a) \(P=\left(\dfrac{x^2-2}{x^2+2x}+\dfrac{1}{x+2}\right):\dfrac{x+1}{x}\left(x\ne\left\{0;-2;-1\right\}\right)\\ =\left[\dfrac{x^2-2}{x\left(x+2\right)}+\dfrac{1}{x+2}\right].\dfrac{x}{x+1}\\ =\dfrac{x^2-2+x}{x\left(x+2\right)}.\dfrac{x}{x+1}\\ =\dfrac{\left(x+2\right)\left(x-1\right)}{x\left(x+2\right)}.\dfrac{x}{x+1}\\ =\dfrac{x-1}{x+1}\)

b) \(P=\dfrac{5}{2}\Rightarrow\dfrac{x-1}{x+1}=\dfrac{5}{2}\\ \Rightarrow2\left(x-1\right)=5\left(x+1\right)\\ \Leftrightarrow2x-2=5x+5\\ \Leftrightarrow5x-2x=-2-5\\ \Leftrightarrow3x=-7\\ \Leftrightarrow x=-\dfrac{7}{3}\left(TMDK\right)\)

Vậy : x=-7/3 thì P=5/2

c) \(P=\dfrac{x-1}{x+1}=\dfrac{x+1-2}{x+1}\\ =1-\dfrac{2}{x+1}\)

Để P nhận gt nguyên => 2/x+1 đạt gt nguyên

=> 2 chia hết cho x+1

=> x+1 thuộc Ư(2)={1;-1;2;-2}

=> x thuộc {0;-2;1;-3}

Đối chiếu đk : x khác {0;-2;-1}

Kết luận : x thuộc {1;-3} là 2 giá trị nguyên của x thỏa mãn P nguyên

a: 

b: Phương trình hoành độ giao điểm là:

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

=>\(\dfrac{5}{2}x=-3\)

=>\(x=-3:\dfrac{5}{2}=-\dfrac{6}{5}\)=-1,2

Thay x=-1,2 vào y=2x+6, ta được:

\(y=2\cdot\left(-1,2\right)+6=3,6\)

vậy: C(-1,2;3,6)

c: Tọa độ A là:

\(\left\{{}\begin{matrix}y=0\\2x+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=0\end{matrix}\right.\)

Tọa độ B là:

\(\left\{{}\begin{matrix}y=0\\-\dfrac{1}{2}x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=0\end{matrix}\right.\)

vậy: A(-3;0); B(6;0); C(-1,2;3,6)

\(AB=\sqrt{\left(6+3\right)^2+\left(0-0\right)^2}=9\)

\(AC=\sqrt{\left(-1,2+3\right)^2+\left(3,6-0\right)^2}=\dfrac{9\sqrt{5}}{5}\)

\(BC=\sqrt{\left(-1,2-6\right)^2+\left(3,6-0\right)^2}=\dfrac{18\sqrt{5}}{5}\)

Vì \(AC^2+BC^2=AB^2\)

nên ΔABC vuông tại C

=>\(S_{CAB}=\dfrac{1}{2}\cdot CA\cdot CB=\dfrac{1}{2}\cdot\dfrac{9}{\sqrt{5}}\cdot\dfrac{18}{\sqrt{5}}=\dfrac{81}{5}\)

d: (d2): y=-1/2x+3

=>\(-\dfrac{1}{2}x-y+3=0\)

\(d\left(M;\left(d2\right)\right)=\dfrac{\left|0\cdot\left(-\dfrac{1}{2}\right)+\left(-3\right)\cdot\left(-1\right)+3\right|}{\sqrt{\left(-\dfrac{1}{2}\right)^2+\left(-1\right)^2}}=6:\dfrac{\sqrt{5}}{2}=\dfrac{12}{\sqrt{5}}\)

Xét ΔADB vuông tại D có DH là đường cao

nên \(AH\cdot AB=AD^2\left(1\right)\)

Xét ΔADC vuông tại D có DK là đường cao

nên \(AK\cdot AC=AD^2\left(2\right)\)

Từ (1) và (2) suy ra \(AH\cdot AB=AK\cdot AC\)

=>\(\dfrac{AH}{AC}=\dfrac{AK}{AB}\)

Xét ΔAHK vuông tại A và ΔACB vuông tại A có

\(\dfrac{AH}{AC}=\dfrac{AK}{AB}\)

Do đó: ΔAHK~ΔACB

a) \(A=\left(\dfrac{3x^2+3}{x^3-1}-\dfrac{x-1}{x^2+x+1}-\dfrac{1}{x-1}\right):\dfrac{2x^2-5x+5}{x-1}\left(x\ne1\right)\)

\(=\left[\dfrac{3x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\cdot\dfrac{x-1}{2x^2-5x+5}\)

\(=\dfrac{3x^2+3-\left(x-1\right)^2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{2x^2-5x+5}\)

\(=\dfrac{2x^2-x+2-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{2x^2-5x+5}\)

\(=\dfrac{2x^2-x+2-x^2+2x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{2x^2-5x+5}\)

\(=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{2x^2-5x+5}\)

\(=\dfrac{1}{2x^2-5x+5}\)

b) Ta có: \(A=\dfrac{1}{2x^2-5x+5}=\dfrac{1}{2\left(x^2-\dfrac{5}{2}x+\dfrac{5}{2}\right)}\)

\(=\dfrac{1}{2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}+\dfrac{5}{2}\right)}\)

