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a)\(\sqrt{7.63}\)=21
b)\(\sqrt{2,5.30.48}\)=60
c)\(\sqrt{0,4.6,4}\)=1,6
d)\(\sqrt{2,7.5.1,5}\)=4,5
TRẢ LỜI :
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}\)
c) √20 - √45 + 3√18 + √72
= √4.5 - √9.5 + 3√9.2 + √36.2
= 2√5 - 3√5 + 9√2 + 6√2
= -√5 + 15√2
+ Ta có:
2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)
=2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5
=2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).
+ Ta có:
3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)
=3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7
=3(√10−√7)3=√10−√7=3(10−7)3=10−7.
+ Ta có:
1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)
=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y
+ Ta có:
2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)
=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
a) 2 \sqrt{6}, \sqrt{29}, 4 \sqrt{2}, 3 \sqrt{5} ;26,29,42,35;
b) \sqrt{38}, 2 \sqrt{14}, 3 \sqrt{7}, 6 \sqrt{2}38,214,37,62
a) \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
b) \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)
a) (\(\sqrt{3}\)-1)2=3-2\(\sqrt{3}\)+1= 4-2\(\sqrt{3}\) (ĐPCM)
b) \(\sqrt{4-2\sqrt{3}}\)=\(\sqrt{3}\)-1 >0
Bình phương 2 vế, ta có:
4-2\(\sqrt{3}\)=3-2\(\sqrt{3}\)+1= 4-2\(\sqrt{3}\) (ĐPCM)
a) \(\left(\sqrt{3}-1\right)^2\)=\(\left(\sqrt{3}\right)^2\)- 2\(\sqrt{3}\) +1= 3- 2\(\sqrt{3}\) +1=4-2\(\sqrt{3}\)
b) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\) = \(\sqrt{\left(\sqrt{3}-1\right)^2}\) - \(\sqrt{3}\)= \(|\sqrt{3}-1|\)-\(\sqrt{3}\)=\(\sqrt{3}\)-1-\(\sqrt{3}\)=-1
a
căn có nghĩa
\(\Leftrightarrow\frac{a}{3}\ge0\)
\(\Leftrightarrow a\ge0\)
b
căn có nghĩa
\(\Leftrightarrow-5a\ge0\)
\(\Leftrightarrow b\le0\left(-5\le0\right)\)
c
căn có nghĩa
\(\Leftrightarrow4-a\ge0\)
\(\Leftrightarrow-a\ge0-4\)
\(\Leftrightarrow-a\ge-4\)
\(\Leftrightarrow a\le4\)
d
căn có nghĩa
\(\Leftrightarrow3a+7\ge0\)
\(\Leftrightarrow a\ge-\frac{7}{3}\)
a) \sqrt{-9a}-\sqrt{9+12 a+4 a^{2}}−9a−9+12a+4a2
=\sqrt{-9 a}-\sqrt{3^{2}+2.3 .2 a+(2 a)^{2}}=−9a−32+2.3.2a+(2a)2
=\sqrt{3^{2} \cdot(-a)}-\sqrt{(3+2 a)^{2}}=32⋅(−a)−(3+2a)2
=3 \sqrt{-a}-|3+2 a|=3−a−∣3+2a∣
Thay a=-9a=−9 ta được:
3 \sqrt{9}-|3+2 \cdot(-9)|=3.3-15=-639−∣3+2⋅(−9)∣=3.3−15=−6.
b) Điều kiện: m \neq 2m=2
a) \(\sqrt{x^2}\)=7
=> x2=49
=> x={-7;7}
b) \(\sqrt{x^2}\)=|-8|=8
=> x2=64
=>x={-8;8}
c) \(\sqrt{4x^2}\)=6
4x2=36
=>x2=9
=> x={-3;3}
d)\(\sqrt{9x^2}\)=|-12|=12
=> 9x2=144
=> x2=16
=> x={-4;4}
a)x=+7 hoặc x= -7
b) x=8 hoặc x= -8
c)x=3 hoặc x =-3
d) x=4 hoặc x= -4
a, \(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)do \(0,1>0\)
b, \(\sqrt{\left(-0,3\right)^2}=\sqrt{\left(0,3\right)^2}=\left|0,3\right|=0,3\)do \(0,3>0\)
c, \(-\sqrt{\left(-1,3\right)^2}=-\sqrt{\left(1,3\right)^2}=-\left|1,3\right|=-1,3\)do \(1,3>0\)
d, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\sqrt{\left(0,4\right)^2}=-0,4.\left|0,4\right|=-0,4.0,4=-0,14\)
