Câu 2:

ĐKXĐ: x<>0

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

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

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

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

\(=-\left(\dfrac{1}{x}+\dfrac{1}{2}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}\forall x< >0\)

Dấu '=' xảy ra khi 1/x+1/2=0

=>1/x=-1/2

=>x=-2

Bài `1`

\(\sqrt{4-2\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{\sqrt{3}-3}{\sqrt{3}-1}\\ =\sqrt{3-2\sqrt{3}+1}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}+1-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}+1-\sqrt{3}\\ =-\sqrt{3}\)

2:

a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

b: B=5

=>\(5\left(\sqrt{x}+3\right)=\sqrt{x}+8\)

=>\(5\sqrt{x}+15=\sqrt{x}+8\)

=>\(4\sqrt{x}=-7\)(loại)

Vậy: \(x\in\varnothing\)

\(\sqrt{36}+\sqrt{9}-\sqrt{49}\)

\(=6+3-7\)

\(=2\)

\(\sqrt{2}\cdot\left(\sqrt{50}-3\sqrt{2}\right)\)

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

\(=\sqrt{2}\cdot2\sqrt{2}\)

\(=4\)

a) \(T=\sqrt{36}+\sqrt{9}-\sqrt{49}\)

    \(=6+3-7\)

      \(=2\)

b)  \(B=\sqrt{2\left(\sqrt{50}-3\sqrt{2}\right)}\)

         \(=\sqrt{10\sqrt{2}-6\sqrt{2}}\)

           \(=\sqrt{\left(10-6\right)\sqrt{2}}\)

           \(=\sqrt{4\sqrt{2}}\)

            \(\approx2,39\)

\(=\dfrac{x+5}{2x}+\dfrac{x-6}{x-5}-\dfrac{3x^2-2x-9}{2x\left(x-5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x-5\right)+2x\left(x-6\right)-3x^2+2x+9}{2x\left(x-5\right)}\)

\(=\dfrac{x^2-25+2x^2-12x-3x^2+2x+9}{2x\left(x-5\right)}\)

\(=\dfrac{-10x-16}{2x\left(x-5\right)}=\dfrac{-5x-8}{x\left(x-5\right)}\)

2:

a: x+201=351

=>x=351-201

=>x=150

b: \(8\cdot5^2-27:25\)

\(=8\cdot25-\dfrac{27}{25}\)

\(=200-1,08=198,92\)

d: \(2023-23:\left[9+2\left(2^3-0,21\right)\right]\)

\(=2023-23:\left[9+16-0,42\right]\)
\(=2023-\dfrac{23}{25-0,42}\)

\(=2023-\dfrac{1150}{1229}=\dfrac{2485117}{1229}\)

b: 2(x-21)=84

=>x-21=84/2=42

=>x=42+21=63

c: 135-4(81-x)=55

=>4(81-x)=135-55=80

=>81-x=20

=>x=61

Ta có: \(2\left(x-1\right)^2+3\ge3\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=1\)

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

Dấu "=" xảy ra \(\Leftrightarrow x=1\)

Vậy \(B_{max}=\dfrac{1}{3}\Leftrightarrow x=1\)

a. Ta có: \(3x2xy-\frac{2}{3}x^2y-4x^2.\frac{1}{3}y=6x^2y-\frac{4}{3}x^2y=\left(6-\frac{2}{3}-\frac{4}{3}\right)x^2y=4x^2y.\)

b. Thay \(x=-2,y=\frac{1}{8}\)vào đơn thức \(4x^2y\), ta được: \(4x^2y=4\left(-2\right)^2.\frac{1}{8}=2\).

Vậy, giá trị của biểu thức \(x=-2,y=\frac{1}{8}\rightarrow=2\) 

Ta có: 1 b − 2 − 1 b = 2 b ( b − 2 )  

Áp dụng tính được  B = 2018 b ( b − 2018 )