1: Để A>0 thì x-1<0

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

1) Để A > 0 thì:

\(x-1< 0\Leftrightarrow x< 1\)

\(\Rightarrow0\le x< 1\) và \(x\ne1\)

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

Để A<1 thì \(\dfrac{2}{\sqrt{x}-1}< 0\)

\(\Rightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)

Mà x\(\ge0,x\ne1\)

\(\Rightarrow0\le x< 1\)

Bài 2: 

Để A<1 thì A-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

a: ĐKXĐ: x<>0; x<>1

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

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

b: |2x+1|=3

=>x=1(loại); x=-2(nhận)

Khi x=-2 thì P=4/-3=-4/3

c: P=-1/2

=>x^2/x-1=-1/2

=>2x^2=-x+1

=>2x^2+x-1=0

=>2x^2+2x-x-1=0

=>(x+1)(2x-1)=0

=>x=1/2; x=-1

a: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{\sqrt{x}-1}\)

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

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

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

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

a) A xác định \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

b) Để \(A=-\dfrac{1}{2}\)

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

\(\Leftrightarrow2x^2=-\left(x+1\right)\)

\(\Leftrightarrow2x^2+x+1=0\)

\(\Delta=1-8=-7< 0\)

Nên phương trình trên vô nghiệm \(\left(x\in\varnothing\right)\)

c) Để \(A< 1\) 

\(\Leftrightarrow\dfrac{x^2}{x+1}< 1\)

\(\Leftrightarrow x^2< x+1\left(x\ne-1\right)\)

\(\Leftrightarrow x^2-x-1< 0\)

\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1< 0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}< 0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2< \dfrac{5}{4}\)

\(\Leftrightarrow-\dfrac{\sqrt[]{5}}{2}< x-\dfrac{1}{2}< \dfrac{\sqrt[]{5}}{2}\)

\(\Leftrightarrow\dfrac{-\sqrt[]{5}+1}{2}< x< \dfrac{\sqrt[]{5}+1}{2}\)

d) Để A nguyên

\(\Leftrightarrow\dfrac{x^2}{x+1}\in Z\)

\(\Leftrightarrow x^2⋮x+1\)

\(\Leftrightarrow x^2-x\left(x+1\right)⋮x+1\)

\(\Leftrightarrow x^2-x^2+x⋮x+1\)

\(\Leftrightarrow x⋮x+1\)

\(\Leftrightarrow x-x-1⋮x+1\)

\(\Leftrightarrow-1⋮x+1\)

\(\Leftrightarrow x+1\in\left\{-1;1\right\}\)

\(\Leftrightarrow x\in\left\{-2;0\right\}\left(x\in Z\right)\)

!ERROR 404!

d. Áp dụng BĐT Caushy Schwartz ta có:

\(x+y+\dfrac{1}{x}+\dfrac{1}{y}\le x+y+\dfrac{\left(1+1\right)^2}{x+y}=x+y+\dfrac{4}{x+y}\le1+\dfrac{4}{1}=5\)

-Dấu bằng xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

c. Bạn kiểm tra lại đề nhé.

b. \(5x\left(2-x\right)=-5x\left(x-2\right)=-5\left(x^2-2x\right)=-5\left(x^2-2x+1-1\right)=-5\left(x-1\right)^2+5\le5\)-Dấu bằng xảy ra \(\Leftrightarrow x=1\)

a.

\(\left(80-2x\right)\left(50-2x\right)x=\dfrac{2}{3}\left(40-x\right)\left(50-2x\right)3x\le\dfrac{2}{3}\left(\dfrac{40-x+50-2x+3x}{3}\right)^3=18000\)

Dấu "=" xảy ra khi \(40-x=50-2x=3x\Leftrightarrow x=10\)

b.

\(5x\left(2-x\right)=5.x\left(2-x\right)\le\dfrac{5}{4}\left(x+2-x\right)^2=5\)

Dấu "=" xảy ra khi \(x=2-x\Rightarrow x=1\)

c.

Biểu thức này chỉ có min, ko có max

d.

\(x+y\le1\Rightarrow-\left(x+y\right)\ge-1\)

\(x+y+\dfrac{1}{x}+\dfrac{1}{y}=\left(4x+\dfrac{1}{x}\right)+\left(4y+\dfrac{1}{y}\right)-3\left(x+y\right)\ge2\sqrt{\dfrac{4x}{x}}+2\sqrt{\dfrac{4y}{y}}-3.1=5\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

9.

\(A>1\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}-1}>1\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-1>0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2-\sqrt{x}+1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\dfrac{-1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp với điều kiện giả thiết.

10.

\(P< 1\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}< 1\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-1< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp với điều kiện giả thiết.

Bài 9:

Để A>1 thì A-1>0

\(\Leftrightarrow\dfrac{\sqrt{x}-2-\sqrt{x}+1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\dfrac{-1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

Bài 10:

Để P<1 thì P-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

a: \(A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}=\dfrac{2x-3\sqrt{x}+1}{x-1}\)

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

b: Để A nguyên thì \(2\sqrt{x}+2-3⋮\sqrt{x}+1\)

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

=>x=0 hoặc x=4

c: Để A<1 thì A-1<0

=>\(\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\)

=>căn x-2<0

=>0<=x<4

a: \(A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}=\dfrac{2x-3\sqrt{x}+1}{x-1}\)

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

b: Để A nguyên thì \(2\sqrt{x}+2-3⋮\sqrt{x}+1\)

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

=>x=0 hoặc x=4

c: Để A<1 thì A-1<0

=>\(\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\)

=>căn x-2<0

=>0<=x<4