Cho mình hỏi \(\sqrt{4}\),\(\sqrt{9}\),\(\sqrt{16}\)
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Em kéo xuống trang 40, mục số 3:
Một số mẹo nhỏ với Casio.pdf - Google Drive
nhưng mà em dùng casio 580vnx nên hơi khó để tách á thầy
\(\sqrt{16-6\sqrt{7}}-\sqrt{32+10\sqrt{7}}.\)
\(=\sqrt{9-6\sqrt{7}+7}-\sqrt{25+10\sqrt{7}+7}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\sqrt{7}^2}-\sqrt{5^2+2.5.\sqrt{7}+\sqrt{7^2}}\)
\(\sqrt{\left(3-\sqrt{7}\right)^2}-\sqrt{\left(5+\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-5-\sqrt{7}=-2-2\sqrt{7}\)
\(\sqrt{17-4}.\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{13}.\sqrt{5+4\sqrt{5}+4}\)
\(=\sqrt{13}\left(\sqrt{5}+2\right)\)
\(=\sqrt{65}+2\sqrt{13}\)
\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)+9\sqrt{x}-4+\left(4x-4\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\)
\(=\dfrac{x+4\sqrt{x}+4x\sqrt{x}+16x-4x-16\sqrt{x}}{x-16}\)
\(=\dfrac{13x+4x\sqrt{x}-12\sqrt{x}}{x-16}\)
\(a,B=4\sqrt{x=1}-3\sqrt{x+1}+2\)\(\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
\(b,\)đưa về \(\sqrt{x+1}=4\Rightarrow x=15\)
a, Với \(x\ge-1\)
\(\Rightarrow B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
b, Ta có B = 16 hay
\(4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\)bình phương 2 vế ta được
\(\Leftrightarrow x+1=16\Leftrightarrow x=15\)
a) \(\sqrt{16}+\sqrt{225}.\sqrt{9}=4+15.3=4+45=49\)
b) \(\sqrt{\dfrac{10000}{400}}+\sqrt{\left(-3\right)^2}.\sqrt{6^4}=\dfrac{100}{20}+\sqrt{9}.\sqrt{36^2}=5+3.36=5+108=113\)
1: \(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
\(=7\cdot\sqrt3+3\sqrt6-4\cdot3\sqrt3\)
\(=3\sqrt6+7\sqrt3-12\sqrt3=3\sqrt6-5\sqrt3\)
2: \(\sqrt{28}-4\cdot\sqrt{63}+7\cdot\sqrt{112}\)
\(=2\sqrt7-4\cdot3\sqrt7+7\cdot4\sqrt7\)
\(=2\sqrt7-12\sqrt7+28\sqrt7=18\sqrt7\)
3: \(\sqrt{49}-5\cdot\sqrt{28}+\frac12\cdot\sqrt{63}\)
\(=7-5\cdot2\sqrt7+\frac12\cdot3\sqrt7=7-10\sqrt7+1,5\sqrt7=7-8,5\cdot\sqrt7\)
4: \(\left(2\sqrt6-4\sqrt3-\frac14\sqrt8\right)\cdot3\sqrt6\)
\(=6\cdot\sqrt{36}-12\sqrt{18}-\frac14\cdot2\cdot\sqrt2\cdot3\sqrt6\)
\(=36-36\sqrt2-\frac32\sqrt{12}=36-36\sqrt2-3\sqrt3\)
6: \(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt3=\sqrt{\frac{48}{3}}-3\cdot\sqrt{\frac{27}{3}}-\sqrt{\frac{147}{3}}\)
\(=\sqrt{16}-3\cdot\sqrt9-\sqrt{49}=4-3\cdot3-7\)
=-3-9
=-12
5: \(\left(2\cdot\sqrt{1\frac{9}{16}}-5\cdot\sqrt{5\frac{1}{16}}\right):\sqrt{16}\)
\(=\left(2\cdot\sqrt{\frac{25}{16}}-5\cdot\sqrt{\frac{81}{16}}\right):\sqrt{16}\)
\(=\left(2\cdot\frac54-5\cdot\frac94\right):4=\frac{10-45}{4\cdot4}=\frac{-35}{16}\)
