1

ĐK: \(x\ge1\)

Đặt \(t=\sqrt{x-1}\left(t\ge0\right)\Rightarrow x=t^2+1\)

Khi đó: 

\(x-2\sqrt{x-1}=16\)

\(\Leftrightarrow t^2-2t+1=16\\ \Leftrightarrow\left(t-1\right)^2=4^2\\ \Leftrightarrow t-1=4\\ \Leftrightarrow t=4+1=5\left(tm\right)\)

\(\Leftrightarrow\sqrt{x-1}=5\)

\(\Leftrightarrow x-1=5^2=25\\ \Leftrightarrow x=25+1=26\left(tm\right)\)

Vậy PT có nghiệm duy nhất x = 26.

2 ĐK: \(3\le x\le1\)

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

Từ điều kiện và bài giải ta kết luận PT vô nghiệm.

3 ĐK: \(x\ge4\)

\(\Leftrightarrow\sqrt{x-4}=7-2=5\\ \Leftrightarrow x-4=5^2=25\\ \Leftrightarrow x=25+4=29\left(tm\right)\)

Vậy PT có nghiệm duy nhất x = 29.

4

ĐK: \(x\ge1\)

Đặt \(t=\sqrt{x-1}\left(t\ge0\right)\Rightarrow x=t^2+1\)

Khi đó:

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

Trường hợp 1:

Với \(0\le t< 1\) thì:

\(\left(1\right)\Leftrightarrow t^2+1-\left(1-t\right)=0\\ \Leftrightarrow t^2+t=0\\ \Leftrightarrow t\left(t+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=0\Rightarrow\sqrt{x-1}=0\Rightarrow x=1\left(nhận\right)\\t=-1\left(loại\right)\end{matrix}\right.\)

Trường hợp 2:

Với \(t\ge1\) thì:

\(\left(1\right)\Leftrightarrow t^2+1-\left(t-1\right)=0\\ \Leftrightarrow t^2-t+2=0\)

\(\Delta=\left(-1\right)^2-4.2=-7< 0\)

=> Loại trường hợp 2.

Vậy PT có nghiệm duy nhất x = 1.

5

ĐK: \(x\ge2\)

Đặt \(\sqrt{x-2}=t\left(t\ge0\right)\Rightarrow x=t^2+2\)

Khi đó:

\(\sqrt{x-2}-\sqrt{x^2-2x}=0\\ \Leftrightarrow\sqrt{x-2}-\sqrt{x}.\sqrt{x-2}=0\\ \Leftrightarrow\sqrt{t^2+2-2}-\sqrt{t^2+2}.\sqrt{t^2+2-2}=0\\ \Leftrightarrow\sqrt{t^2}-\sqrt{t^2+2}.\sqrt{t^2}=0\\ \Leftrightarrow t-\sqrt{t^2+2}.t=0\\ \Leftrightarrow t\left(1-\sqrt{t^2+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=0\Rightarrow\sqrt{x-2}=0\Rightarrow x=2\left(tm\right)\\\sqrt{t^2+2}=1\Rightarrow t^2+2=1\Rightarrow t^2=-1\left(loại\right)\end{matrix}\right.\)

Vậy phương trình có nghiệm duy nhất x = 2.

6 Không có ĐK vì đưa về tổng bình lên luôn \(\ge0\)

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

Trường hợp 1:

Với \(x\ge-\sqrt{2}\) thì:

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

Với \(x< -\sqrt{2}\) thì:

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

Vậy phương trình có 2 nghiệm \(x=-1\) hoặc \(x=-1-2\sqrt{2}\)

c) \(\sqrt[]{8+\sqrt[]{x}}+\sqrt{5-\sqrt[]{x}}=5\)

\(\Leftrightarrow\left(\sqrt[]{8+\sqrt[]{x}}+\sqrt{5-\sqrt[]{x}}\right)^2=25\left(1\right)\left(đkxđ:0\le x\le25\right)\)

Áp dụng Bất đẳng thức Bunhiacopxki cho 2 cặp số dương \(\left(1;\sqrt[]{8+\sqrt[]{x}}\right);\left(1;\sqrt{5-\sqrt[]{x}}\right)\)

\(\left(1.\sqrt[]{8+\sqrt[]{x}}+1.\sqrt{5-\sqrt[]{x}}\right)^2\le\left(1^2+1^2\right)\left(8+\sqrt[]{x}+5-\sqrt[]{x}\right)=26\)

\(\left(1\right)\Leftrightarrow26=25\left(vô.lý\right)\)

Vậy phương trình đã cho vô nghiệm

b) \(\sqrt[]{1+4x}+2\sqrt[]{2-x}+2\sqrt[]{\left(1+4x\right)\left(2-x\right)}=3\)  \(\left(đkxđ:-\dfrac{1}{4}\le x\le2\right)\)

\(\)\(\Leftrightarrow\sqrt[]{1+4x}+2\sqrt[]{2-x}=3-2\sqrt[]{\left(1+4x\right)\left(2-x\right)}\)

