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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. bài 1: a)\(\left(3-\sqrt{2}\right)\sqrt{7+4\sqrt{3}}\) \(=\left(3-\sqrt{2}\right)\sqrt{\left(2+\sqrt{3}\right)^2}\) \(=\left(3-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)\(do2>\sqrt{3}\) \(=6+3\sqrt{3}-2\sqrt{2}-\sqrt{6}\) b) \(\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\) \(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\) \(=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)do\sqrt{5}>\sqrt{2}\) \(=\sqrt{15}-\sqrt{6}+5-\sqrt{10}\) c)\(\left(2+\sqrt{5}\right)\sqrt{9-4\sqrt{5}}\) \(=\left(2+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-2\right)^2}\) \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)do\sqrt{5}>2\) \(=5-4\) \(=1\left(hđt.3\right)\) d)\(\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}\) \(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\sqrt{4-\sqrt{15}}\) \(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{8-2\sqrt{15}}\) \(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\) \(=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)do\sqrt{5}>\sqrt{3}\) \(=5-3\) \(=2\) e)\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\) \(=\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+9\sqrt{2}\right)\) \(=2\left(2-4+9\right)\) \(=2.7=14\) f)\(\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\) \(=2-\sqrt{6-2\sqrt{5}}\) \(=2-\sqrt{\left(\sqrt{5}-1\right)^2}\) \(=2-\left(\sqrt{5}-1\right)\) \(=2-\sqrt{5}+1\) \(=3-\sqrt{5}\) g)\(\sqrt{3}-\sqrt{2}\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\) \(=\sqrt{3}-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)\) \(=\sqrt{3}-\sqrt{6}-2\) h) \(\left(\sqrt{2}-\sqrt{3+\sqrt{5}}\right)\sqrt{2}+2\sqrt{5}\) \(=\left(2-\sqrt{6+2\sqrt{5}}\right)+2\sqrt{5}\) \(=\left(2-\sqrt{\left(\sqrt{5}+1\right)^2}\right)+2\sqrt{5}\) \(=2-\left(\sqrt{5}+1\right)+2\sqrt{5}\left(do\sqrt{5}>1\right)\) \(=2-\sqrt{5}-1+2\sqrt{5}\) \(=1-\sqrt{5}\) bài 2) a) \(\sqrt{4x^2-4x+1}=5\) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\) \(\Leftrightarrow2x-1=5\)hoặc \(\Leftrightarrow2x-1=-5\) \(\Leftrightarrow x=3\)hoặc \(\Leftrightarrow x=-2\) Vậy x = 3 hoặc x = -2 Bài 1: a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\) \(=5-\sqrt{3}-2+\sqrt{3}=3\) b; \(B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\) \(=\dfrac{\left(2-\sqrt{3}\right)\cdot\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\) \(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}\) \(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\) c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\) \(=\sqrt{\sqrt{5}-\sqrt{3+3-2\sqrt{5}}}\) \(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=1\) d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\) \(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\) Bài 1 ) a)\(3\sqrt{\frac{1}{3}}-\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{2}\) b)\(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}=\left(\sqrt{3}+1\right)-\left|1-\sqrt{3}\right|=\left(\sqrt{3}+1\right)-\sqrt{3}+1=2\) Bài 2) a)\(\sqrt{36x^2-12x+1}=5\) \(\Leftrightarrow36x^2-12x+1=25\) \(\Leftrightarrow36x^2-12x+1=25\) \(\Leftrightarrow\left(6x\right)^2-2.6x+1=25\) \(\Leftrightarrow\left(6x-1\right)^2=25\) \(\Rightarrow6x-1=5\) \(\Leftrightarrow6x=6\) \(\Rightarrow x=1\) b)\(\sqrt{x-5}-2\sqrt{4x-20}-\frac{1}{3}\sqrt{9x-45}=12\) \(\Leftrightarrow\sqrt{x-5}-2\sqrt{4.\left(x-5\right)}-\frac{1}{3}\sqrt{9.