7: \(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt2-\sqrt{162}:\sqrt2\)
\(=\left(5\sqrt2-21\right):\sqrt2-9\sqrt2:\sqrt2\)
\(=5-\frac{21}{\sqrt2}-9=-4-\frac{21\sqrt2}{2}=\frac{-8-21\sqrt2}{2}\)
8: \(\left(2\cdot\sqrt{1\frac{9}{10}}-\sqrt{5\frac{1}{10}}\right):\sqrt{10}=\left(2\cdot\sqrt{\frac{19}{10}}-\sqrt{\frac{51}{10}}:\sqrt{10}\right)\)
\(=2\cdot\frac{\sqrt{19}}{10}-\frac{\sqrt{51}}{10}=\frac{2\sqrt{19}-\sqrt{51}}{10}\)
9: \(2\cdot\sqrt{\frac{16}{3}}-3\cdot\sqrt{\frac{1}{27}}-6\cdot\sqrt{\frac{4}{75}}\)
\(=2\cdot\frac{4}{\sqrt3}-3\cdot\frac{1}{3\sqrt3}-6\cdot\frac{2}{5\sqrt3}\)
\(=\frac{8}{\sqrt3}-\frac{1}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7\sqrt3}{3}-\frac{12\sqrt3}{15}=\frac{35\sqrt3-12\sqrt3}{15}=\frac{23\sqrt3}{15}\)
10: \(2\sqrt{27}-6\cdot\sqrt{\frac43}+\frac35\cdot\sqrt{75}\)
\(=2\cdot3\sqrt3-6\cdot\frac{2}{\sqrt3}+\frac35\cdot5\sqrt3\)
\(=6\sqrt3-4\sqrt3+3\sqrt3=5\sqrt3\)
11: \(\frac{\sqrt{18}}{\sqrt2}-\frac{\sqrt{12}}{\sqrt3}=\sqrt9-\sqrt4=3-2=1\)
Ta có: \(\sqrt9\cdot y^3-\sqrt{16}=7\cdot\sqrt{121}\)
=>\(3y^3-4=7\cdot11=77\)
=>\(3y^3=81\)
=>\(y^3=27=3^3\)
=>y=3
\(\frac{x+25}{9}=\frac{51-y}{16}=\frac{70+z}{25}\)
=>\(\frac{x+25}{9}=\frac{51-3}{16}=\frac{z+70}{25}\)
=>\(\frac{x+25}{9}=\frac{z+70}{25}=\frac{48}{16}=3\)
=>x+25=27; z+70=75
=>x=2; z=5
a.
\(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\left(x\ge-1\right)\)
\(B=\sqrt{16}.\sqrt{x+1}-\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}+\sqrt{x+1}\)
\(B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(B=\left(4-3+2+1\right).\sqrt{x+1}\)
\(B=4.\sqrt{x+1}\)
b.
\(B=16\\\)
\(\Rightarrow4\sqrt{x+1}=16\)
\(\Rightarrow\sqrt{x+1}=\dfrac{16}{4}=4\)
\(\Rightarrow x+1=4^2\)
\(\Rightarrow x+1=16\rightarrow x=16-1=15\) (thỏa mãn)
vậy x=15
\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)
\(=6\cdot\frac{\sqrt{3}}{2}+10\cdot\frac{2\sqrt{3}}{5}-15\cdot\frac{4}{\sqrt{3}}+9\cdot\frac{2}{\sqrt{3}}\)
\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}=-7\sqrt{3}\)
Trả lời:
\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)
\(=6.\frac{\sqrt{3}}{\sqrt{4}}+10.\frac{\sqrt{12}}{\sqrt{25}}-15.\frac{\sqrt{16}}{\sqrt{3}}+9.\frac{\sqrt{4}}{\sqrt{3}}\)
\(=6.\frac{\sqrt{3}}{2}+10.\frac{\sqrt{2^2.3}}{5}-15.\frac{4}{\sqrt{3}}+9.\frac{2}{\sqrt{3}}\)
\(=3\sqrt{3}+10.\frac{2\sqrt{3}}{5}-15.\frac{4\sqrt{3}}{3}+9.\frac{2\sqrt{3}}{3}\)
\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}\)
\(=\left(3+4-20+6\right).\sqrt{3}=-7\sqrt{3}\)
2 , 3 , 4 . k mình nha
2;3;4 NHA BẠN