a) \(H=\left(\dfrac{a-3\sqrt{a}}{a-2\sqrt{a}-3}-\dfrac{2a}{a-1}\right):\dfrac{1-\sqrt{a}}{a-2\sqrt{a}+1}\)

\(H=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}-\dfrac{2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{1-\sqrt{a}}{\left(\sqrt{a}-1\right)^2}\)

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

\(H=\dfrac{a-\sqrt{a}-2a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{-1}{\sqrt{a}-1}\)

\(H=\dfrac{-a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot-\left(\sqrt{a}-1\right)\)

\(H=\dfrac{-\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot-\left(\sqrt{a}-1\right)\)

\(H=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\)

\(H=\sqrt{a}\)

b) Thay x = 2023 vào ta có: 

\(H=\sqrt{2023}\)

Câu 1: 

1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)

\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{7}{x^2+3}\)

Câu 2a đề sai, pt này ko giải được

2b.

\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)

Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)

\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)

3a.

\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)

\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)

3b

Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6

\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6

\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)

Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)

\(\Rightarrow2021^{2022}-1⋮6\)

\(\Rightarrow S-1⋮6\)

2a. 

À nãy mình nhìn lộn dấu trừ bên vế phải thành dấu cộng

ĐKXĐ: ...

\(\Leftrightarrow\dfrac{3x+2022+2x-2021}{\left(2x-2021\right)\left(3x+2022\right)}=\dfrac{10x-2024-\left(15x-2023\right)}{\left(15x-2023\right)\left(10x-2024\right)}\)

\(\Leftrightarrow\dfrac{5x-1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{5x-1}{\left(15x-2023\right)\left(10x-2024\right)}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\Rightarrow x=...\\\dfrac{1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{1}{\left(15x-2023\right)\left(10x-2024\right)}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(2x-2021\right)\left(3x+2022\right)+\left(15x-2023\right)\left(10x-2024\right)=0\)

\(\Leftrightarrow\left[12x-4045-\left(10x-2024\right)\right]\left(3x+2022\right)+\left(12x-4045+3x+2022\right)\left(10x-2024\right)=0\)

\(\Leftrightarrow\left(12x-4045\right)\left(3x+2022\right)-\left(10x-2024\right)\left(3x+2022\right)+\left(12x-4045\right)\left(10x-2024\right)+\left(3x+2022\right)\left(10x-2024\right)=0\)

\(\Leftrightarrow\left(12x-4045\right)\left(13x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{13}\\x=\dfrac{4045}{12}\end{matrix}\right.\)

1,

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

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

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

2.

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

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

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

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

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

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

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

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

b,     B        =                       \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\)  + \(\dfrac{1}{2^3}\) -   \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)

2 \(\times\)  B       =                 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) -  \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)

2 \(\times\) B + B  =                1  -  \(\dfrac{1}{2^{100}}\)

3B             =              ( 1 - \(\dfrac{1}{2^{100}}\)) 

             B =               ( 1 - \(\dfrac{1}{2^{100}}\)) : 3

       A              =          1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\) 

A\(\times\)  3             =   3 +  1 + \(\dfrac{1}{3}\) +  \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+  \(\dfrac{1}{3^{n-1}}\) 

A \(\times\) 3 - A        = 3 - \(\dfrac{1}{3^n}\)

       2A           = 3  - \(\dfrac{1}{3^n}\)

         A           = ( 3 - \(\dfrac{1}{3^n}\)) : 2

C = \(\dfrac{3}{2^2}\) \(\times\) \(\dfrac{8}{3^2}\) \(\times\) \(\dfrac{15}{4^2}\) \(\times\) ...........\(\times\) \(\dfrac{899}{30^2}\)

C = \(\dfrac{1\times3}{2^2}\) \(\times\) \(\dfrac{2\times4}{3^2}\) \(\times\) \(\dfrac{3\times5}{4^2}\) \(\times\)........\(\times\) \(\dfrac{29\times31}{30^2}\)

C = \(\dfrac{1\times2\times\left(3\times4\times5\times....\times29\right)^2\times30\times31}{2^2\times\left(3\times4\times5\times.......\times29\right)^2\times30^2}\)

C =  \(\dfrac{2\times\left(3\times4\times5\times.....\times29\right)^2\times30}{2\times\left(3\times4\times5\times.....\times29\right)^2\times30}\) \(\times\) \(\dfrac{1\times31}{2\times30}\)

C = 1 \(\times\) \(\dfrac{31}{60}\)

C = \(\dfrac{31}{60}\)

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

Bạn ghi lại đề đi bạn. Khó hiểu quá!

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)

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

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

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

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

\(=a-2\sqrt{a}\)

ĐKXĐ: a > 0 và a khác 1

\(P=\dfrac{2\left(a^2+2\right)}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1-a\right)\left(1+a+a^2\right)}\)\(=\dfrac{2a^2+4-\left(1+a+a^2\right).\left(1-\sqrt{a}+1+\sqrt{a}\right)}{1-a^3}\)

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

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