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a) \(\left\{{}\begin{matrix}x\ge0\\-\sqrt{x+7}< 0\\-5x-4\ne0\\-3x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+7>0\\-5x\ne4\\-3x\ne-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>-7\\x\ne\frac{-4}{5}\\x\ne\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne\frac{2}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x\ge0\\x+4\ne0\\x-2\ge0\\-2x-10\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-4\\x\ge2\\-2x\ne10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne-5\end{matrix}\right.\Leftrightarrow x\ge2\)
c) \(\left\{{}\begin{matrix}x\ge0\\-x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-3\\x\ne-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge0\)
d) \(\left\{{}\begin{matrix}2x-7\ge0\\x\ge0\\3x-4\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{7}{2}\\x\ge0\\x\ne\frac{4}{3}\\x\ne3\end{matrix}\right.\Leftrightarrow x\ge\frac{7}{2}\)
a) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\\ \Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)
S = (3;6)
b)\(\sqrt{x^2-4}-2\sqrt{x-2}=0\\ \Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-2\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\x=2\end{matrix}\right.\) S= (2)
c)\(\sqrt{\frac{2x-3}{x-1}}=2\left(đkxđ:x\ne1\right)\Leftrightarrow2\sqrt{x-1}=\sqrt{2x-3}\\ \Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\) S= (1/2)
d) đkxđ : x khác -1
\(\sqrt{\frac{4x+3}{x+1}}=3\Leftrightarrow4x+3=9x+9\Leftrightarrow x=-\frac{6}{5}\) S = (-6/5)
e) đk x >= 3/2
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\Leftrightarrow2x-3=4x-4\Leftrightarrow x=\frac{1}{2}\) (loại) vậy pt vô nghiệm
f) đk x >= -3/4
\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\Leftrightarrow4x+3=9x+9\Leftrightarrow x=-\frac{6}{5}\) (loại) vậy pt vô nghiệm
h)
ĐK: \(\left\{\begin{matrix} 3x-12\geq 0\\ x-5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 4\\ x\neq 5\end{matrix}\right.\)
k)
ĐK: \(\left\{\begin{matrix} x-1\geq 0\\ x-2\neq 0\\ x-3\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\neq 2\\ x\neq 3\end{matrix}\right.\)
m)
ĐK: \(\left\{\begin{matrix} x-2\neq 0\\ x-4\neq 0\\ \frac{2x-3}{x-2}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 2\\ x\neq 4\\ x>2\end{matrix}\right.\) hoặc \(x\leq \frac{3}{2}\)
Lời giải:
a) ĐK: $-4x+16\geq 0\Leftrightarrow x\leq 4$
b) ĐK: \(\left\{\begin{matrix} 2x-1\neq 0\\ \frac{-3}{2x-1}\geq 0\end{matrix}\right.\Leftrightarrow 2x-1< 0\Leftrightarrow x< \frac{1}{2}\)
c) ĐK: $-5x^2\geq 0\Leftrightarrow 5x^2\leq 0$. Mà $5x^2\geq 0$ với mọi $x\in\mathbb{R}$ nên biểu thức có nghĩa khi $5x^2=0\Leftrightarrow x=0$
d) ĐK:
\(\left\{\begin{matrix} -x^2-4x-4\neq 0\\ \frac{-3}{-x^2-4x-4}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -(x+2)^2\neq 0\\ \frac{3}{(x+2)^2}\geq 0\end{matrix}\right.\Leftrightarrow x\neq -2\)
e) ĐK: $\frac{2x-4}{-3}\geq 0\Leftrightarrow 2x-4\leq 0\Leftrightarrow x\leq 2$
f) ĐK: \(\left\{\begin{matrix} 3x-9\geq 0\\ 2x-8>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ x>4\end{matrix}\right.\Leftrightarrow x>4\)
\(a,\sqrt{1-3x}\)
\(< =>1-3x\ge0\)
\(3x\le1\)
\(x\le\frac{1}{3}\)
\(b,-3< 0\)
\(< =>2x-5\ne0;2x-5\le0< =>2x-5< 0\)
\(x< \frac{5}{2}\)
\(c,\sqrt{3x+2}+\sqrt{-2x+3}\)
\(\hept{\begin{cases}3x+2\ge0\\-2x+3\ge0\end{cases}}\)
\(\hept{\begin{cases}x\ge-\frac{2}{3}\\x\le\frac{3}{2}\end{cases}}\)
\(< =>-\frac{2}{3}\le x\le\frac{3}{2}\)
\(d,\frac{x-5}{\sqrt{-4x}}\)
\(\sqrt{-4x}\ge0;\sqrt{-4x}\ne0< =>\sqrt{-4x}>0\)
\(-4x>0\)
\(x< 0\)
\(e,\sqrt{x-2}+\frac{1}{x-3}\)
\(\sqrt{x-2}\ge0;x-3\ne0\)
\(x\ge2;x\ne3\)
\(f,\sqrt{-\left(x-2\right)^2}\)
\(\sqrt{-\left(x-2\right)^2}\ge0\)
\(-\left|x-2\right|\ge0\)
\(-\left|x-2\right|\le0\)
lên chỉ có 1 nghiệm duy nhất là
\(x-2=0< =>x=2\)
\(g,\sqrt{\frac{-2x^2}{3x+2}}\)
\(-2x^2\le0\)
\(\sqrt{\frac{-2x^2}{3x+2}}\ge0< =>3x+2\le0;3x+2\ne0\)
\(x\le-\frac{2}{3};x\ne-\frac{2}{3}< =>x< -\frac{2}{3}\)
a)\(\sqrt{1-3x}\)có nghĩa \(\Leftrightarrow\sqrt{1-3x}\ge0\)
