ĐKXĐ: \(x\ge0;x\ne1;\)

\(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\)

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

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

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

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

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

\(=\frac{\sqrt{x}\left(x-\sqrt{x}-4\right)\left(\sqrt{x}-1\right)}{2}\)

Ta có: \(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\times\frac{\left(1-x\right)^2}{2}\) 

\(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(1-x\right)^2}{2}\) 

\(P=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(1-x\right)^2}{2}\) 

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

\(P=\frac{\sqrt{x}\left(x-4-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{2}\)

ĐKXĐ : \(x^3+x^2+6\ge0\)

\(pt\Leftrightarrow x^2+x+9=6+x^2+x^3\)

    \(\Leftrightarrow x^3-x-3=0\)

Đến đây có lẽ dùng công thức Cardano là ra , nhưng mà không biết bạn học Cardano chưa nhỉ ?

Incursion_03Cardano mình học rồi

M đạt max khi \(\frac{1}{M}\) đạt min

\(\frac{1}{M}=1-\frac{1}{x}+\frac{1}{x^2}=\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}+\frac{3}{4}\)

\(\frac{1}{M}\ge\frac{3}{4}\Rightarrow M\le\frac{4}{3}\)

dấu = xảy ra khi x=1/2

bạn trả lời rõ hơn dc ko bạn

a) \(\sqrt{a}+1>\sqrt{a+1}\)\(\Leftrightarrow\)\(a+2\sqrt{a}+1>a+1\)\(\Leftrightarrow\)\(2\sqrt{a}>0\)( luôn đúng \(\forall x>0\) ) 

b) \(a-1< a\)\(\Leftrightarrow\)\(\sqrt{a-1}< \sqrt{a}\)

c) \(\left(\sqrt{6}-1\right)^2=6-2\sqrt{6}+1>3-2\sqrt{3.2}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)

do \(\sqrt{6}-1>0;\sqrt{3}-\sqrt{2}>0\) nên \(\sqrt{6}-1>\sqrt{3}-\sqrt{2}\) ( đpcm ) 

Ta có \(0< 3y^2+1< 4y^2+4\)

=> \(y^4< y^4+3y^4+1< \left(y^2+2\right)^2\)

=> \(y^4< x^4< \left(y^2+2\right)^2\)

Mà x,y nguyên

=> \(x^2=y^2+1\)

=> \(y^4+2y^2+1=y^4+3y^2+1\)

=> \(y=0\)=> x=0

Vậy (x,y)=(0;0)

MN ƠI GIÚP E VỚI

MAI E ĐI HOK RỒI

a,b,c∈[0,1]⇒b≥b²;c≥c³

Ta có:

a,b,c∈[0,1]⇒(1−a)(1−b)(1−c)≥0

⇔1−a−b−c+ab+bc+ca−abc≥0

⇔a+b+c−ab−bc−ca+abc≤1

⇒a+b²+c³−ab−bc−ca≤1

⇒đpcm

Dấu "=" xảy ra khi trong a,b,ccó 1 số bằng 1, 1 số bằng 0, số còn lại là 1 hoặc 0