\(\sqrt{0,16.0,65.225}\)
\(\sqrt{250.360}\)
\(\sqrt{0,16.0,65.225}\)
\(\sqrt{250.360}\)
a,\(\sqrt{0,16.0,65,225}\)
\(=\sqrt{0,16}.\sqrt{0,65}.\sqrt{225}\)
\(=0,4.25.\sqrt{0,65}=10.\sqrt{0,65}\)
\(=\sqrt{6,5}\)
b, \(\sqrt{250.360}\)
\(=\sqrt{250}.\sqrt{360}=\sqrt{25}.\sqrt{10}.\sqrt{36}.\sqrt{10}\)
\(=5.10.6=300\)
Chúc bạn học tốt!!!
\(\sqrt{20}.\sqrt{72}.\sqrt{4,9}\)
\(\sqrt{20}\cdot\sqrt{72}\cdot\sqrt{4,9}\)
\(=\sqrt{7056}\)
\(=84\)
\(\sqrt{20}.\sqrt{72}.\sqrt{4,9}=\sqrt{20.72.4,9}=\sqrt{7056}=84\)
đề bài: rút gọn
√108x^3 × √3x (x>=0)
\(\sqrt{108x^3}.\sqrt{3x}=\sqrt{108x^3.3x}=\sqrt{324x^4}\)
\(=18x^2\) (do \(x\ge0\))
Chúc bạn học tốt!!!
rút gọn
√108x^3 × √3x (x>=0)
\(\sqrt{108x^3}.\sqrt{3x}=\sqrt{108x^3.3x}=\sqrt{324x^4}=18x^2\)( vì x >= 0)
Chứng minh:
a, \(\sqrt{\sqrt{3}+2\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt{3}-2\sqrt{\sqrt{3}-1}}=2\)
b, \(\sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-4\sqrt{x-4}}=1\)
a,
\(\sqrt{\sqrt{3}+2\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt{3}-2\sqrt{\sqrt{3}-1}}\\ =\sqrt{\sqrt{3}-1+2\sqrt{\sqrt{3}-1}+1}+\sqrt{\sqrt{3}-1-2\sqrt{\sqrt{3}-1}+1}\\ =\sqrt{\left(\sqrt{\sqrt{3}-1}+1\right)^2}+\sqrt{\left(1-\sqrt{\sqrt{3}-1}\right)^2}\\ =\sqrt{\sqrt{3}-1}+1+1-\sqrt{\sqrt{3}-1}\\ =2\)
b.
\(\sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-4\sqrt{x-4}}\\ =\sqrt{x-4-2\sqrt{x-4}+1}-\sqrt{x-4-4\sqrt{x-4}+4}\\ =\sqrt{\left(\sqrt{x-4}-1\right)^2}-\sqrt{\left(\sqrt{x-4}-2\right)^2}\\ =\sqrt{x-4}-1-\sqrt{x-4}+2\\ =1\left(đpcm\right)\)\
a.\(\sqrt{13^2-12^2}\)
b.\(\sqrt{17^2-8^2}\)
c.\(\sqrt{117^2-108^2}\)
d.\(\sqrt{313^2-312^2}\)
a, \(\sqrt{13^2-12^2}=\sqrt{\left(13-12\right)\left(13+12\right)}\)
\(=\sqrt{1.25}=\sqrt{25}=5\)
b, \(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}\)
\(=\sqrt{9.25}=\sqrt{9}.\sqrt{25}=3.5=15\)
c, \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}\)
\(=\sqrt{9.225}=\sqrt{9}.\sqrt{225}=3.15=45\)
d, \(\sqrt{313^2-312^2}=\sqrt{\left(313-312\right)\left(313+312\right)}\)
\(=\sqrt{1.625}=\sqrt{625}=25\)
Chúc bạn học tốt!!!
