khi \(x=y=1;z=3\).

\(Toru\)

a: Xét ΔBAC có AM là phân giác

nên \(\dfrac{BM}{MC}=\dfrac{AB}{AC}\)

=>\(\dfrac{BM}{MC}=\dfrac{a}{b}\)

=>\(\dfrac{BM}{a}=\dfrac{MC}{b}\)

mà BM+MC=BC=a

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{BM}{a}=\dfrac{MC}{b}=\dfrac{BM+MC}{a+b}=\dfrac{a}{a+b}\)

=>\(BM=\dfrac{a\cdot a}{a+b}=\dfrac{a^2}{a+b}\)

Xét ΔBCA có CN là phân giác

nên \(\dfrac{BN}{NA}=\dfrac{BC}{CA}\)

=>\(\dfrac{BN}{NA}=\dfrac{a}{b}\)

=>\(\dfrac{BN}{NA}=\dfrac{BM}{MC}\)

Xét ΔBAC có \(\dfrac{BN}{NA}=\dfrac{BM}{MC}\)

nên MN//AC

b: Xét ΔBAC có MN//AC

nên \(\dfrac{MN}{AC}=\dfrac{BM}{BC}\)

=>\(\dfrac{MN}{b}=\dfrac{a^2}{a+b}:a=\dfrac{a}{a+b}\)

=>\(MN=\dfrac{a\cdot b}{a+b}\)