Các bạn giúp mình câu này với
Cảm ơn các bạn nhiều!
Rút gọn A=\(\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(\dfrac{\sqrt{28}-2\sqrt{3}}{\sqrt{7}-\sqrt{3}}\)
Các bạn giúp mình với. Mình cảm ơn rất rất nhiều !
\(\dfrac{2\sqrt{7}-2\sqrt{3}}{\sqrt{7}-\sqrt{3}}=2\)
Tìm điều kiện xác định của a để các căn sau có nghĩa:
1.
\(\sqrt{\dfrac{-a}{3}}\)
2. \(\sqrt{\dfrac{a^2+1}{1-3a}}\)
3. \(\sqrt{a^2-6a+10}\)
4. \(\sqrt{\dfrac{a-1}{a+2}}\)
Làm ơn giúp mình với. Cảm ơn mọi người nhiều❤
1)Để căn có nghĩa \(\Leftrightarrow\dfrac{-a}{3}\ge0\Leftrightarrow a\le0\)
Vậy...
2)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a^2+1}{1-3a}\ge0\\1-3a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1-3a>0\left(vìa^2+1>0\right)\\1-3a\ne0\end{matrix}\right.\)
\(\Leftrightarrow1-3a>0\Leftrightarrow3a< 1\Leftrightarrow a< \dfrac{1}{3}\)
Vậy...
3)Để căn có nghĩa
\(\Leftrightarrow a^2-6a+10\ge0\Leftrightarrow\left(a^2-6a+9\right)+1\ge0\Leftrightarrow\left(a-3\right)^2+1\ge0\left(lđ;\forall a\right)\)
Vậy căn luôn có nghĩa với mọi a
4)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a-1}{a+2}\ge0\\a+2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+2< 0\end{matrix}\right.\end{matrix}\right.\\a+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a\ge1\\a>-2\end{matrix}\right.\\\left\{{}\begin{matrix}a\le1\\a< -2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a< -2\end{matrix}\right.\)
Vậy...
\(\left\{\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right\}:\left\{1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right\}\)
a, Rút gọn P
b, Tìm các giá trị của x để P = \(\dfrac{6}{5}\)
Giúp mình với ạ, Cảm ơn trước!
\(a,\)
\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :
\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)
\(\Leftrightarrow-3\sqrt{x}+11=0\)
\(\Leftrightarrow-3\sqrt{x}=-11\)
\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)
\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{121}{9}\)
Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)
\(\frac{-5\sqrt{x}+4}{3\sqrt{x}-2}+\frac{6\sqrt{x}+4}{2\sqrt{x}+3}+\frac{29\sqrt{x}-28}{3\left(6x+5\sqrt{x}-6\right)}\)
Các bạn giúp mình câu này với,mình đang cần gấp.Cảm ơn các bạn nhiều!
Rút gọn biểu thức
a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
giúp em với ạ, em cảm ơn!
a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=-1+3\sqrt{5}\)
b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}=2-\sqrt{3}+1+\sqrt{3}=3\)
a: \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)
\(=3\sqrt{5}-1\)
b: \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}+1\)
=3
rút gọn hộ mik con này với
\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\dfrac{5}{\sqrt{3}-2\sqrt{2}}-\dfrac{5}{\sqrt{3}+\sqrt{8}}\)
\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}+\dfrac{5}{2\sqrt{2}-3}-\dfrac{5}{\sqrt{3}+\sqrt{8}}\)
\(=\sqrt{3}+1+\sqrt{3}-1+2\sqrt{2}+3-2\sqrt{2}+3\)
\(=6+2\sqrt{3}\)
