Lời giải:

Áp dụng BĐT AM-GM:

$A=x(1-x^2)=x(1-x)(1+x)=(1+\sqrt{3})x.(2+\sqrt{3})(1-x)(1+x).\frac{1}{(\sqrt{3}+1)(\sqrt{3}+2)}$

$=(x+x\sqrt{3})[2+\sqrt{3}-(2+\sqrt{3})x](1+x).\frac{1}{(\sqrt{3}+1)(\sqrt{3}+2)}$

\(\leq \left[\frac{x+x\sqrt{3}+2+\sqrt{3}-(2+\sqrt{3})x+1+x}{3}\right]^3.\frac{1}{(1+\sqrt{3})(\sqrt{3}+2)}\\ =\frac{(\sqrt{3}+1)^3}{3\sqrt{3}}.\frac{1}{(\sqrt{3}+1)(\sqrt{3}+2)}=\frac{2}{3\sqrt{3}}\)

Vậy $A_{\max}=\frac{2}{3\sqrt{3}}$. Giá trị này đạt tại $x=\frac{1}{\sqrt{3}}$

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc hiểu đề và hỗ trợ bạn nhanh hơn nhé.

a, Với \(x\ge0;x\ne1\):

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

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

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

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

b, Thay \(x=7-4\sqrt{3}\) vào P, ta được:

\(P=\sqrt{7-4\sqrt{3}}-\left(7-4\sqrt{3}\right)\)

\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.2+2^2}+4\sqrt{3}-7\)

\(=\sqrt{\left(\sqrt{3}-2\right)^2}+4\sqrt{3}-7\)

\(=\left|\sqrt{3}-2\right|+4\sqrt{3}-7\)

\(=2-\sqrt{3}+4\sqrt{3}-7\) (vì \(\sqrt{3}< 2\))

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

$Toru$

a) \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\left(x\ge0,x\ne1\right)\\ =\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(x-1\right)^2}{2}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\left(x-1\right)^2}{2}\\ \)

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

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

b) \(x=7-4\sqrt{3}\left(TMDK\right)\)

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

Thay vào biểu thức P, ta được:

\(P=-\left(7-4\sqrt{3}\right)+2-\sqrt{3}=-5+3\sqrt{3}\)

Phương trình không có dấu = thì sao giải đc bạn ? 

Đề sai kìa 

\(B=\dfrac{x-1}{\sqrt{x}-1}-\dfrac{x+3\sqrt{x}}{\sqrt{x}}\)

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

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

\(B=\dfrac{x-1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}+x}{\sqrt{x}}\left(x>0,x\ne1\right)\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(3+\sqrt{x}\right)}{\sqrt{x}}\\ =\sqrt{x}+1-\left(\sqrt{x}+3\right)\\ =\sqrt{x}+1-\sqrt{x}-3\\ =-2\)

\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot\left(-m\right)=4m+36\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>4m+36>0

=>m>-9

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=6\\x_1x_2=\dfrac{c}{a}=-m\end{matrix}\right.\)

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=36+4m\)

=>\(x_1-x_2=\pm2\sqrt{m+9}\)

\(x_1^2-x_2^2=24\)

=>\(\left(x_1+x_2\right)\left(x_1-x_2\right)=24\)

=>\(\pm2\sqrt{m+9}=4\)

=>\(\pm\sqrt{m+9}=2\)

=>\(\left[{}\begin{matrix}\sqrt{m+9}=2\left(nhận\right)\\\sqrt{m+9}=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow m+9=4\)

=>m=-5(nhận)

\(\left\{{}\begin{matrix}x^2+y^2+2x=4\\x+2y+xy=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+2x=4\\2x+4y+2xy=8\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+2x+2x+4y+2xy=12\)

\(\Leftrightarrow\left(x+y\right)^2+4\left(x+y\right)-12=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=2\\x+y=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2-x\\y=-6-x\end{matrix}\right.\)

TH1: \(y=2-x\). Thế vào pt thứ 2 của hệ, ta có:

\(x+2\left(2-x\right)+x\left(2-x\right)=4\)

\(\Leftrightarrow x+4-2x+2x-x^2-4=0\)

\(\Leftrightarrow x-x^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=2\\x=1\Rightarrow y=1\end{matrix}\right.\)

TH2: \(y=-x-6\). Thay vào pt thứ 2 của hệ, ta có:

\(x+2\left(-x-6\right)+x\left(-x-6\right)=4\)

\(\Leftrightarrow x-2x-12-x^2-6x-4=0\)

\(\Leftrightarrow-x^2-7x-16=0\) (vô nghiệm vì \(-x^2-7x-16< 0\) với mọi \(x\))

Vậy hệ phương trình đã cho có 2 cặp nghiệm \(\left(x;y\right)\) là \(\left(1;1\right)\) và \(\left(0;2\right)\)

Lời giải:
Áp dụng BĐT Bunhiacopxky:

$(\sqrt{2a+b}+\sqrt{2b+c}+\sqrt{2c+a})^2\leq [(2a+b)+(2b+c)+(2c+a)](1+1+1)=3(a+b+c).3=9(a+b+c)=81$

$\Rightarrow \sqrt{2a+b}+\sqrt{2b+c}+\sqrt{2c+a}\leq 9$

Vậy ta có đpcm

Dấu "=" xảy ra khi $a=b=c=3$

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

\(\Leftrightarrow\left(\dfrac{1}{2}x+\dfrac{1}{2}\right)\left(3-x\right)-\left(3-x\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(\dfrac{1}{2}x+\dfrac{1}{2}-1\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(\dfrac{1}{2}x-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(3-x\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy phương trình đã cho có tập nghiệm là \(S=\left\{3;1\right\}\).

$Toru$

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

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

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

=>\(-x^2+2x+3+2x=6\)

=>\(-x^2+4x-3=0\)

=>\(\left(x-1\right)\left(x-3\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Hình vẽ đâu bạn ơi?