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a) \(5^{3n+2}=25^{n+3}\)
\(\Leftrightarrow5^{3n+2}=5^{2\left(n+3\right)}\)
\(\Leftrightarrow3n+2=2\left(n+3\right)\)
\(\Leftrightarrow3n+2=2n+6\)
\(\Leftrightarrow n=4\)
b) \(6.5^{n-2}+10^2=2.5^3\left(n>1\right)\)
\(\Leftrightarrow6.5^{n-2}=2.5^3-10^2\)
\(\Leftrightarrow6.5^{n-2}=2.5^3-2^2.5^2\)
\(\Leftrightarrow6.5^{n-2}=2.5^2\left(5-2\right)\)
\(\Leftrightarrow6.5^{n-2}=2.5^2.3\)
\(\Leftrightarrow5^{n-2}=5^2\)
\(\Leftrightarrow n-2=2\)
\(\Leftrightarrow n=4\)
b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)
\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)
\(=4\left(7+3\sqrt{5}\right)\)
\(=28+12\sqrt{5}\)
Lời giải:
a.
$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$
$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$
$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$
$=2\sqrt{5}-5\sqrt{10}$
$\Rightarrow A=\sqrt{10}-5\sqrt{5}$
b.
$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$
$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$
$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$
$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$
$\Rightarrow B=28+12\sqrt{5}$
c.
$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$
$=(7-5)(6-\sqrt{35})$
$=2(6-\sqrt{35})=12-2\sqrt{35}$
1. \(=\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-2\sqrt{4\sqrt{7}}=\frac{7}{2}+\frac{3}{2}+\frac{7}{2}-\frac{3}{2}-2\sqrt{4\sqrt{7}}\)
\(=7-2\sqrt{4\sqrt{7}}\)
cho hỏi tại sao có số \(\frac{7}{2};\frac{3}{2}\)zậy chỉ với
50) \(\sqrt{98-16\sqrt{3}}=4\sqrt{6}-\sqrt{2}\)
51) \(\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
52) \(\sqrt{4+\sqrt{15}}=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
53) \(\sqrt{5-\sqrt{21}}=\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)
54) \(\sqrt{6-\sqrt{35}}=\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{10}}{2}\)
55) \(\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
56) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
Ngu vl sử dụng công thức để tính trên mtct gà
a) \(\sqrt{11+2\sqrt[]{18}}\)
\(=\sqrt{11+6\sqrt[]{2}}\)
\(=\sqrt{9+2.3\sqrt[]{2}+2}\)
\(=\sqrt{\left(3+\sqrt[]{2}\right)^2}=\left|3+\sqrt[]{2}\right|=3+\sqrt[]{2}\)
b) \(\sqrt[]{7+2\sqrt[]{10}}\)
\(=\sqrt[]{7+2\sqrt[]{5}.\sqrt[]{2}}\)
\(=\sqrt[]{5+2\sqrt[]{5}.\sqrt[]{2}+2}\)
\(=\sqrt[]{\left(\sqrt[]{5}+\sqrt[]{2}\right)^2}=\left|\sqrt[]{5}+\sqrt[]{2}\right|=\sqrt[]{5}+\sqrt[]{2}\)
c) \(\sqrt[]{7+4\sqrt[]{3}}\)
\(=\sqrt[]{4+2.2\sqrt[]{3}+3}\)
\(=\sqrt[]{\left(2+\sqrt[]{3}\right)^2}=\left|2+\sqrt[]{3}\right|=2+\sqrt[]{3}\)
d) \(\sqrt[]{16-2\sqrt[]{55}}\) \(\left(12\rightarrow16\right)\)
\(=\sqrt[]{11-2\sqrt[]{5}.\sqrt[]{11}+5}\)
\(=\sqrt[]{\left(\sqrt[]{11}-\sqrt[]{5}\right)^2}==\left|\sqrt[]{11}-\sqrt[]{5}\right|=\sqrt[]{11}-\sqrt[]{5}\left(\sqrt[]{11}>\sqrt[]{5}\right)\)
a) \(\left(-\dfrac{1}{3}\sqrt{63}\right)^2=\dfrac{1}{9}\cdot63=7\)
\(\left(-2\sqrt{2}\right)^2=8\)
mà 7<8
nên \(-\dfrac{1}{3}\sqrt{63}>-2\sqrt{2}\)
b) Ta có: \(\left(2\sqrt{55}\right)^2=4\cdot55=220\)
\(\left(\dfrac{3}{5}\sqrt{750}\right)=\dfrac{9}{25}\cdot750=270\)
mà 220<270
nên \(2\sqrt{55}< \dfrac{3}{5}\sqrt{750}\)
hay \(-2\sqrt{55}< -\dfrac{3}{5}\sqrt{750}\)
1012
ai thấy đúng ủng hộ vài **** đê ~~ ☺☺
1012, 272727 chứ
ai thấy đúng ủng hộ vài **** đê ~~ ☺☺