\(=\dfrac{1}{2\left[\left(x^2-2\cdot\dfrac{5}{4}+\dfrac{25}{16}\right)+\dfrac{15}{16}\right]}\)

\(=\dfrac{1}{2\left(x-\dfrac{5}{4}\right)^2+\dfrac{15}{8}}\)

Mà: \(2\left(x-\dfrac{5}{4}\right)^2\ge0\forall x\ne1\)

\(\Rightarrow2\left(x-\dfrac{5}{4}\right)^2+\dfrac{15}{8}\ge\dfrac{15}{8}\forall x\ne1\)

\(\Rightarrow\dfrac{1}{2\left(x-\dfrac{5}{4}\right)^2+\dfrac{15}{8}}\le\dfrac{8}{15}\forall x\ne1\) 

Dấu "=" xảy ra khi: \(\left(x-\dfrac{5}{4}\right)^2=0\Leftrightarrow x=\dfrac{5}{4}\)

Vậy: \(A_{max}=\dfrac{8}{15}\Leftrightarrow x=\dfrac{5}{4}\)

a) \(\dfrac{1+\dfrac{1}{x}}{x-\dfrac{1}{x}}=\dfrac{\dfrac{x+1}{x}}{\dfrac{x^2-1}{x}}=\dfrac{x+1}{x^2-1}=\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{1}{x-1}\left(x\ne0;x\ne1;x\ne-1\right)\) 

b) \(\left(\dfrac{1}{x^2+4x+4}-\dfrac{1}{x^2-4x+4}\right):\left(\dfrac{1}{x+2}+\dfrac{1}{x-2}\right)\left(x\ne\pm2\right)\) 

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

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

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

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

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

\(=\dfrac{-4}{x^2-4}\) 

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

\(\left(\dfrac{x}{x+1}+1\right):\left(1-\dfrac{3x^2}{1-x^2}\right)\)

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

\(=\dfrac{2x+1}{x+1}\cdot\dfrac{x^2-1}{4x^2-1}\)

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

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

d:

ĐKXĐ: x<>1

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

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

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

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

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

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

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

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

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

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

Lời giải:
Áp dụng BĐT Cô-si:

$\frac{x^3}{y^2}+y+y\geq 3\sqrt[3]{x^3}=3x$

$\frac{y^3}{z^2}+z+z\geq 3\sqrt[3]{y^3}=3y$

$\frac{z^3}{x^2}+x+x\geq 3\sqrt[3]{z^3}=3z$

Cộng 3 BĐT trên theo vế và thu gọn thì:

$P+2(x+y+z)\geq 3(x+y+z)$

$\Rightarrow P\geq x+y+z=2023$

Vậy $P_{\min}=2023$. Giá trị này đạt tại $x=y=z=\frac{2023}{3}$

Bạn lưu ý lần sau gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn và hỗ trợ tốt hơn nhé.

SOS mai nộp bài r ạ mong mn giúp đỡ

Câu 6:

a: ĐKXĐ: \(x\notin\left\{0;3;2;-2\right\}\)

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

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

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

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

\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2+8x}{x+2}\cdot\dfrac{x}{x-3}\)

\(=\dfrac{4x^2}{x-3}\)

b: Để A>0 thì \(\dfrac{4x^2}{x-3}>0\)

=>x-3>0

=>x>3

c: |x-7|=4

=>\(\left[{}\begin{matrix}x-7=4\\x-7=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Thay x=11 vào A, ta được:

\(A=\dfrac{4\cdot11^2}{11-3}=\dfrac{4\cdot121}{8}=\dfrac{121}{2}\)

I: Trắc nghiệm
Câu 1: C

Câu 2: A

Câu 3: D

Câu 4: A

II: Tự luận

Câu 5:

a: ĐKXĐ: x<>-1/2

\(\dfrac{5-3x}{2x+1}-\dfrac{-2+5x}{2x+1}\)

\(=\dfrac{5-3x+2-5x}{2x+1}\)

\(=\dfrac{-8x+7}{2x+1}\)

b: ĐKXĐ: x<>-1

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

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

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

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

Câu 6(Đề 4)

a: \(A=\left(\dfrac{3x^2+3}{x^3-1}-\dfrac{x-1}{x^2+x+1}-\dfrac{1}{x-1}\right):\dfrac{2x^2-5x+5}{x-1}\)

\(=\left(\dfrac{3x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x-1}{x^2+x+1}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{2x^2-5x+5}\)

\(=\dfrac{3x^2+3-\left(x-1\right)^2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{2x^2-5x+5}\)

\(=\dfrac{2x^2-x+2-x^2+2x-1}{x^2+x+1}\cdot\dfrac{1}{2x^2-5x+5}\)

\(=\dfrac{x^2+x+1}{x^2+x+1}\cdot\dfrac{1}{2x^2-5x+5}=\dfrac{1}{2x^2-5x+5}\)

b: \(2x^2-5x+5=2\left(x^2-\dfrac{5}{2}x+\dfrac{5}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{15}{16}\right)\)

\(=2\left(x-\dfrac{5}{4}\right)^2+\dfrac{15}{8}>=\dfrac{15}{8}\forall x\)

=>\(A=\dfrac{1}{2x^2-5x+5}< =1:\dfrac{15}{8}=\dfrac{8}{15}\forall x\)

Dấu '=' xảy ra khi x=5/4