1)so sánh các số sau:
a)0,5\(\sqrt{100}\)-\(\sqrt{\dfrac{4}{25}}\) và (\(\sqrt{1\dfrac{1}{9}}\)-\(\sqrt{\dfrac{9}{16}}\)):5
b)\(\sqrt{25+9}\) và \(\sqrt{25}+\sqrt{9}\)
2) CMR: Với a,b dương thì \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
K
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Những câu hỏi liên quan
24 tháng 5 2016
\(0,5\sqrt{100}-\sqrt{\frac{4}{25}}=0,5.10-\frac{\sqrt{4}}{\sqrt{25}}=5-\frac{2}{5}=\frac{23}{5}=\frac{138}{30}\)
\(\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5=\left(\sqrt{\frac{10}{9}-\frac{3}{4}}\right):5=\sqrt{\frac{13}{36}}:5=\frac{\sqrt{13}}{6}:5=\frac{\sqrt{13}}{30}\)
Vì 13 < 138 nên \(\sqrt{13}< 138\Rightarrow\frac{\sqrt{13}}{30}< \frac{138}{30}\)
Vậy \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}>\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5\).
22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
9 tháng 9 2020
Bài 1: Tính
a) Ta có: \(\left(\sqrt{3}+2\right)^2\)
\(=\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot2+2^2\)
\(=3+4\sqrt{3}+4\)
\(=7+4\sqrt{3}\)
b) Ta có: \(-\left(\sqrt{2}-1\right)^2\)
\(=-\left[\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2\right]\)
\(=-\left(2-2\sqrt{2}+1\right)\)
\(=-\left(3-2\sqrt{2}\right)\)
\(=2\sqrt{2}-3\)
Bài 2: Tính
a) Ta có: \(0.5\cdot\sqrt{100}-\sqrt{\frac{25}{4}}\)
\(=\frac{1}{2}\cdot10-\frac{5}{2}\)
\(=5-\frac{5}{2}\)
\(=\frac{5}{2}\)
b) Ta có: \(\left(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}}-\frac{3}{4}\right)\cdot\frac{1}{5}\)
\(=\left(\frac{5}{4}-\frac{3}{4}\right)\cdot\frac{1}{5}\)
\(=\frac{2}{4}\cdot\frac{1}{5}\)
\(=\frac{1}{10}\)
Bài 3: So sánh
a) Ta có: \(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{12}\)
mà \(\sqrt{18}>\sqrt{12}\)(Vì 18>12)
nên \(3\sqrt{2}>2\sqrt{3}\)
\(\Leftrightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
b) Ta có: \(\left(15-2\sqrt{10}\right)^2\)
\(=225-2\cdot15\cdot2\sqrt{10}+\left(2\sqrt{10}\right)^2\)
\(=225-60\sqrt{10}+40\)
\(=265-60\sqrt{10}\)
\(=135+130-60\sqrt{10}\)
Ta có: \(\left(3\sqrt{15}\right)^2=3^2\cdot\left(\sqrt{15}\right)^2=9\cdot15=135\)
Ta có: \(130-60\sqrt{10}\)
\(=\sqrt{16900}-\sqrt{36000}< 0\)(Vì 16900<36000)
\(\Leftrightarrow130-60\sqrt{10}+135< 135\)(cộng hai vế của BĐT cho 135)
\(\Leftrightarrow\left(15-2\sqrt{10}\right)^2< \left(3\sqrt{15}\right)^2\)
\(\Leftrightarrow15-2\sqrt{10}< 3\sqrt{15}\)
\(\Leftrightarrow\frac{15-2\sqrt{10}}{3}< \frac{3\sqrt{15}}{3}=\sqrt{15}\)
hay \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)
KH
21 tháng 9 2017
a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)
=1+3+5+7+9
=25
b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)
