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A = \(x^2+3x-7=x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\)
\(\Rightarrow\)min A = \(-\frac{37}{4}\Leftrightarrow x=-\frac{3}{2}\)
B = \(x-5\sqrt{x}-1\) ĐKXĐ: \(x\ge0\)
\(=x-2\sqrt{x}\frac{5}{2}+\frac{25}{4}-\frac{29}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
\(\Rightarrow\)min B = \(-\frac{29}{4}\Leftrightarrow x=\frac{25}{4}\)( thỏa mãn)
C = \(\frac{-4}{\sqrt{x}+7}\) ĐKXĐ:\(x\ge0\)
Ta có: \(\sqrt{x}+7\ge7\Rightarrow\frac{4}{\sqrt{x}+7}\le\frac{4}{7}\)\(\Leftrightarrow\frac{-4}{\sqrt{x}+7}\ge-\frac{4}{7}\)
\(\Rightarrow\)min C = \(-\frac{4}{7}\Leftrightarrow x=0\)
D = \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=1-\frac{2}{\sqrt{x}+3}\ge1-\frac{2}{3}=\frac{1}{3}\)
\(\Rightarrow\)min D = \(\frac{1}{3}\Leftrightarrow x=0\)
E = \(\frac{x+7}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-6\ge2\sqrt{16}-6=2\)
\(\Rightarrow\)min E = \(2\Leftrightarrow x=1\)(thỏa mãn)
F = \(\frac{x^2+3x+5}{x^2}\) ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow\)\(x^2\left(F-1\right)-3x-5=0\)
△ = \(3^2+20\left(F-1\right)\ge0\)\(\Leftrightarrow F\ge\frac{11}{20}\)
\(\Rightarrow\)min F = \(\frac{11}{20}\Leftrightarrow x=-\frac{10}{3}\)( thỏa mãn)
6.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2\end{matrix}\right.\)
4.
ĐKXĐ: \(x\ge4\)
Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)
\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)
\(\Leftrightarrow3t^2-7t+34=0\)
Phương trình vô nghiệm
5.
ĐKXĐ: ...
- Với \(x=0\) ko phải nghiệm
- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:
\(\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)
\(\Leftrightarrow2x=4\Rightarrow x=2\)
1)\(\sqrt{2x^2-2x+\frac{1}{2}}=\frac{1}{\sqrt{2}}\left(ĐKXĐ:x^2-x+\frac{1}{4}\ge0\right)\)
\(2x^2-2x+\frac{1}{2}=\frac{1}{2}\)
\(2x^2-2x=0\)
\(2x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
2)\(\sqrt{9x-9}-2\sqrt{\frac{x-1}{4}}=6\left(ĐKXĐ:x\ge1\right)\)
\(\sqrt{9\left(x-1\right)}-2.\frac{\sqrt{x-1}}{2}=6\)
\(3\sqrt{x-1}-\left(\sqrt{x-1}\right)=6\)
\(2\sqrt{x-1}=6\)
\(\sqrt{x-1}=3=\sqrt{9}\)
\(\Rightarrow x=10\)
4)\(1-3x+\sqrt{x^2-6x+9}=0\)
\(1-3x+\sqrt{\left(x-3\right)^2}=0\)
\(1-3x+x-3=0\)
\(x=-1\)
5)\(\frac{1}{2}\sqrt{\frac{3x+9}{4}}+\sqrt{x+3}=\sqrt{1-x}\)
\(\frac{1}{2}.\frac{\sqrt{3x+9}}{2}+\sqrt{x+3}=\sqrt{1-x}\)
\(\frac{\sqrt{3x+9}}{4}+\sqrt{x+3}=\sqrt{1-x}\)
\(\frac{\sqrt{3x+9}+4\sqrt{x+3}}{4}=\frac{4\sqrt{1-x}}{4}\)
\(\Rightarrow\sqrt{3}.\sqrt{x+3}+4\sqrt{x+3}=4\sqrt{1-x}\)
\(\Rightarrow\left(\sqrt{3}+4\right)\left(\sqrt{x+3}\right)=\sqrt{2-2x}\)
6)\(\sqrt{4x^2-9}.\left(\sqrt{x+1}+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x^2-9=0\\\sqrt{x+1}+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x^2=9\\\sqrt{x+1}=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-1\end{cases}}\)
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
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