\(\Leftrightarrow\left(\sqrt[]{1+4x}+2\sqrt[]{2-x}\right)^2=\left[3-2\sqrt[]{\left(1+4x\right)\left(2-x\right)}\right]^2\left(1\right)\)

Áp dụng Bất đẳng thức Bunhiacopxki :

\(\left(1.\sqrt[]{1+4x}+2\sqrt[]{2-x}\right)^2\le\left(1^2+2^2\right)\left(1+4x+2-x\right)=5\left(3x+3\right)\)

Áp dụng Bất đẳng thức Cauchy :

\(2\sqrt[]{\left(1+4x\right)\left(2-x\right)}\le1+4x+2-x=3x+3\)

Dấu "=" xảy ra khi và chỉ khi

\(1+4x=2-x\)

\(\Leftrightarrow x=\dfrac{1}{5}\left(thỏa.đk\right)\)

\(pt\left(1\right)\Leftrightarrow5\left(4x+3\right)=4x+3\)

\(\Leftrightarrow4\left(4x+3\right)=0\)

\(\Leftrightarrow x=-\dfrac{3}{4}\left(k.thỏa.x=\dfrac{1}{5}.vô.lý\right)\)

Vậy phương trình đã cho vô nghiệm

#)Hỏi j đi bn, bn ph hỏi cái j chứ làm lun rùi còn để cộng đồng ngắm ak ???

Bó cả tay lẫn chân !!! Bất lực như gặp cực hình !

Chắc là bạn ấy hỏi bạn ấy làm có đúng ko ha gì đó ?

\(ĐK:0< x\le4\)

Đặt \(\sqrt{2+\sqrt{x}}=a>0;\sqrt{2-\sqrt{x}}=b>0\)

\(\Rightarrow a^2+b^2=2+\sqrt{x}+2-\sqrt{x}=4\)

\(PT\Leftrightarrow\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\\ \Leftrightarrow\dfrac{a^2\sqrt{2}-a^2b+b^2\sqrt{2}+ab^2}{2+\sqrt{2}\left(a-b\right)-ab}=\sqrt{2}\\ \Leftrightarrow\sqrt{2}\left(a^2+b^2\right)+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-\sqrt{2}ab\\ \Leftrightarrow4\sqrt{2}-ab\left(a-b\right)=2\sqrt{2}+2\left(a-b\right)-\sqrt{2}ab\\ \Leftrightarrow\left(2+ab\right)\left(a-b\right)=2\sqrt{2}+\sqrt{2}ab\\ \Leftrightarrow\left(2+ab\right)\left(a-b\right)-\sqrt{2}\left(2+ab\right)=0\\ \Leftrightarrow\left(a-b-\sqrt{2}\right)\left(2+ab\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}ab=-2\\a-b=\sqrt{2}\end{matrix}\right.\)

Xét \(\left\{{}\begin{matrix}ab=-2\\a^2+b^2=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=8\\\left(a+b\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=\pm2\sqrt{2}\\a+b=0\end{matrix}\right.\left(loại.vì.a>0;b\ge0\right)\)

Xét \(\left\{{}\begin{matrix}a-b=\sqrt{2}\\a^2+b^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+2\sqrt{2}b+2+b^2=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\2b^2+2\sqrt{2}b-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+b\sqrt{2}-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left[{}\begin{matrix}b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\b=\dfrac{-\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\left(sd.\Delta\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\left(b\ge0\right)\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)

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

Tới đây dễ r nha

\(\left(\dfrac{2}{2-\sqrt{x}}+\dfrac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right)\) (ĐK: \(x\ne4;x>0\)) 

\(=\left[\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]:\left[\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]\)

\(=\dfrac{-2\sqrt{x}+\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\left[\dfrac{-\left(\sqrt{x}+2\right)^2+\left(\sqrt{x}-2\right)^2-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]\)

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

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

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

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

\(=-\dfrac{3\sqrt{x}+6-x-2\sqrt{x}}{4x}\)

\(=-\dfrac{\sqrt{x}-x+6}{4x}\)

\(\left(\dfrac{\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\) (ĐK: \(x\ge0;x\ne1;x\ne\dfrac{1}{9}\)) 

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

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

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

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

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

a: =>\(x^2\cdot2\sqrt{2}+x\left(2+2\sqrt{2}\right)+4=0\)

\(\text{Δ}=\left(2\sqrt{2}+2\right)^2-4\cdot2\sqrt{2}\cdot4=12-24\sqrt{2}< 0\)

=>PTVN

b: 

\(\Leftrightarrow2x^2+2x+\sqrt{3}-x^2+2\sqrt{3}x+\sqrt{3}=0\)

=>\(x^2+x\left(2\sqrt{3}+2\right)+2\sqrt{3}=0\)

\(\text{Δ}=\left(2\sqrt{3}+2\right)^2-4\cdot2\sqrt{3}=16>0\)

PT có hai nghiệm là;

\(\left\{{}\begin{matrix}x_1=\dfrac{-2\sqrt{3}-2-4}{2}=-\sqrt{3}-3\\x=\dfrac{-2\sqrt{3}-2+4}{2}=-\sqrt{3}+1\end{matrix}\right.\)