\(\Leftrightarrow1-3x\ge0\)
\(\Leftrightarrow-3x\ge-1\)
\(\Leftrightarrow x\ge\frac{1}{3}\)
b)\(\sqrt{\frac{-3}{2x-5}}\)có nghĩa \(\Leftrightarrow\sqrt{\frac{-3}{2x-5}}\ge0\)
\(\Leftrightarrow\frac{-3}{2x-5}\ge0\)
\(\Leftrightarrow2x-5>0\)
\(\Leftrightarrow2x>5\)
\(\Leftrightarrow x>\frac{5}{2}\)
c)\(\sqrt{3x+2}+\sqrt{-2x+3}\)có nghĩa \(\sqrt{3x+2}+\sqrt{-2x+3}\ge0\)
\(\Leftrightarrow3x+2-2x+3\ge0\)
\(\Leftrightarrow x+5\ge0\)
\(\Leftrightarrow x\ge-5\)
d)\(\frac{x-5}{\sqrt{-4x}}\)có nghĩa \(\Leftrightarrow\frac{x-5}{\sqrt{-4x}}\ge0\)
\(\Leftrightarrow\frac{x-5}{\sqrt{-\left(2x\right)^2}}\ge0\)
\(\Leftrightarrow\frac{x-5}{-2x}\ge0\)
\(\Leftrightarrow-2x>0\)
\(\Leftrightarrow x>2\)(Câu này không chắc làm đúng không, chắc sai goi)
f)\(\sqrt{-x^2+4x-4}\)có nghĩa \(\Leftrightarrow\sqrt{-x^2+4x-4}\ge0\)
\(\Leftrightarrow-x^2+4x-4\ge0\)
\(\Leftrightarrow-\left(x-2\right)^2\ge0\)
không có z thỏa mãn
g)\(\sqrt{\frac{-2x^2}{3x+2}}\)có nghĩa \(\Leftrightarrow\sqrt{\frac{-2x^2}{3x+2}}\ge0\)
\(\Leftrightarrow\frac{-2x^2}{3x+2}\ge0\)
\(\Leftrightarrow3x+2>0\)
\(\Leftrightarrow3x>-2\)
\(\Leftrightarrow x>\frac{-2}{3}\)
@Cừu
a) ĐK: \(\left\{{}\begin{matrix}x\ne-1\\\frac{4-x}{x+1}\ge0\end{matrix}\right.\). Lập bảng xét dấu sẽ được \(-1< x\le4\)
b) Tương tự
c)(em ko chắc) ĐK: \(\left\{{}\begin{matrix}x^2-4\ge0\left(1\right)\\\frac{x-2}{x+1}\ge0\left(2\right)\\x\ne-1\end{matrix}\right.\). Giải (1) ta được \(x\le-2\text{hoặc }x\ge2\)
Giải (2) được \(x\le-1\text{ hoặc }x\ge2\)
Kết hợp lại ta được: \(x\le-2\text{hoặc }x\ge2\)
\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)
\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)
\(\Rightarrow3x+2=2\left(x+2\right)\)
\(\Rightarrow3x+2=2x+4\)
\(\Rightarrow3x-2x=4-2\)
\(\Rightarrow x=2\)
\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)
\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)
\(\Rightarrow2\sqrt{x-2}=4\)
\(\Rightarrow\sqrt{x-2}=2\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)
\(\Rightarrow x+4=1-2x+2\sqrt{\left(1-2x\right)\left(1-x\right)}+1-x\)
\(\Rightarrow x+4=2-3x+2\sqrt{1-3x+2x^2}\)
\(\Rightarrow x+4-2+3x=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)
\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)
\(\Rightarrow2x^2+7x=0\)
\(\Rightarrow x\left(2x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)
\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)
\(\frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{2x\left(\sqrt{3}-1\right)}{3-1}=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+\sqrt{3}x-\sqrt{3x}+x=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)
\(\Rightarrow x=1\)
a)\(\sqrt{\frac{2x-3}{x-1}}=2\RightarrowĐk:\frac{2x-3}{x-1}\ge0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge\frac{3}{2}\\x< 1\end{array}\right.\)
\(\sqrt{\frac{2x-3}{x-1}}=2\Rightarrow\frac{2x-3}{x-1}=4\)
\(\Leftrightarrow2x-3=4\left(x-1\right)\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)(nhận)
b)\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\RightarrowĐk:\begin{cases}2x-3\ge0\\x-1>0\end{cases}\)
\(\Leftrightarrow x\ge\frac{3}{2}\)
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\Leftrightarrow\sqrt{2x-3}=2\sqrt{x-1}\)
\(\Leftrightarrow2x-3=4x-4\)\(\Leftrightarrow x=\frac{1}{2}\)(loại)
c)\(\sqrt{\frac{4x+3}{x+1}}=3\RightarrowĐk:\frac{4x+3}{x+1}\ge0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge\frac{-3}{4}\\x< -1\end{array}\right.\)
\(\sqrt{\frac{4x+3}{x+1}}=3\Rightarrow\frac{4x+3}{x+1}=9\)
\(\Leftrightarrow4x+3=9\left(x+1\right)\Leftrightarrow4x+3=9x+9\)
\(\Leftrightarrow5x=-6\Leftrightarrow x=\frac{-6}{5}\)(nhận)
c)\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\RightarrowĐk:\begin{cases}4x+3\ge0\\x+1>0\end{cases}\)
\(\Rightarrow x\ge\frac{-3}{4}\)
\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\Rightarrow\sqrt{4x+3}=3\sqrt{x+1}\)
\(\Leftrightarrow4x+3=9\left(x+1\right)\Leftrightarrow4x+3=9x+9\)
\(\Leftrightarrow x=\frac{-6}{5}\)(loại)