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\(a,\sqrt{2x-1}\)
\(\sqrt{2x-1}\) có nghĩa khi:
\(2x-1\ge0\\ \Leftrightarrow2x\ge1\\ \Leftrightarrow x\ge\dfrac{1}{2}\)
\(b,\sqrt{\dfrac{3}{x^{ }+1}}\)
\(\sqrt{\dfrac{3}{x+1}}\) có nghĩa khi:
\(x+1\ge0\\ \Leftrightarrow x\ge-1\)
\(c,\sqrt{3x^2}\)
\(\forall x\in Rvì3x^2\ge0\)
\(d,\sqrt{\dfrac{3}{x^2}}\\ \forall x\in Rvìx^2\ge0\)
\(e,\sqrt{\dfrac{-1}{x^2+2}}\)
Không có nghĩa \(\forall x\in R\)
\(f,\sqrt{\dfrac{2}{3}x-\dfrac{1}{5}}\)
\(\sqrt{\dfrac{2}{3}x-\dfrac{1}{5}}\) có nghĩa khi:
\(\dfrac{2}{3}x-\dfrac{1}{5}\ge0\\ \)
\(\Leftrightarrow\)\(\dfrac{2}{3}x\ge\dfrac{1}{5}\\ \)
\(x\ge\dfrac{1}{10}\)
a/ \(x^2+4x-5>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -5\end{matrix}\right.\)
b/ \(\left\{{}\begin{matrix}2x-1\ge0\\x-\sqrt{2x-1}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\\left\{{}\begin{matrix}x>0\\x^2>2x-1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\end{matrix}\right.\)
c/ \(\left\{{}\begin{matrix}x^2-3\ge0\\1-\sqrt{x^2-3}\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{matrix}\right.\\x\ne\pm2\end{matrix}\right.\)
d/ \(\left\{{}\begin{matrix}x+\dfrac{1}{x}\ge0\\-2x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>0\\x\le0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn
e/ \(\left\{{}\begin{matrix}3x-1\ge0\\5x-3\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\ge\dfrac{3}{5}\end{matrix}\right.\) \(\Rightarrow x\ge\dfrac{3}{5}\)
1)
a) \(6=\sqrt{36}< \sqrt{40}\)
b) \(3=\sqrt{9}< \sqrt{10}\)
c) \(2\sqrt{3}< 2\sqrt{4}=4\)
d) \(3\sqrt{2}=\sqrt{18}< \sqrt{36}=6\)
e) \(7=\sqrt{49}< \sqrt{50}\)
2)
a) \(x\ge0\)
b) \(-2x+1\ge0\Leftrightarrow-2x\ge-1\Leftrightarrow x\le\dfrac{1}{2}\)
c) \(5-a\ge0\Leftrightarrow a\le5\)
d) \(2x-3>0\Leftrightarrow2x>3\Leftrightarrow x>\dfrac{3}{2}\)
e) \(-3< x< 1\)
f) \(-3x\ge-4\Leftrightarrow x\le\dfrac{4}{3}\)
g) \(x^2-2x-3\ge0\Leftrightarrow\left(x+1\right)\left(x-3\right)\ge0\Leftrightarrow-1\le x\le3\)
a/ đkxđ: \(x+3\ge0\Leftrightarrow x\ge-3\)
b/ \(\left\{{}\begin{matrix}4x-1\ge0\\x\ne\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{4}\\x\ne\dfrac{1}{2}\end{matrix}\right.\)
c/ \(2-x^2>0\Leftrightarrow x^2< 2\Leftrightarrow-\sqrt{2}< x< \sqrt{2}\)
d/ \(6-x-x^2>0\Leftrightarrow\left(x+3\right)\left(2-x\right)>0\Leftrightarrow\left(x+3\right)\left(x-2\right)< 0\Leftrightarrow-3< x< 2\)
a: \(=\dfrac{2x+1-x-\sqrt{x}-1}{x\sqrt{x}-1}=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
c: \(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{x-1}=\dfrac{-x+\sqrt{x}+2}{x-1}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{x-1}=\dfrac{-\sqrt{x}+2}{\sqrt{x}-1}\)
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
3) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{4x-20}=4\)
\(\Leftrightarrow4x-20=16\)
\(\Leftrightarrow4x=36\)
\(\Leftrightarrow x=9\)
vậy ...
1)
\(A=\dfrac{\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}\right)^2-2^2}\\ A=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)
\(B=\dfrac{x^2-2x\sqrt{2}+2}{x^2-2}=\dfrac{x^2-2x\sqrt{2}+\left(\sqrt{2}\right)^2}{x^2-\sqrt{2}}\\ B=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{\left(x-\sqrt{2}\right)}{\left(x+\sqrt{2}\right)}\)
\(C=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+\left(\sqrt{5}\right)^2}\\ C=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
\(D=\dfrac{\sqrt{a}-2a}{2\sqrt{a}-1}=\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)}{2\sqrt{a}-1}=\sqrt{a}\)
\(E=\dfrac{x^2-2}{x-\sqrt{2}}=\dfrac{x^2-\left(\sqrt{2}\right)^2}{x-\sqrt{2}}\\ E=\dfrac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{x-\sqrt{2}}=x+\sqrt{2}\)
\(F=\dfrac{\sqrt{x}-3}{x-9}=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}\right)^2-3^2}\\ F=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ F=\dfrac{1}{\sqrt{x}+3}\)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
Lời giải:
a) ĐK: \(\left\{\begin{matrix} x-2\neq 0\\ x-2\geq 0\end{matrix}\right.\Leftrightarrow x-2>0\Leftrightarrow x>2\)
b) ĐK: \(\left\{\begin{matrix} x+2\neq 0\\ x-2\geq 0\end{matrix}\right.\Leftrightarrow x\geq 2\)
c) ĐK: \(\left\{\begin{matrix} x^2-4\neq 0\\ x-2\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x-2)(x+2)\neq 0\\ x\geq 2\end{matrix}\right.\Leftrightarrow x>2\)
d) ĐK: \(3-2x>0\Leftrightarrow x< \frac{3}{2}\)
e) ĐK: \(2x+3>0\Leftrightarrow x> \frac{-3}{2}\)
f) ĐK: \(x+1< 0\Leftrightarrow x< -1\)