1/ \(\sqrt{x-2}-\sqrt{1-3x}=0\\ đk:\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)

=> pt vô no

2/ \(\sqrt{15-x}+\sqrt{3-x}=6\\ đk\left\{{}\begin{matrix}15-x\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le15\\x\le3\end{matrix}\right.\Leftrightarrow x\le3\)

\(pt\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)

\(\Leftrightarrow2\sqrt{\left(15-x\right)\left(3-x\right)}=2x+36\)

\(\Leftrightarrow4\left(15-x\right)\left(3-x\right)=\left(2x+18\right)^2\left(đk:x\ge-9\right)\)

\(\Leftrightarrow-144x=144\Leftrightarrow x=-1\left(nhan\right)\)

Câu 1: ĐKXĐ: \(\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại x thỏa mãn ĐKXĐ \(\Rightarrow\) pt vô nghiệm

Câu 2:

ĐKXĐ: \(x\le3\)

\(\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)

\(\Leftrightarrow x+9=\sqrt{x^2-18x+45}\) (\(x\ge-9\))

\(\Leftrightarrow x^2+18x+81=x^2-18x+45\)

\(\Leftrightarrow36x=-36\Rightarrow x=-1\)

Câu 3:

ĐKXĐ: \(x\ge1\)

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

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

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

Phương trình vô nghiệm

chủ yếu là bình phương hai vế,đặt ĐK rồi chuyển thành phương trình bậc hai rồi giải

1.\(ĐKXĐ:x\ge0\)

\(PT\Leftrightarrow x^2+x=x^2\Leftrightarrow x=0\)(t/m)

Vậy pt có nghiêm duy nhất là x=0

2.ĐKXĐ:\(1-x^2\ge0\Leftrightarrow-1\le x\le1\)

\(PT\Leftrightarrow1-x^2=x^2-2x+1\left(x\ge1\right)\)

\(\Leftrightarrow2x^2-2x=0\)

\(\Leftrightarrow2x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loai,vi,x\ge1\right)\\x=1\left(chon\right)\end{matrix}\right.\)

Vậy phương trình có nghiệm duy nhất là x=1

3.ĐKXĐ:\(x^2-4x+3\ge0\)

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

\(\Leftrightarrow x^2-4x+3=x^2-4x+4\left(x\ge2\right)\)

\(\Leftrightarrow0=1\left(Sai\right)\)

Vậy pt đã cho vô nghiệm

4.ĐKXĐ:\(x^2-1\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)

\(\sqrt{x^2-1}-x^2+1=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=0\\1-\sqrt{x^2-1}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm1\left(tm\right)\\\sqrt{x^2-1}=1\left(\cdot\right)\end{matrix}\right.\)

Giải (*): \(\left(\cdot\right)\Leftrightarrow x^2-1=1\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\left(tm\right)\)

Kết luận: tập nghiệm của pt là:\(S=\left\{\pm1;\pm\sqrt{2}\right\}\)

5.ĐKXĐ:\(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)

\(\sqrt{x^2-4}-x+2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-\sqrt{x-2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=\sqrt{x-2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x+2=x-2\Leftrightarrow2=-2\left(vo,li,nen,loai\right)\end{matrix}\right.\)

Vậy pt đã cho có nghiệm duy nhất là x=2

6.ĐKXĐ:\(1-2x^2\ge0\Leftrightarrow-\frac{\sqrt{2}}{2}\le x\le\frac{\sqrt{2}}{2}\)

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

\(\Leftrightarrow1-2x^2=x^2-2x+1\left(x\ge1\right)\)

\(\Leftrightarrow3x^2-2x=0\)

\(\Leftrightarrow x\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loai\right)\\x=\frac{2}{3}\left(loai\right)\end{matrix}\right.\)

Kết luận: PT đã cho vô nghiệm

cái này có phải bình phương hai vế nên ko nhỉ?

Câu 6 có sai ko?

đúng rồi bạn nhé

Tacó \(\Delta\)=(-7)²-4x1x2=41>0 =>\(\sqrt{_{ }x1}\)=\(\dfrac{7+\sqrt{41}}{2}\)=>\(_{x1}\)=\(\dfrac{\left(7+\sqrt{41}\right)^2}{4}\)=\(\dfrac{45+7\sqrt{41}}{2}\) =>\(\sqrt{_{ }x2}\)=\(\dfrac{7-\sqrt{41}}{2}\)=>\(_{x_2}\)=\(\dfrac{\left(7-\sqrt{41^{ }}\right)^2}{4}\)=\(\dfrac{45-7\sqrt{41}}{2}\) so sánh với điều kiện X>_0

a, ĐKXĐ : \(\left[{}\begin{matrix}x\ge0\\ y>0\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x>0\\y\ge0\end{matrix}\right.\)

Ta có :\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

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

= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)

= \(\left(x-\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\)

= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)

= \(\sqrt{xy}\)

\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}\) \(=\sqrt{\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}}\)\(=\sqrt{\frac{a^2-1}{b^2-1}}\) (*)

Thay a=7,25 và b= 3,25 vào (*) ta có:

\(\sqrt{\frac{7,25^2-1}{3,25^2-1}}\) \(=\frac{5\sqrt{33}}{4}:\frac{3\sqrt{17}}{4}=\frac{5\sqrt{33}}{3\sqrt{17}}=\frac{5\sqrt{561}}{51}\)

I) xd mọi x

\(\sqrt{x^2-8x+16}+\sqrt{x^2-10x+25}=9\)

\(\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-5\right)^2}=9=>\left|x-4\right|+\left|x-5\right|=9\)

\(\left[{}\begin{matrix}x< 4\Rightarrow4-x+5-x=>x=0\left(n\right)\\4\le x< 5\Rightarrow x-4+5-x=9\left(vn\right)\\x\ge5\Rightarrow x-4+x-5=9\Rightarrow x=9\left(n\right)\\\end{matrix}\right.\)

kết luận

\(\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)