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a)\(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=\sqrt{\left(2a-6\right)^2}=2a-6\)
b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=\sqrt{\left[3\left(b-2\right)\right]^2}=3b-6\)
c) bạn xem lại đề
d)
\(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\sqrt{\left(15a\right)^2}-3a=15a-3a=12a\)
e) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{\sqrt{16}}{\sqrt{x^2}}=\dfrac{4}{x}\)
\(A=\sqrt{9.3.3.16\left(1-a^2\right)}=3.3.4.\left|1-a\right|=36\left(a-1\right)\)
\(B=\frac{1}{a-b}a^2.\left|a-b\right|=\frac{a^2\left(a-b\right)}{a-b}=a^2\)
\(C=\sqrt{5.45.a^2}-3a=\sqrt{5^2.3^2.a^2}-3a=15\left|a\right|-3a=15a-3a=12a\)
\(D=\left(3-a\right)^2-\sqrt{\frac{2.180}{10}a^2}=\left(3-a\right)^2-6\left|a\right|\)
a,\(\sqrt{4\left(a-5\right)^2}=\sqrt{4}.\sqrt{\left(a-5\right)^2}=2.\left|a-5\right|=2\left(a-5\right)\left(a\ge5\right)\)
b,\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3=-1}\)
c,Mạn phép sửa đề ,nếu ko thì kết quả ko đẹp
\(\sqrt{8+2\sqrt{15}}-\sqrt{5}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{5}=\sqrt{5}+\sqrt{3}-\sqrt{5}=\sqrt{3}\)
d,\(\sqrt{\left(3-2\sqrt{3}\right)^2}-\sqrt{\left(3+2\sqrt{3}\right)^2}=2\sqrt{3}-3-3-2\sqrt{3}=-6\)
e,\(\sqrt{24\left(b-3\right)}^2=\sqrt{24^2}.\sqrt{\left(b-3\right)^2}=24.\left(3-b\right)\left(b< 3\right)\)
a/ \(\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}\)
\(=\sqrt{\frac{2a}{3}\cdot\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\sqrt{\frac{a^2}{2^2}}=\sqrt{\left(\frac{a}{2}\right)^2}=\left|\frac{a}{2}\right|\)
mak ta có \(a\ge0\)
\(\Rightarrow\left|\frac{a}{2}\right|=\frac{a}{2}\)\(\Rightarrow\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}=\frac{a}{2}\)
b/ \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}\)
\(=\sqrt{13a\cdot\frac{52}{a}}=\sqrt{\frac{13a\cdot52}{a}}=\sqrt{13\cdot52}=\sqrt{13\cdot13\cdot4}=\sqrt{13^2\cdot2^2}=\sqrt{\left(13\cdot2\right)^2}=13\cdot2=26\)
c/ \(\sqrt{5a}\cdot\sqrt{45}-3a\)
\(=\sqrt{5a\cdot45a}-3a=\sqrt{5a\cdot5a\cdot9}-3a\)
\(=\sqrt{5^2\cdot a^2\cdot3^2}-3a=\left|5\cdot a\cdot3\right|-3a\)
\(=15\left|a\right|-3a\)
Có \(a\ge0\Rightarrow\left|a\right|=a\)
\(\Rightarrow15\left|a\right|-3a=15a-3a=12a\)
\(\Rightarrow\sqrt{5a}\cdot\sqrt{45}-3a=12a\)
d/ \(\left(3-a\right)^2-\sqrt{0,2}\cdot\sqrt{180a^2}\)
\(=\left(3-a\right)^2-\sqrt{0,2\cdot180a^2}\)
\(=\left(3-a\right)^2-\sqrt{0,2\cdot9\cdot2\cdot10\cdot a^2}\)
\(=\left(3-a\right)^2-\sqrt{4\cdot9\cdot a^2}\)
\(=\left(3-a\right)^2-\sqrt{2^2\cdot3^2\cdot a^2}\)
\(=\left(3-a\right)^2-\left|2\cdot3\cdot a\right|\)
\(=\left(3-a\right)^2-6\left|a\right|=9-6a+a^2-6\left|a\right|\)
Chia làm 2 Trường Hợp:
+ TH1 : \(9-6a+a^2-6a=9-12a+a^2\left(a\ge0\right)\)
+ TH2 : \(9-6a+a^2-\left(-6a\right)=9+a^2\left(a< 0\right)\)
a) \(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=2\sqrt{\left(a-3\right)^2}=2.\left|a-3\right|=2\left(a-3\right)=2a-6\) (Vì \(a\ge3\) )
