\(\left(a\right):2x-7\sqrt{x}+3=0\left(x\ge0\right)\\ < =>\left(2x-6\sqrt{x}\right)-\left(\sqrt{x}-3\right)=0\\ < =>2\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\\ < =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ =>\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(TM\right)\\x=9\left(TM\right)\end{matrix}\right.\)

\(\left(b\right):3\sqrt{x}+5< 6\\ < =>3\sqrt{x}< 1\\ < =>\sqrt{x}< \dfrac{1}{3}\\ < =>0\le x< \dfrac{1}{9}\)

\(\left(c\right):x-3\sqrt{x}-10< 0\\ < =>\left(x-5\sqrt{x}\right)+\left(2\sqrt{x}-10\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}-5\right)+2\left(\sqrt{x}-5\right)< 0\\ < =>\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}+2>0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}0\le x< 25\\x\ge0\end{matrix}\right.< =>0\le x< 25\)

\(\left(d\right):x-5\sqrt{x}+6=0\left(x\ge0\right)\\ < =>\left(x-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)=0\\ < =>\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\\ < =>\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\left(TM\right)\)

\(\left(e\right):x+5\sqrt{x}-14< 0\\ < =>\left(x+7\sqrt{x}\right)-\left(2\sqrt{x}+14\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)< 0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}+7>0\\\sqrt{x}-2< 0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.< =>0\le x< 4\)

Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)

\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

Bạn đăng từng câu 1 nhé

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

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

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

b: \(=\dfrac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)

\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)

a: Thay \(x=6-2\sqrt{5}\) vào A, ta được:

\(A=1-\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=1-\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\)

b: Ta có: P=A:B

\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{x-5\sqrt{x}+6}\right)\)

\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

\(a.x=3-2\sqrt{2}\\ \Rightarrow\sqrt{x}=\sqrt{3-2\sqrt{2}}\\ =\sqrt{2-2\sqrt{2}+1}\\ =\sqrt{\left(\sqrt{2}-1\right)^2}\\ =\left|\sqrt{2}-1\right|\\ =\sqrt{2}-1\left(vì\sqrt{2}>1\right)\)

Thay \(\sqrt{x}=\sqrt{2}-1\) vào A ta được

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

\(b.B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\\ B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ B=\dfrac{x-3\sqrt{x}-\sqrt{x}+3-x+4-10+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{1}{\sqrt{x}-2}\)

\(c,P=A:B\\ P=\dfrac{\sqrt{x}}{1+\sqrt{x}}:\dfrac{1}{\sqrt{x}-2}\\ P=\dfrac{x-2\sqrt{x}}{1+\sqrt{x}}\)

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

Có: \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}+1\ge1\left(I\right)\)

Lại có: \(\sqrt{x}\ge0\)

\(\Rightarrow-\sqrt{x}\le0\\ \Rightarrow-\sqrt{x}+2\le2\)

mà \(-\sqrt{x}\le0\)

\(\Rightarrow-\sqrt{x}\left(-\sqrt{x}+2\right)\ge2\)

Kết hợp với \(\left(I\right)\) \(\Rightarrow\) \(P=\dfrac{-\sqrt{x}\left(-\sqrt{x}+2\right)}{\sqrt{x}+1}\ge2\)

Vậy gtnn của P = \(2\) khi \(x=10+4\sqrt{6}\)

a: Khi \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\) thì 

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

b: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)

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

Xét VT ta có : \(x\ge0\)

mà \(\sqrt{x};\sqrt{x+3};\sqrt{x+2},x\ge0\forall x\)

=> VT = B+ 10 >0

=> \(VT\ne VT\)

=> pt vô nghiệm

a.

ĐKXĐ: \(x\ge3\)

(Tốt nhất bạn kiểm tra lại đề cái căn đầu tiên của \(\sqrt{x-3}\) là căn bậc 2 hay căn bậc 3). Vì nhìn ĐKXĐ thì thấy căn bậc 2 là không hợp lý rồi đó

Pt tương đương:

\(\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)=0\)

Do \(x\ge3\Rightarrow x-2>0\Rightarrow\left(x+1\right)\left(x-2\right)>0\)

\(\Rightarrow\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)>0\)

Pt vô nghiệm

b.

