\(A=\dfrac{3x}{x-2}\cdot\sqrt{x^2-4x+4}\)

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

=3x

\(B=\dfrac{-5y}{x+3}\cdot\sqrt{x^2+6x+9}\)

\(=\dfrac{-5y}{x+3}\cdot\left|x+3\right|\)

\(=\pm5y\)

\(A=\frac{x^2+x+1-\frac{3}{4}x^2-\frac{3}{2}-\frac{3}{4}+\frac{3}{4}\left(x^2+2x+1\right)}{x^2+2x+1}=\frac{\frac{1}{4}\left(x^2-2x+1\right)+\frac{3}{4}\left(x^2+2x+1\right)}{x^2+2x+1}\)

    \(=\frac{1}{4}.\frac{\left(x-1\right)^2}{\left(x+1\right)^2}+\frac{3}{4}\ge\frac{3}{4}\)

Vậy GTNN cùa A là \(\frac{3}{4}khix=1\)

Ta có:

\(B=\frac{x^4+x^2+5-\frac{19}{20}x^4-\frac{19}{10}x-\frac{19}{20}+\frac{19}{20}\left(x^4+2x^2+1\right)}{x^4+2x^2+1}=\frac{\frac{1}{20}\left(x^4-18x^2+81\right)+\frac{19}{20}\left(x^4+2x^2+1\right)}{x^4+2x^2+1}\)

    \(=\frac{1}{20}.\frac{\left(x^2-9\right)^2}{\left(x^2+1\right)^2}+\frac{19}{20}\ge\frac{19}{20}\)

Vậy GTLN của B là 19/20 khi x = -3 hoăc x = 3.

Ai tìm được GTLN nào?

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: ...

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

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

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

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

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

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

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

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

Bài 1:

a, \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-1}{\sqrt{x}+3}\)

b, \(6-2x-\sqrt{9-6x+x^2}\)

\(=6-2x-\sqrt{\left(3-x\right)^2}\)

\(=6-2x-3+x\left(x< 3\right)\)

\(=3-x\)

Bài 2:

\(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\sqrt{\left(1-6x\right)^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

+) Xét \(x\ge\dfrac{1}{6}\) có:
\(6x-1=5\Leftrightarrow x=1\)

+) Xét \(x< \dfrac{1}{6}\) có:

\(1-6x=5\)

\(\Leftrightarrow x=\dfrac{-2}{3}\)

Vậy \(\left[{}\begin{matrix}x=1\\x=\dfrac{-2}{3}\end{matrix}\right.\)

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

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

1) Thay x=16 vào biểu thức ta có:

 \(A=\dfrac{\sqrt{x}}{\sqrt{x+3}}=\dfrac{\sqrt{16}}{\sqrt{16}+3}=\dfrac{4}{4+3}=\dfrac{4}{7}\)

2) \(A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\\ \Rightarrow A+B=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\Rightarrow A+B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\Rightarrow A+B=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\Rightarrow A+B=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\Rightarrow A+B=\dfrac{3}{\sqrt{x}+3}\)

1: Thay x=16 vào A, ta được:

\(A=\dfrac{4}{4+3}=\dfrac{4}{7}\)

a: \(A=\sqrt{x}+\dfrac{\sqrt{x}\left(1+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\sqrt{x}+\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

Khi x=4 thì \(A=2+\dfrac{2\cdot2+1}{2+1}=2+\dfrac{5}{3}=\dfrac{11}{3}\)

b: Khi x=(2-căn 3)^2 thì \(A=2-\sqrt{3}+\dfrac{2\left(2-\sqrt{3}\right)+1}{2-\sqrt{3}+1}\)

\(=2-\sqrt{3}+\dfrac{4-2\sqrt{3}+1}{3-\sqrt{3}}\)

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

\(=\dfrac{\left(2-\sqrt{3}\right)\left(3-\sqrt{3}\right)+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{6-2\sqrt{3}-3\sqrt{3}+3+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{14-7\sqrt{3}}{3-\sqrt{3}}\)

d: A=2

=>\(\dfrac{x+\sqrt{x}+2\sqrt{x}+1}{\sqrt{x}+1}=2\)

=>\(x+3\sqrt{x}+1=2\left(\sqrt{x}+1\right)=2\sqrt{x}+2\)

=>\(x+\sqrt{x}-1=0\)

=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{-1+\sqrt{5}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{-1-\sqrt{5}}{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{6-2\sqrt{5}}{4}=\dfrac{3-\sqrt{5}}{2}\)

\(A=\dfrac{1}{3}x+x-\dfrac{4}{3}x=0\)

Do đó: Khi x=2011/2012 thì A=0