b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=3\sqrt{\left(b-2\right)^2}=3\left|b-2\right|=3\left(2-b\right)\)
\(=6-3b\) (vì b < 2 )
b) \(\sqrt{27.48\left(1-a\right)^2}=\sqrt{27.3.16.\left(1-a\right)^2}=\sqrt{81.16.\left(1-a\right)^2}\)
\(=\sqrt{9^2.4^2.\left(1-a\right)^2}=9.4\sqrt{\left(1-a\right)^2}=36.\left|1-a\right|=36\left(1-a\right)=36-36a\) (vì a > 1)
a) \(\sqrt{\left(\sqrt{7-2}\right)^2}=\sqrt{5}\)
b)\(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-3\sqrt{2}\right)^2}\)
=\(\sqrt{2}-1-2+3\sqrt{2}=4\sqrt{2}-3\)
c)\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
=\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=2\sqrt{3}\)
d) hình như bn ghi sai
e)\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}\right):\sqrt{2}\)
=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{3}+1}\right):\sqrt{2}\)
=\(\dfrac{\sqrt{2+\sqrt{3}}\left(\sqrt{3}+1\right)+\sqrt{2-\sqrt{3}}\left(\sqrt{3}-1\right)}{2\sqrt{2}}\)
=\(\dfrac{\sqrt{6+3}+\sqrt{2+\sqrt{3}}+\sqrt{6-3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{2+\sqrt{3}}+\sqrt{3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)
=\(\dfrac{3+\sqrt{3}}{2\sqrt{2}}\)
f) \(\sqrt{9a^2}+3a-7=-3a+3a-7=-7\)
g)\(\dfrac{\sqrt{4x^2-4x+1}}{4x-2}+3x+2\)
=\(\dfrac{\sqrt{\left(2x-1\right)^2}}{4x-2}+3x+2=\dfrac{2x-1}{2\left(2x-1\right)}+3x+2\)
=\(\dfrac{1}{2}+3x+2=\dfrac{5}{2}+3x\)
h)\(\sqrt{\left(5a-1\right)^2}+2a-3\)
nếu a<0 :\(-5a+1+2a-3=-3a-2\)
nếu a>0 : \(5a-1+2a-3=7a-4\)
i)\(\sqrt{\dfrac{2a}{5}}.\sqrt{\dfrac{5a}{18}}+2\left(a-1\right)\)
=\(\sqrt{\dfrac{10a^2}{90}}+2a-2=\sqrt{\dfrac{a^2}{9}}+2a-2\)
=\(\dfrac{a}{3}+2a-2=\dfrac{7a}{3}-2\)
\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)
2)
\(\sqrt{12,1.360}=\sqrt{12,1}.\sqrt{36}.\sqrt{10}\)
\(=\sqrt{12,1.36.10}\)
= \(\sqrt{121.36}\)
\(=\sqrt{4356}\)
\(=66\)
3)
\(\sqrt{5a}.\sqrt{45a}-3a\)
\(=\sqrt{5.45a^2}-3a\)
\(=\sqrt{225a^2}-3a\)
\(=\sqrt{\left(15a\right)^2}-3a\)
\(=-15a-3a\) ( vì \(a\le0\))
\(=-18a\)
5)
\(\sqrt{0,36a^2}\)
\(=\sqrt{\left(0,6a\right)^2}\)
\(=-0,6a\) ( vì \(a< 0\) )
Để tối mình rảnh lên coi có làm tiếp được nữa hông thì mình làm ha.
Chúc bạn học tốt!
1)
\(\sqrt{3a^3}.\sqrt{12}\)
\(=\sqrt{3}.\sqrt{a^3}.\sqrt{12}\)
\(=\sqrt{3.12}.\sqrt{a^3}\)
\(=6\sqrt{a^3}\)
4)
\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)
\(=9.6a.a^2-\sqrt{0,2}.\sqrt{18}.\sqrt{10}.\sqrt{a^2}\)
\(=54a^3-\sqrt{2}.\sqrt{18}.\sqrt{a^2}\)
\(=34a^3-\sqrt{2.18}.\sqrt{a^2}\)
\(=54a^3-6\sqrt{a^2}\)
\(=54a^3-6a^2\) ( vì a<0)
6)
\(\sqrt{a^4.\left(3-a^{ }\right)^2}\)
\(=\sqrt{\left(a^2\right)^2.\left(3-a\right)^2}\)
\(=\sqrt{\left(a^2\right)^2}.\sqrt{\left(3-a\right)^2}\)
\(=\left|a^2\right|\left|3-a\right|\) ( vì a>3 => a>3 nên 3-a<0)
Mà\(\left|3-a\right|=-\left(-3-a\right)=-3+a=a-3\)
\(=a^2\left(a-3\right)\)
\(=a^3-3a^2\)
Còn lại bạn làm tương tự nha, trể quá rùi :)))))