do \(0,4>0\)
\(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)
\(\sqrt{\left(-0,3\right)^2}=\left|-0,3\right|=0,3\)
\(-\sqrt{\left(-1,3\right)^2}=-\left|-1,3\right|=-1,3\)
\(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,16\)
\(\sqrt{\left(0.1\right)^2^{ }}=0.1\)
45
\(\sqrt{\left(0.1\right)^2=0.1}\) TT TA CÓ 0.3 ;1.3 -0.16
a) \(\sqrt{\left(0,1\right)^2}\) = \(|0,1|\) = 0,1
b) \(\sqrt{\left(-0,3\right)^2}\) = \(|\)-0,3\(|\)
c) \(-\sqrt{\left(-1,3\right)^2}\) = -\(|\)-1,3\(|\) = -1,3
d) \(-0,4\sqrt{\left(-0,4\right)^2}\) = -0,4\(|\)-0,4\(|\)=-0,14
a) √(0,1)^2 = | 0,1| = 0,1
b) √(-0,3)^2 = |-0,3|= 0,3
c) -√(-1,3)^2 =| -1,3| = 1,3
d) -0,4√(-0,4)^2 = | -√0,4. (-0,4) | = -2√10/25
a) \(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)
b) \(\sqrt{\left(-0,3\right)^2}=\left|-0,3\right|=0,3\)
c) \(-\sqrt{\left(-1,3\right)^2}=-\left|-1,3\right|=-1,3\)
d) \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\left|-0,4\right|=-0,4.0,4=-0,16\)
a) 0,1
b) 0,3
c) -1,3
d) -0,16
a, vì 0,1 > 0 => A=0,1
b, vì -0.3 < 0 => B= /-0,3/ = 0,3
c, vì -1,3 <0 => B= -/-1,3/ = 1,3
d, vì -0,4 < 0 => D = -0,4./0,4/ = -0,14
a) \(\sqrt{\left(0,1\right)^2}\) = \(|0,1|=0,1\)
b) \(\sqrt{\left(-0,3\right)^2}=|-0,3|=0,3\)
c) \(-\sqrt{\left(-1,3\right)^2}=-|-1,3|=-1,3\)
d) \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4|-0,4|=-0,4.0,4=-0,16\)
a) 0,1; b) 0,3; c) -1,3; d) -0,16
a)\(\sqrt{\left(0,1\right)^2}=|0,1|=0,1 \)
b)\(\sqrt{\left(-0,3\right)^2}=|-0,3|=0,3\)
c)\(-\sqrt{\left(1,3\right)^2}=|-1,3|=1,3\)
d)\(-0,4\sqrt{\left(-0,4\right)^2}=-0,4.|-0,4|=-0,16\)
a.0,1
b.0,3
c.-1,3
d.-0,16
a) \sqrt{(0,1)^2}=|0,1|=0,1(0,1)2=∣0,1∣=0,1.
b) \sqrt{(-0,3)^2}=|-0,3|=0,3(−0,3)2=∣−0,3∣=0,3.
c) -\sqrt{(-1,3)^2}=-|-1,3|=-1,3−(−1,3)2=−∣−1,3∣=−1,3.
d) -0,4\sqrt{(-0,4)^2}=-0,4|-0,4|=-0,4.0,4=-0,14−0,4(−0,4)2=−0,4∣−0,4∣=−0,4.0,4=−0,14.
a, \(\sqrt{\left(0,1\right)^2}\) = \(|\)0,1\(|\) = 0,1
b, \(\sqrt{\left(-0,3\right)^2}\) = \(|-0,3|\) = 0,3
c, -\(\sqrt{\left(-1,3\right)^2}\) = -\(\left|-1,3\right|\) = -1,3
d, -0,4\(\sqrt{\left(-0,4\right)^2}\) = -0,4.\(\left|-0,4\right|\) = -0,4. 0,4 = -0,16
a)\(\sqrt{\left(0,1\right)^2}=0,1\)
b√(−0,3)2(−0,3)2 =0,3
c) -1,3
d)-0,16
a) 0,1 b) 0,3 c) -1,3 d) -0,16
a. \(\sqrt{\left(0,1\right)^2}\) = |0,1| = 0,1
b.\(\sqrt{\left(-0,3\right)^2}\) = |-0,3| = 0,3
c.\(-\sqrt{\left(-1,3\right)^2}\) = -|-1,3| = -1,3
d. \(-0,4\sqrt{\left(-0,4\right)^2}\) = -0,4|-0,4| = -0,4.0,4 = -0,14
a, 0,1
b,0,3
c, -1,3
d, -0,6
a) \(\sqrt{\left(0,1\right)^2}\) =\(\)∣0,1∣=0,1
b)\(\sqrt{\left(-0,3\right)^2}\)=∣−0,3∣=0,3
c)-\(\sqrt{\left(-1,3\right)}^2\) =−∣−1,3∣=−1,3
d)-0,4\(\sqrt{\left(-0,4\right)^2}\)=−0,4∣−0,4∣=−0,4.0,4=−0,14
a,0,1 b,0,3 c,-1,3 d,-0,16
a) 0.1
b) 0.3
c) -1.3
d) -0.16
a) √(0,1)2=|0,1|=0,1(0,1)2=|0,1|=0,1.
b) √(−0,3)2=|−0,3|=0,3(−0,3)2=|−0,3|=0,3.
c) −√(−1,3)2=−|−1,3|=−1,3−(−1,3)2=−|−1,3|=−1,3.
d) −0,4√(−0,4)2=−0,4|−0,4|=−0,4.0,4=−0,14−0,4(−0,4)2=−0,4|−0,4|=−...