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

Pt: \(2x+3-\sqrt{2x+3}-\left(4x^2-6x+2\right)=0\)

Đặt \(\sqrt{2x+3}=t\ge0\) ta được:

\(t^2-t-\left(4x^2-6x+2\right)=0\)

\(\Delta=1+4\left(4x^2-6x+2\right)=\left(4x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t_1=\dfrac{1+4x-3}{2}=2x-1\\t_2=\dfrac{1-4x+3}{2}=2-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=2x-1\left(x\ge\dfrac{1}{2}\right)\\\sqrt{2x+3}=2-2x\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4x^2-4x+1\left(x\ge\dfrac{1}{2}\right)\\2x+3=4x^2-8x+4\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{17}}{4}\\x=\dfrac{5-\sqrt{21}}{4}\end{matrix}\right.\)

c.

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow2\left(x^2-4x+4\right)+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}=0\)

\(\Leftrightarrow2\left(x-2\right)^2-5\left(x-2\right)\sqrt{x+1}+2\left(x+1\right)=0\)

Đặt \(\left\{{}\begin{matrix}x-2=a\\\sqrt{x+1}=b\end{matrix}\right.\) ta được:

\(2a^2-5ab+2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=x-2\left(x\ge2\right)\\\sqrt{x+1}=2x-4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-4x+4\\x+1=4x^2-16x+16\end{matrix}\right.\) (\(x\ge2\))

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\) (đã loại nghiệm)

f: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=2\sqrt{6}+2\sqrt{5}\)

l: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\sqrt{10}-\sqrt{7}\)

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

Đk:\(x\ge2\)

\(pt\Leftrightarrow\left(\sqrt{x+1}-2\right)-\left(\sqrt{x-2}-1\right)=0\)

\(\Leftrightarrow\dfrac{x+1-4}{\sqrt{x+1}+2}-\dfrac{x-2-1}{\sqrt{x-2}+1}=0\)

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

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

Dễ thấy:\(\dfrac{1}{\sqrt{x+1}+2}-\dfrac{1}{\sqrt{x-2}+1}< 0\)

Nên \(x-3=0\Rightarrow x=3\)

b)\(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)

Đk:\(x\ge1\)

\(pt\Leftrightarrow\sqrt{x-1}=\sqrt{5x-1}+\sqrt{3x-2}\)

\(\Leftrightarrow x-1=5x-1+3x-2+2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)

\(\Leftrightarrow2-7x=2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)

\(\Leftrightarrow49x^2-28x+4=4\left(5x-1\right)\left(3x-2\right)\)

\(\Leftrightarrow49x^2-28x+4=60x^2-52x+8\)

\(\Leftrightarrow-11x^2+24x-4=0\Leftrightarrow\left(2-x\right)\left(11x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{11}\end{matrix}\right.\) (loại hết)

c)\(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0\)

Đk:\(x\ge0\)

\(pt\Leftrightarrow\sqrt{x}-\left(\sqrt{x+1}+1\right)-\left(\sqrt{x+4}+2\right)+\left(\sqrt{x+9}-3\right)=0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{x+1-1}{\sqrt{x+1}+1}-\dfrac{x+4-4}{\sqrt{x+4}+2}+\dfrac{x+9-9}{\sqrt{x+9}-3}=0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{x}{\sqrt{x+1}+1}-\dfrac{x}{\sqrt{x+4}+2}+\dfrac{x}{\sqrt{x+9}-3}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}+1}-\dfrac{1}{\sqrt{x+4}+2}+\dfrac{1}{\sqrt{x+9}-3}\right)=0\)

Dễ thấy:\(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}+1}-\dfrac{1}{\sqrt{x+4}+2}+\dfrac{1}{\sqrt{x+9}-3}>0\)

Nên \(x=0\)