\left(x-5\right)}=12\) \(\Leftrightarrow\sqrt{x-5}-4\sqrt{\left(x-5\right)}-\sqrt{\left(x-5\right)}=12\) \(\Leftrightarrow-4\sqrt{\left(x-5\right)}=12\) \(\Rightarrow\)ko tồn tại giá trị nào của x trong biểu thức này P/s tham khảo nha 1a) \(3\sqrt{\frac{1}{3}}-\frac{1}{\sqrt{3}+\sqrt{2}}\) =\(3\sqrt{\frac{3}{3^2}}-\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\) =\(3\frac{\sqrt{3}}{\sqrt{3^2}}-\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}\) =\(3\frac{\sqrt{3}}{3}-\frac{\sqrt{3}-\sqrt{2}}{3-2}\) =\(\sqrt{3}-\left(\sqrt{3}-\sqrt{2}\right)\) =\(\sqrt{3}-\sqrt{3}+\sqrt{2}\)=\(\sqrt{2}\) b)\(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\) =\(|\sqrt{3}+1|-|1-\sqrt{3}|\) =\(\sqrt{3}+1-\left(-\left(1-\sqrt{3}\right)\right)\) =\(\sqrt{3}+1+1-\sqrt{3}\) =\(1+1\)=\(2\) 2) a) \(\sqrt{36x^2-12x+1}=5\) <=>\(\sqrt{\left(6x\right)^2-2.6x.1+1^2}=5\) <=>\(\sqrt{\left(6x-1\right)^2}=5\) <=>\(|6x-1|=5\) Nếu \(6x-1>=0\)=> \(6x>=1\)=>\(x>=\frac{1}{6}\) Nên \(|6x-1|=6x-1\) Ta có \(|6x-1|=5\) <=> \(6x-1=5\) <=> \(6x=6\) <=> \(x=1\)(thỏa) Nếu \(6x-1< 0\)=> \(6x< 1\)=>\(x< \frac{1}{6}\) Nên \(|6x-1|=-\left(6x-1\right)=1-6x\) Ta có \(|6x-1|=5\) <=> \(1-6x=5\) <=> \(-6x=4\) <=> \(x=\frac{4}{-6}=\frac{-2}{3}\)(thỏa) Vậy \(x=1\)và \(x=\frac{-2}{3}\) b) \(\sqrt{x-5}-2\sqrt{4x-20}-\frac{1}{3}\sqrt{9x-45}=12\) <=>\(\sqrt{x-5}-2\sqrt{4\left(x-5\right)}-\frac{1}{3}\sqrt{9\left(x-5\right)}=12\) <=>\(\sqrt{x-5}-2.2\sqrt{x-5}-\frac{1}{3}.3\sqrt{x-5}=12\) <=>\(\sqrt{x-5}-4\sqrt{x-5}-\sqrt{x-5}=12\) <=>\(-4\sqrt{x-5}=12\) <=> \(\sqrt{x-5}=-3\) <=> \(\left(\sqrt{x-5}\right)^2=\left(-3\right)^2\) <=>\(x-5=9\) <=>\(x=14\) Vậy x=14 Kết bạn với mình nhá a) DK: x\(\ge\)-2,x\(\ge\)\(\dfrac{1}{2}\) =>\(\sqrt{4\left(x+2\right)}-\sqrt{2x-1}+\sqrt{9\left(x+2\right)}=0\) \(\Leftrightarrow2\sqrt{x+2}-\sqrt{2x-1}+3\sqrt{x+2}=0\) \(\Leftrightarrow5\sqrt{x+2}-\sqrt{2x-1}=0\) \(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\) <=>25x+50=2x-1 =>23x=-51 =>x=\(-\dfrac{51}{23}\)(ko thỏa mãn dk) => phương trình vô nghiệm.. b) ĐKXĐ:\(x\ge1,x\ge-1\) \(\Leftrightarrow\sqrt{\left(x+1\right)\left(x-1\right)}-3\sqrt{x-1}=0\) \(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}-3\right)=0\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)(nhận) Vậy S={1;8} c) ĐKXĐ: \(x\ge0\) \(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\) \(\Leftrightarrow-11\sqrt{2x}=-11\) \(\Leftrightarrow\sqrt{2x}=1\) \(\Leftrightarrow2x=1\\
\Leftrightarrow x=\dfrac{1}{2}\) Câu a :\(\sqrt{4x+8}-2\sqrt{2x-1}+\sqrt{9x+18}=0\) ( ĐK : \(x\ge\dfrac{1}{2}\) ) \(\Leftrightarrow\sqrt{4x+8}+\sqrt{9x+18}=\sqrt{2x-1}\) \(\Leftrightarrow2\sqrt{x+2}+3\sqrt{x+2}=\sqrt{2x-1}\) \(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\) \(\Leftrightarrow25\left(x+2\right)=2x-1\) \(\Leftrightarrow25x+50=2x-1\) \(\Leftrightarrow23x=-51\) \(\Leftrightarrow x=-\dfrac{51}{23}< -\dfrac{1}{2}\) Vậy phương trình vô nghiệm . Câu b : \(\sqrt{x^2-1}-\sqrt{9\left(x-1\right)}=0\) ( ĐK : \(x\ge1\) ) \(\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-3\sqrt{\left(x-1\right)}=0\) \(\Leftrightarrow\sqrt{\left(x-1\right)}\left(\sqrt{x+1}-3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\) Vậy \(S=\left\{1;8\right\}\) Câu c : \(\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5\) ( ĐK : \(x\ge\dfrac{5}{6}\) ) \(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\) \(\Leftrightarrow-11\sqrt{2x}+11=0\) \(\Leftrightarrow-11\left(\sqrt{2x}-1\right)=0\) \(\Leftrightarrow\sqrt{2x}-1=0\) \(\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\) Vậy \(S=\left\{\dfrac{1}{2}\right\}\) Chúc bạn học tốt Bài 2 : ĐKXĐ : \(\left\{{}\begin{matrix}x\le3\\x\le5\end{matrix}\right.\) => \(x\le3\) Ta có : \(\sqrt{3-x}+\sqrt{5-x}=2\) => \(\sqrt{3-x}=2-\sqrt{5-x}\) => \(3-x=4-4\sqrt{5-x}+5-x\) => \(-4\sqrt{5-x}=-6\) => \(\sqrt{5-x}=\frac{3}{2}\) => \(x=2,75\) ( TM ) Ta có : \(A=\sqrt{3-2,75}-\sqrt{5-2,75}=-1\) Vậy ...
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Bảo Nguyễn Lê Gia
4 tháng 7 2019
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