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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)
b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)
\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)
c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
11:
a: Số cần tìm là 3;2;1;0
b: 0
10:
\(2\dfrac{1}{2}=2.5>2.25\)
\(6>5\)
nên căn 6>căn 5
-căn 5<0
0<căn 5
=>-căn 5<căn 5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[-\sqrt{2,25}+4\sqrt{\left(-2,15\right)^2}-\left(3\sqrt{\dfrac{7}{6}}\right)^2\right]\sqrt{1\dfrac{9}{16}}\)
\(=\left[-1,5+4\sqrt{2,15^2}-9\cdot\dfrac{7}{6}\right]\sqrt{\dfrac{25}{16}}\)
\(=\left[4\cdot\dfrac{43}{20}-10,5-1,5\right]\cdot\dfrac{5}{4}\)
\(=\left[\dfrac{43}{5}-12\right]\cdot\dfrac{5}{4}\)
\(=\dfrac{43}{5}\cdot\dfrac{5}{4}-12\cdot\dfrac{5}{4}\)
\(=\dfrac{43}{4}-15=\dfrac{-17}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)
b: \(M=1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)
\(=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(=1-5\left(\dfrac{1}{14}+\dfrac{1}{84}+\dfrac{1}{204}+\dfrac{1}{374}\right)\)
\(=1-5\left(\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+\dfrac{1}{12\cdot17}+\dfrac{1}{17\cdot22}\right)\)
\(=1-\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+\dfrac{5}{12\cdot17}+\dfrac{5}{17\cdot22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)
\(=1-\dfrac{11-1}{22}=1-\dfrac{10}{22}=\dfrac{12}{22}=\dfrac{6}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)
b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)
=10+3/64
=643/64
c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Xét hiệu:
$\frac{2022}{\sqrt{2023}}+\frac{2023}{\sqrt{2022}}-(\sqrt{2022}+\sqrt{2023})$
$=(\frac{2022}{\sqrt{2023}}-\sqrt{2023})+(\frac{2023}{\sqrt{2022}}-\sqrt{2022})$
$=\frac{2022-2023}{\sqrt{2023}}+\frac{2023-2022}{\sqrt{2022}}$
$=\frac{1}{\sqrt{2022}}-\frac{1}{\sqrt{2023}}>0$
$\Rightarrow \frac{2022}{\sqrt{2023}}+\frac{2023}{\sqrt{2022}}>\sqrt{2022}+\sqrt{2023}$
\(\sqrt{\dfrac{169}{64}}=\sqrt{\dfrac{13^2}{8^2}}=\dfrac{13}{8}\)
\(\dfrac{\sqrt{169}}{\sqrt{64}}=\dfrac{\sqrt{13^2}}{\sqrt{8^2}}=\dfrac{13}{8}\)
Vậy \(\sqrt{\dfrac{169}{64}}=\dfrac{\sqrt{169}}{\sqrt{64}}\)
Tương tự
Bị đao không hai căn bậc bằng nhau hết mà tính làm gì nhìn vô là biết bằng roy :V