Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)
\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)
\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)
a: \(A=\dfrac{\left(\sqrt{7}+\sqrt{3}\right)^2+\left(\sqrt{7}-\sqrt{3}\right)^2}{4}\)
\(=\dfrac{10+2\sqrt{21}+10-2\sqrt{21}}{4}=\dfrac{20}{4}=5\)
b: \(B=6\sqrt{3}+\sqrt{3}-1-2\sqrt{2}\)
\(=7\sqrt{3}-2\sqrt{2}-1\)
a: Ta có: \(A=\sqrt{8}-2\sqrt{18}+3\sqrt{50}\)
\(=2\sqrt{2}-6\sqrt{2}+15\sqrt{2}\)
\(=11\sqrt{2}\)
b: Ta có: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
\(=5\sqrt{5}-\sqrt{5}+\sqrt{5}-1\)
\(=5\sqrt{5}-1\)
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn.
a) \(4\sqrt{2x+1}-\sqrt{8x+4}+\dfrac{1}{2}\sqrt{32x+16}=12\) (ĐK: \(x\ge-\dfrac{1}{2}\))
\(\Leftrightarrow4\sqrt{2x+1}-\sqrt{4\left(2x+1\right)}+\dfrac{1}{2}\cdot4\sqrt{2x+1}=12\)
\(\Leftrightarrow4\sqrt{2x+1}-2\sqrt{2x+1}+2\sqrt{2x+1}=12\)
\(\Leftrightarrow4\sqrt{2x+1}=12\)
\(\Leftrightarrow\sqrt{2x+1}=\dfrac{12}{4}\)
\(\Leftrightarrow2x+1=3^2\)
\(\Leftrightarrow2x=9-1\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=\dfrac{8}{2}\)
\(\Leftrightarrow x=4\left(tm\right)\)
b) \(\sqrt{4x^2-4x+1}=5\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)
\(\Leftrightarrow\left|2x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\left(x\ge\dfrac{1}{2}\right)\\2x-1=-5\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{2}\\x=-\dfrac{4}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
c) \(\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}=-\dfrac{1}{2}\)(ĐK: \(x\ge0;x\ne1\))
\(\Leftrightarrow-\left(\sqrt{x}-1\right)=2\left(2\sqrt{x}-3\right)\)
\(\Leftrightarrow-\sqrt{x}+1=4\sqrt{x}-6\)
\(\Leftrightarrow4\sqrt{x}+\sqrt{x}=1+6\)
\(\Leftrightarrow5\sqrt{x}=7\)
\(\Leftrightarrow\sqrt{x}=\dfrac{7}{5}\)
\(\Leftrightarrow x=\dfrac{49}{25}\left(tm\right)\)
bài 1:
a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)
\(=-33\sqrt{2}\)
b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
\(=10-2\sqrt{21}+14\sqrt{21}\)
\(=12\sqrt{21}+10\)
Bài 2:
a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)
\(\Leftrightarrow\left|2x+3\right|=8\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}=8\)
hay x=4
c: Ta có: \(\sqrt{9x-9}+1=13\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow x-1=16\)
hay x=17
1: =>x^2-x=3-x
=>x^2=3
=>x=căn 3 hoặc x=-căn 3
2: =>x^2-4x+3=x^2-4x+4 và x>=2
=>3=4(vô lý)
3: =>2|x-1|=6
=>|x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2 hoặc x=4
4: =>|2x-3|=|x-2|
=>2x-3=x-2 hoặc 2x-3=-x+2
=>x=1 hoặc x=5/3
5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)
=>x+2=0
=>x=-2