\(=\sqrt{3+2\sqrt{2}+1}+\sqrt{3-2\sqrt{2}+1}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}\\ =\left|\sqrt[]{3}+1\right|+\left|\sqrt{3}-1\right|-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{5}-\dfrac{5\left(\sqrt{3}-2\sqrt[]{2}\right)}{5}\\ =\sqrt{3}+1+\sqrt{3}-1-\sqrt{3}-2\sqrt{2}-\sqrt[]{3}+2\sqrt{2}\\ =0\)
rút gọn các biểu thức sau
\(\dfrac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}\)+\(\dfrac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)
\(\dfrac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}\)+\(\dfrac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
a: \(=\dfrac{6+4\sqrt{2}}{\sqrt{2}+2+\sqrt{2}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}-2+\sqrt{2}}\)
\(=\dfrac{6+4\sqrt{2}}{2+2\sqrt{2}}+\dfrac{6-4\sqrt{2}}{2\sqrt{2}-2}\)
\(=\dfrac{3+2\sqrt{2}}{\sqrt{2}+1}+\dfrac{3-2\sqrt{2}}{\sqrt{2}-1}\)
=căn 2+1+căn 2-1=2căn 2
b: \(=\dfrac{\sqrt{3}+\sqrt{3+\sqrt{3}}+\sqrt{3}-\sqrt{3+\sqrt{3}}}{1-\sqrt{3}-1}=\dfrac{-2\sqrt{3}}{\sqrt{3}}=-2\)
\(\dfrac{8}{\sqrt{5}-1}-\dfrac{22}{4+\sqrt{5}}+\dfrac{\sqrt{15}+2\sqrt{5}}{2+\sqrt{3}}\)
rút gọn giúp mình với
\(\dfrac{8}{\sqrt{5}-1}-\dfrac{22}{4+\sqrt{5}}+\dfrac{\sqrt{15}+2\sqrt{5}}{2+\sqrt{3}}\)
\(=\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{22\left(4-\sqrt{5}\right)}{\left(\sqrt{5}+4\right)\left(4-\sqrt{5}\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}+2\right)}{2+\sqrt{3}}\)
\(=\dfrac{8\sqrt{5}+8}{5-1}-\dfrac{88-22\sqrt{5}}{16-5}+\sqrt{5}\)
\(=\dfrac{8\sqrt{5}+8}{4}-\dfrac{88-22\sqrt{5}}{11}+\sqrt{5}\)
\(=2\sqrt{5}+2-8+2\sqrt{5}+\sqrt{5}=5\sqrt{5}-6\)
Rút gọn rồi tính các biểu thức sau:
a)\(A=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\) với \(a^2=6-3\sqrt{3};b^2=2+\sqrt{3}\)
b)\(B=\dfrac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}\)với \(x=1+\sqrt{5}\)
Câu a, bạn coi lại đề xem $a^2=6-3\sqrt{3}$ hay $a=6-3\sqrt{3}$???
b.
\(B=\frac{\sqrt{(x-2)+(x+2)+2\sqrt{(x-2)(x+2)}}}{\sqrt{x^2-4}+x+2}\)
\(=\frac{\sqrt{(\sqrt{x-2}+\sqrt{x+2})^2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}=\frac{1}{\sqrt{x+2}}\)
\(=\frac{1}{\sqrt{3+\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{(\sqrt{5}+1)^2}}=\frac{\sqrt{2}}{\sqrt{5}+1}\)
Nguyễn Hoàng trung: Chả qua nếu $a=6-3\sqrt{3}; b=2+\sqrt{3}$ thì kết quả sẽ đẹp hơn. Còn như đề thì vẫn rút gọn được.
\(A=\frac{a-\sqrt{ab}+b}{(\sqrt{a}-\sqrt{b})^2}=\frac{a-\sqrt{ab}+b}{a-2\sqrt{ab}+b}\)
\(2a^2=12-6\sqrt{3}=(3-\sqrt{3})^2\Rightarrow a=\frac{3-\sqrt{3}}{\sqrt{2}}\) (do $a\geq 0$)
\(2b^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\Rightarrow b=\frac{\sqrt{3}+1}{\sqrt{2}}\) (do $b\geq 0$)
\(\Rightarrow a+b=2\sqrt{2}; ab=\frac{\sqrt{3}(\sqrt{3}-1)(\sqrt{3}+1)}{2}=\sqrt{3}\)
Do đó: $A=\frac{2\sqrt{2}-\sqrt[4]{3}}{2\sqrt{2}-2\sqrt[4]{3}}$
Rút gọn : ( giúp với )
a) \(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)
b) \(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
d) \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}\)
a. \(\dfrac{\sqrt{2}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}.\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)
d. \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{5-2\sqrt{5}+1}}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}-1}=\sqrt{5}-1\)