=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)
=\(\dfrac{15}{12}\)
c) =0,2+0.3+0,4
= 0.9
d) =9-8+7
=8
j) =1,2-1,3+1.4
= (-0,1)+1,4
=1,4
g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)
= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)
= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)
=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)
= \(\dfrac{71}{20}\)
Nhớ tick cho mk nha~
27 tháng 7 2021
1) \(\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)
2) \(\dfrac{\sqrt{12.5}}{0.5}=\sqrt{\dfrac{12.5}{0.25}}=5\sqrt{2}\)
3) \(\sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\)
4) \(\dfrac{\sqrt{230}}{\sqrt{2.3}}=\sqrt{\dfrac{230}{2.3}}=\sqrt{100}=10\)
5) \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=0\cdot\sqrt{6}=0\)
5 tháng 10 2023
`#3107.101107`
a)
`2/5 \sqrt{25} - 1/2 \sqrt{4}`
`= 2/5 * \sqrt{5^2} - 1/2 * \sqrt{2^2}`
`= 2/5*5 - 1/2*2`
`= 2 - 1`
`= 1`
b)
`0,5*\sqrt{0,09} + 5*\sqrt{0,81}`
`= 0,5*\sqrt{(0,3)^2} + 5*\sqrt{(0,9)^2}`
`= 0,5*0,3 + 5*0,9`
`= 0,15 + 4,5`
`= 4,65`
c)
`2/5\sqrt{25/36} - 5/2\sqrt{4/25}`
`= 2/5*\sqrt{(5^2)/(6^2)} - 5/2*\sqrt{(2^2)/(5^2)}`
`= 2/5*5/6 - 5/2*2/5`
`= 1/3 - 1`
`= -2/3`
d)
`-2 \sqrt{(-36)/(-16)} + 5 \sqrt{(-81)/(-25)}`
`= -2*\sqrt{36/16} + 5*\sqrt{81/25}`
`= -2*\sqrt{(6^2)/(4^2)} + 5*\sqrt{(9^2)/(5^2)}`
`= -2*6/4 + 5*9/5`
`= -3 + 9`
`= 6`
KV
26 tháng 12 2023
a) \(4.\left(-\dfrac{1}{2}\right)^3-2.\left(-\dfrac{1}{2}\right)^2+3.\left(-\dfrac{1}{2}\right)+1\)
\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}+3.\left(-\dfrac{1}{2}\right)+1\)
\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1\)
\(=-\dfrac{3}{2}\)
b) \(8.\sqrt{9}-\sqrt{64}\)
\(=8.3-8\)
\(=24-8\)
\(=16\)
c) \(\sqrt{\dfrac{9}{16}}+\dfrac{25}{46}:\dfrac{5}{23}-\dfrac{7}{4}\)
\(=\dfrac{3}{4}+\dfrac{5}{2}-\dfrac{7}{4}\)
\(=-1+\dfrac{5}{2}\)
\(=\dfrac{3}{2}\)
1.
a. \(0,5\sqrt{100}-\sqrt{\dfrac{4}{25}}=5-\dfrac{2}{5}=\dfrac{23}{5}>1\)
\(\dfrac{\left(\sqrt{1\dfrac{1}{9}}-\sqrt{\dfrac{9}{16}}\right)}{5}=\dfrac{\dfrac{\sqrt{10}}{3}-\dfrac{3}{4}}{5}=\dfrac{-9+4\sqrt{10}}{60}\approx0,06< 1\)
\(\Rightarrow0,5\sqrt{100}-\sqrt{\dfrac{4}{25}}>\dfrac{\left(\sqrt{1\dfrac{1}{9}}-\sqrt{\dfrac{9}{16}}\right)}{5}\)
2.
Ta có:
\(\left(\sqrt{a+b}\right)^2=a+b\)
\(\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}\right)^2+2\sqrt{ab}+\left(\sqrt{b}\right)^2=a+2\sqrt{ab}+b\)
=> \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
1b.
Áp dụng công thức trên
=> \(\sqrt{25+9}< \sqrt{25}+\sqrt{9}\)
2.
\(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\\ \Rightarrow a+b< a+2\sqrt{ab}+b\\ \Rightarrow2\sqrt{ab}>0\\ \Rightarrow\sqrt{ab}>0\)
Luôn đúng với mọi a;b dươn g
=> đpcm