P=0

A) \(P=\frac{-3x+7}{x+4}\)

B)\(P=\frac{10}{3}\)

Từ \(x + y + z = 0\) suy ra \(x + y = - z\)

\(x^{2} + 2 x y + y^{2} = z^{2}\)

\(x^{2} + y^{2} - z^{2} = - 2 x y\).

Tương tự ta có: \(y^{2} + z^{2} - x^{2} = - 2 y z\) và \(z^{2} + x^{2} - y^{2} = - 2 z x\).

Do đó \(A = \frac{x y}{- 2 x y} + \frac{y z}{- 2 y z} + \frac{z x}{- 2 z x} = - \frac{1}{2} - \frac{1}{2} - \frac{1}{2} = - \frac{3}{2}\).

Vậy \(A = - \frac{3}{2}\).

Lời giải:

Áp dụng định lý Pitago: $AC=\sqrt{AB^2+BC^2}=\sqrt{3^2+4^2}=5$ (cm) 

Áp dụng tính chất tia phân giác:

$\frac{IA}{IC}=\frac{AB}{BC}=\frac{3}{4}$

Mà $IA+IC=AC=5$

$\Rightarrow IA=5:(3+4).3=\frac{15}{7}; IC=5:(3+4).4=\frac{20}{7}$ (cm)

Hình vẽ:

(2x+3)²-(2x+5)²

= 4x²+12x+3²-4x²+20x+5²

=12x+9-12x-8x-25

=9-8x-25

=(-16)-8x

Lời giải:
$(2x+3)^2+(2x+5)^2=4x^2+12x+9+4x^2+20x+25$

$=8x^2+32x+34$

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

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

\(=\left(x^2+y\right)^2-2\cdot\left(x^2+y\right)\cdot1+1^2\)

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

a: ĐKXĐ: \(\begin{cases}3x<>0\\ x+1<>0\\ 2-4x<>0\\ \end{cases}=>\begin{cases}x<>0\\ x<>-1\\ x<>\frac12\end{cases}\)

b: \(\frac{x+2}{3x}+\frac{2}{x+1}-3\)

\(=\frac{\left(x+2\right)\left(x+1\right)+2\cdot3x-3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\)
\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}=\frac{-8x^2+2}{3x\left(x+1\right)}=\frac{-2\left(4x^2-1\right)}{3x\left(x+1\right)}\)

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

Ta có: \(P=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)
\(=\frac{-2\left(2x-1\right)\left(2x+1\right)}{3x\left(x+1\right)}\cdot\frac{x+1}{-2\left(2x-1\right)}+\frac{x^2-3x-1}{3x}\)

\(=\frac{2x+1}{3x}+\frac{x^2-3x-1}{3x}=\frac{x^2-x}{3x}=\frac{x-1}{3}\)

Khi x=2023 thì \(P=\frac{2023-1}{3}=\frac{2022}{3}=674\)

c: Để P là số nguyên thì x-1⋮3

=>x-1=3k(k∈Z)

=>x=3k+1(k∈Z)

a: Xét tứ giác ABKD có \(\hat{BAD}=\hat{ADK}=\hat{BKD}=90^0\)

nên ABKD là hình chữ nhật

=>AB=DK và BK=AD

AB=DK

mà AB=4cm

nên DK=4cm

Ta có: DK+KC=DC

=>KC=DC-DK=9-4=5(cm)

ΔBKC vuông tại K

=>\(BK^2+KC^2=BC^2\)

=>\(BK^2=13^2-5^2=144=12^2\)

=>BK=12(cm)

mà BK=AD

nên AD=12cm

M là trung điểm của AD

=>\(AM=MD=\frac{AD}{2}=\frac{12}{2}=6\left(\operatorname{cm}\right)\)

b: Xét ΔABM vuông tại A và ΔDMC vuông tại D có

\(\frac{AB}{DM}=\frac{AM}{DC}\left(\frac46=\frac69=\frac23\right)\)

Do đó: ΔABM~ΔDMC

c: ΔABM~ΔDMC

=>\(\hat{ABM}=\hat{DMC}\)

mà \(\hat{ABM}+\hat{AMB}=90^0\) (ΔAMB vuông tại A)

nên \(\hat{DMC}+\hat{AMB}=90^0\)

Ta có: \(\hat{AMB}+\hat{BMC}+\hat{CMD}=180^0\)

=>\(\hat{BMC}=180^0-90^0=90^0\)

câu a, \(\dfrac{x}{x+1}\); \(\dfrac{x^2}{1-x}\); \(\dfrac{1}{x^2-1}\)  (đk \(x\)≠ -1; 1)

          \(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)

          \(\dfrac{x}{x+1}\) = \(\dfrac{x.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\);

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

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

b, \(\dfrac{10}{x+2}\); \(\dfrac{5}{2x-4}\); \(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)

    2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\)  - 2)

   \(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)

c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\)  đk \(x\) ≠ 2; -2

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

  \(\dfrac{1}{2x+4}\) = \(\dfrac{1}{2.\left(x+2\right)}\) = \(\dfrac{\left(x-2\right)}{2.\left(x+2\right).\left(x-2\right)}\)

\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\)  = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)

Lời giải:
a. ĐKXĐ: \(\left\{\begin{matrix} 3x\neq 0\\ x+1\neq 0\\ 2-4x\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\neq -1\\ x\neq \frac{1}{2}\end{matrix}\right.\)

b.

\(P=\left[\frac{(x+2)(x+1)+3x.2}{3x(x+1)}-3\right].\frac{x+1}{2(1-2x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{x^2+3x+2+6x-9x(x+1)}{3x(x+1)}.\frac{x+1}{2(1-2x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{-8x^2+2}{3x(x+1)}.\frac{x+1}{2(1-2x)}-\frac{3x-x^2+1}{3x}\)

\(=\frac{-2(2x-1)(2x+1)(x+1)}{6x(x+1)(1-2x)}-\frac{3x-x^2+1}{3x}=\frac{1+2x}{3x}-\frac{3x-x^2+1}{3x}=\frac{x^2-x}{3x}=\frac{x-1}{3}\)

Tại $x=2023$ thì:

$P=\frac{2023-1}{3}=\frac{2022}{3}=674$

c.

Để $P$ nguyên thì $x-1\vdots 3$
$\Rightarrow x=3k+1$ với $k$ nguyên bất kỳ. 

Kết hợp với ĐKXĐ thì $x=3k+1$ với $k\in\mathbb{Z}$