Tìm x để căn thức sau xác định
a)A=\(\sqrt{x-3}-\sqrt{\dfrac{1}{4-x}}\)
b)B=\(\dfrac{1}{\sqrt{x-1}}+\dfrac{2}{\sqrt{x^2-4x+4}}\)
Tìm điều kiện để các biểu thức sau xác định
a)\(\sqrt{x+1}-\dfrac{1}{2}\)
b)\(2.\sqrt{1-2x}-\dfrac{\sqrt{3}-1}{4}\)
c)\(\sqrt{x+1}-\sqrt{x-2}\)
d)\(\sqrt{2-3x}-\sqrt{1-2x}\)
e)\(2.\sqrt{\sqrt{3}-2x}+\dfrac{1}{x-1}\)
f)\(\dfrac{1}{2}.\sqrt{x-\dfrac{\sqrt{3}}{2}}-\dfrac{1}{\sqrt{x}-1}\)
g)\(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+2}\)
h)\(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x^2+2}}\)
a, \(x+1\ge0\Leftrightarrow x\ge-1\)
b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)
d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)
e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)
f, \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{3}}{2}\ge0\\x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ge0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\)
g, \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+2\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Tìm x để mỗi căn thức sau có nghĩa:
a. \(\sqrt{3-2x}\) b. \(\sqrt{x+1}+\sqrt{3-x}\) c. \(\dfrac{\sqrt{4x-2}}{x^2-4x+3}\) d. \(\dfrac{\sqrt{4x^2-2x+1}}{\sqrt{3-5x}}\)
ĐKXĐ: \(3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\)
b) ĐKXĐ: \(-1\le x\le3\)
c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\\x\ne3\end{matrix}\right.\).
d) ĐKXĐ: \(x< \dfrac{3}{5}\).
tìm giá x để biểu thức xác định
a,\(\sqrt{\dfrac{-5}{x^2+6}}\)
b,\(\sqrt{\dfrac{3x-2}{x^2-2x+4}}\)
\(a,ĐK:\dfrac{-5}{x^2+6}\ge0\Leftrightarrow x\in\varnothing\)
( Do \(-5< 0;x^2+6>0\Leftrightarrow\dfrac{-5}{x^2+6}< 0,\forall x\))
\(b,ĐK:\dfrac{3x-2}{\left(x-1\right)^2+3}\ge0\\ \Leftrightarrow3x-2\ge0\left[\left(x-1\right)^2+3>0\right]\\ \Leftrightarrow x\ge\dfrac{2}{3}\)
a) ĐKXĐ: \(x^2+6< 0\left(VLý.do.x^2+6\ge6>0\right)\)
Vậy biểu thức k xác định với mọi x
b) \(\sqrt{\dfrac{3x-2}{x^2-2x+4}}=\sqrt{\dfrac{3x-2}{\left(x-1\right)^2+3}}\)
ĐKXĐ: \(\left\{{}\begin{matrix}3x-2\ge0\\\left(x-1\right)^2+3\ne0\left(đúng\forall x\right)\end{matrix}\right.\)
\(\Leftrightarrow x\ge\dfrac{2}{3}\)
Tìm ĐK để căn thức sau xác định:
a) \(\sqrt{x^2+3x-10}\)
b) \(\sqrt{\dfrac{4x-4-x^2}{5}}\)
c) \(\sqrt{x-4\sqrt{x-4}}\)
a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-5\end{matrix}\right.\)
b: ĐKXĐ: \(x=2\)
c: ĐKXĐ: \(x\ge4\)
tìm x để căn thức sau được xác định
1)\(\sqrt{\dfrac{-2}{2x-2}}\)
2)\(\sqrt{\dfrac{2}{3x-1}}\)
3)\(\sqrt{\dfrac{2x-2}{-2}}\)
4)\(\sqrt{\dfrac{3x-2}{5}}\)
5)\(\sqrt{\dfrac{x-2}{x+3}}\)
1: ĐKXĐ: -2/2x-2>=0
=>2x-2<0
=>x<1
2: ĐKXĐ: 2/3x-1>=0
=>3x-1>0
=>x>1/3
3: ĐKXĐ: 2x-2/(-2)>=0
=>2x-2<=0
=>x<=1
4: ĐKXĐ: (3x-2)/5>=0
=>3x-2>=0
=>x>=2/3
5: ĐKXĐ: (x-2)/(x+3)>=0
=>x>=2 hoặc x<-3
1.
a. Tìm điều kiện để căn thức bậc hai có nghĩa \(\sqrt{\dfrac{x^2}{2x-1}}\)
b. \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}.\sqrt[3]{\dfrac{1}{27}}\)
* Giải phương trình
a. \(\sqrt{\left(x+1\right)^2}=3\)
b. \(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)
1)Tìm x để căn thức sau có nghĩa
a)\(\sqrt{2x-4}\) b)\(\sqrt{\dfrac{-7}{4-x}}\)
2) Tính
A=\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}
\)
B=\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
Helpppp
1)
a) \(\sqrt{2x-4}\) có nghĩa khi:
\(2x-4\ge0\)
\(\Leftrightarrow2x\ge4\)
\(\Leftrightarrow x\ge\dfrac{4}{2}\)
\(\Leftrightarrow x\ge2\)
b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi
\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)
\(\Rightarrow4-x\le0\)
\(\Leftrightarrow x\ge4\)
2)
a) \(A=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}\right)^2+2\cdot2\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\cdot\sqrt{5}+2^2}\)
\(A=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(A=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\)
\(A=\sqrt{5}+2-\sqrt{5}+2\)
\(A=4\)
\(B=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-5}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{5}-\sqrt{7}}\)
\(B=\left(-\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}-\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(B=\left[-\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(B=\left(-\sqrt{7}-\sqrt{5}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)
\(B=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(B=-\left(7-5\right)\)
\(B=-2\)
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right).\dfrac{\sqrt{x}-1}{x^2}\)
\(C=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right).\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(E=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{9+x}{9-x}\right).\left(3\sqrt{x}-x\right)\)
help
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right).\dfrac{\sqrt{x}-1}{x^2}\)
\(C=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right).\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(E=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{9+x}{9-x}\right).\left(3\sqrt{x}-x\right)\)
help
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)