a, \(\sqrt{13^2-12^2}=\sqrt{\left(13-12\right)\left(13+12\right)}=\sqrt{25}=5\)
b, \(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}=\sqrt{9.25}=15\)
c, \(\sqrt{117^2-108^2}=\sqrt{\left(117-108\right)\left(117+108\right)}\)
\(=\sqrt{9.225}=45\)
d, \(\sqrt{313^2-312^2}=\sqrt{\left(313-312\right)\left(313+312\right)}=\sqrt{625}=25\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
\(=\dfrac{\left(10+2\sqrt{10}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\)
\(=\dfrac{10\sqrt{5}-10\sqrt{2}+2\sqrt{50}-2\sqrt{20}}{3}+\left[-2\left(1+\sqrt{5}\right)\right]\)
\(=\dfrac{10\sqrt{5}-10\sqrt{2}+10\sqrt{2}-2\sqrt{20}}{3}+\left[-2-2\sqrt{5}\right]\)
\(=\dfrac{10\sqrt{5}-4\sqrt{5}}{3}-2-2\sqrt{5}\)
\(=\dfrac{6\sqrt{5}}{3}-2-2\sqrt{5}\)
\(=2\sqrt{5}-2-2\sqrt{5}\)
\(=-2\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}=2\sqrt{5}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=2\sqrt{5}-2-2\sqrt{5}=-2\)
Vậy \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=-2\)
A)\(\left(3-2\sqrt{2}\right).\left(3+2\sqrt{2}\right)\) B) \(\sqrt{\left(\sqrt{3}-2\right)}^2-\sqrt{\left(\sqrt{3}+2\right)}^2\) C)\(\sqrt{3-2\sqrt[]{2}}-\sqrt{3+2\sqrt{2}}\)
D)\(\left(1+\sqrt{3}-\sqrt{2}\right).\left(1+\sqrt{3}+2\right)\)
E) \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\) F)\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
H)\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
b và c.... ok!
b) \(\sqrt{\left(\sqrt{3}-2\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}=\left(\sqrt{3}-2\right)-\left(\sqrt{3}+2\right)=-4\)
nãy nhìn không kĩ nên mới nói là bình phương lên,sorry nhak
c) Đặt \(C=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
ta có: \(C^2=3-2\sqrt{2}+3+2\sqrt{2}-2=4\)
=> \(C=-\sqrt{2}\) (vì \(\sqrt{3-2\sqrt{2}}< \sqrt{3+2\sqrt{2}}\))
a) hằng đẳng thức số 3 (hiệu 2 bình phương)
b) bình phương cả cái biểu thức đó lên, tính bình thường
c) bình phương cả lên như câu b
d) giống câu a
e) hẳng đẳng thức số 1
f) phá căn ra (biến đổi biểu thức trong căn thành hằng đẳng thức số 1 hoặc 2)
h) nghi là hằng đẳng thức số 1 hoặc số 2, từ từ lát nữa tớ xem
khó hiểu chỗ nào thì hỏi nhé
câu h tớ chỉ biết thế này thôi, mà ko biết là làm vậy có được ko nữa ^^!
\(H=\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{8\sqrt{3}}\)\(=5\sqrt{8\sqrt{3}}-2\sqrt{50\sqrt{3}}\)
A)\(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\)
B)\(\left(\sqrt{2}+1^{ }\right)^3-\left(\sqrt{2}-1\right)^3\) C)\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\) D)\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\) E)\(\dfrac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\) F)\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
a) \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{2-\sqrt{3}}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+2\sqrt{12}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+4\sqrt{3}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(8+4\sqrt{3}\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}\)
\(=\sqrt{\left(4-3\right)\cdot4}\)
\(=\sqrt{1\cdot4}\)
\(=\sqrt{4}\)
\(=2\)
b) \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(5\sqrt{2}-7\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-5\sqrt{2}+7\)
\(=0+14\)
\(=14\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
dài quá ==' cả d, e, f nữa ==' có j rảnh lm cho nhé :D
\(\dfrac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
\(\dfrac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
\(=\dfrac{5+4\sqrt{5}+4-8\sqrt{5}}{2\sqrt{5}-4}\)
\(=\dfrac{9-4\sqrt{5}}{2\sqrt{5}-4}\)
\(=\dfrac{\left(9-4\sqrt{5}\right)\left(2\sqrt{5}+4\right)}{4}\)
\(=\dfrac{\left(9-4\sqrt{5}\right)\cdot2\left(\sqrt{5}+2\right)}{4}\)
\(=\dfrac{\left(9-4\sqrt{5}\right)\left(\sqrt{5}+2\right)}{2}\)
\(=\dfrac{9\sqrt{5}+18-20-8\sqrt{5}}{2}\)
\(=\dfrac{\sqrt{5}-2}{2}\)