\((\frac{\sqrt{x}+1}{2\sqrt{x}-2}-\frac{\sqrt{x}-1}{2\sqrt{x}+2}-\frac{\sqrt{x}+1}{1-x})\div\frac{x+2\sqrt{x}}{x+\sqrt{x}}\)

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

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

Khai triển ra nhé, mk làm như trên thì lâu lắm nên bn tự lm nhé

\(=\frac{2\left(3\sqrt{x}+1\right)}{2(x-1)}\times\frac{\sqrt{x}+1}{\sqrt{x}+2}\)

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

come mon

\(x^3+2x^2+2x+3=\left(3x+1\right)\sqrt{x^3+3}\)(ĐKXĐ: \(x\ge-\sqrt[3]{3}\))

Đặt \(\sqrt{x^3+3}=y\)(\(y\ge0\)) \(\Leftrightarrow x^3+3=y^2\). Khi đó pt cho mang dạng:

\(y^2+2x^2+2x=\left(3x+1\right)y\)

\(\Leftrightarrow y^2-3xy+2x^2+2x-y=0\)

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

\(\Leftrightarrow y\left(y-x\right)-2x\left(y-x\right)-\left(y-2x\right)=0\)

\(\Leftrightarrow\left(y-x\right)\left(y-2x\right)-\left(y-2x\right)=0\)

\(\Leftrightarrow\left(y-2x\right)\left(y-x-1\right)=0\Leftrightarrow\orbr{\begin{cases}y-2x=0\\y-x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=2x\\y=x+1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x^3+3}=2x\left(1\right)\\\sqrt{x^3+3}=x+1\left(2\right)\end{cases}}\)

Ta có: \(\left(1\right)\Leftrightarrow\hept{\begin{cases}x\ge0\\x^3+3=4x^2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x^3-x^2-3x^2+3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2\left(x-1\right)-3\left(x-1\right)\left(x+1\right)=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge0\\\left(x-1\right)\left(x^2-3x-3\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\\left(x-1\right)\left[\left(x-\frac{3}{2}\right)^2-\frac{21}{4}\right]=0\end{cases}}\)\(\Leftrightarrow x=1\) hoặc \(\orbr{\begin{cases}x=\frac{\sqrt{21}+3}{2}\\x=\frac{3-\sqrt{21}}{2}\end{cases}}\) 

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{\sqrt{21}+3}{2}\end{cases}}\) (loại TH \(x=\frac{3-\sqrt{21}}{2}< 0\))

Lại có: \(\left(2\right)\Leftrightarrow\hept{\begin{cases}x+1\ge0\\x^3+3=x^2+2x+1\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x^3-x^2-2x+2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-1\\x^2\left(x-1\right)-2\left(x-1\right)=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge-1\\\left(x-1\right)\left(x^2-2\right)=0\end{cases}}\)

\(\Leftrightarrow x=1\) hoặc \(x=\pm\sqrt{2}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=\sqrt{2}\end{cases}}\)(t/m ĐK) (loại TH \(x=-\sqrt{2}\) vì \(-\sqrt{2}< -1\))

Vậy tập nghiệm của pt là \(S=\left\{1;\sqrt{2};\frac{\sqrt{21}+3}{2}\right\}.\)

ĐKXĐ: \(x>0\)

Ta có:

\(-\sqrt{x}-2\left(x-\frac{1}{x}\right)=\frac{1}{2x^3}-\frac{1}{2x\sqrt{x}}\)

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

\(\frac{\Leftrightarrow1}{2x\sqrt{x}}-\sqrt{x}=2\left(x-\frac{1}{x}+\frac{1}{4x^3}\right)\)

Đặt : \(\frac{1}{2x\sqrt{x}}-\sqrt{x}=a\Rightarrow a^2=x-\frac{1}{x}+\frac{1}{4x^3}\)

Khi đó pt đã cho trở thành:

\(a=2a^2\Leftrightarrow\orbr{\begin{cases}a=0\\a=\frac{1}{2}\end{cases}}\)

+) a = 0\(\Rightarrow x=\frac{1}{\sqrt{2}}\)

Tương tự

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)ĐKXĐ : \(x>1\)

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

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

\(A=\frac{x+2}{\sqrt{x}}\)

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

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

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

\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\)

\(b)\) Áp dụng Cosi với hai số dương ta có : 

\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{2}{\sqrt{x}}}=2\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x}=\frac{2}{\sqrt{x}}\)

\(\Leftrightarrow\)\(x=2\)

Vậy GTNN của \(A\) là \(2\sqrt{2}\) khi \(x=2\)

Chúc bạn học tốt ~ 

PS : mới lớp 8 ko chắc nhé :v 

Ta có:

\(B=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2\)

\(=4\left(x^2+xy+xz\right)\left(x^2+xy+yz+xz\right)+y^2z^2\)

Đặt \(x^2+xy+xz=a\)

Khi đó: B trở thành:

\(4a\left(a+yz\right)+y^2z^2\)

\(=\left(yz+2a\right)^2\)

Hay \(B=\left(2x^2+2xy+2xz+yz\right)^2\)là số chính phương

ap dung bdt co si ta co:\(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}>=3\sqrt[3]{xyz}\)

=>\(3>=3\sqrt[3]{xyz}\)

=>\(1>=\sqrt[3]{xyz}\)

=>\(1>=xyz\)

dau bang xay ra khi \(\frac{xy}{z}=\frac{yz}{x}=\frac{xz}{y}\)=>x=y=z=1

vay x=y=z=1

Áp dụng hệ thức giữa cạnh và góc, ta có:

\(HB=AH.cotgB;HC=AH.cotgC\)

\(\Rightarrow AH=\frac{HB+HC}{cotgB+cotgC}=\frac{20}{cotg40^o+cotg30^o}\approx6,84\left(cm\right)\)

Lại áp dụng hệ thức giữa cạnh và góc, ta có:

\(AB=\frac{AH}{sinB}\approx\frac{6,84}{sin40^o}\approx10,64\left(cm\right)\)

\(AC=\frac{AH}{sinC}\approx\frac{6,84}{sin30^o}=13,68\left(cm\right)\)