Bạn cần viết đề bài bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn. Viết như hiện tại rất khó nhìn.
a.\(\sqrt{5}\)-5
b.x+2\(\sqrt{x}\)
c.12-6\(\sqrt{3}\)
d.28+6\(\sqrt{3}\)
e.4+2\(\sqrt{15}\)
Tính thu gọn :
a , \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
b , \(\sqrt{27+12\sqrt{5}}-\sqrt{27-12\sqrt{5}}\)
c , \(\sqrt{15-6\sqrt{6}}+\sqrt{15+\sqrt{6\sqrt{6}}}\)
d , \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
e , \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
f , \(\sqrt{5+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
Giải phương trình:
a. \(3\sqrt{8x}-\sqrt{32x}+\sqrt{50x}=21\)
b. \(\sqrt{25x+50}+3\sqrt{4x+8}-2\sqrt{16x+32}=15\)
c. \(\sqrt{\left(x-2\right)^2}=12\)
d. \(\sqrt{x^2-6x+9}-3=5\)
e.\(\sqrt{\left(2x-1\right)^2}-x=3\)
f. \(\sqrt{3x-6}-x=-2\)
h. \(\sqrt{3-2x}-2=x\)
Đề bài : RÚT GỌN
√15 - √12 1
a) A = ----------------- - -----------
√5 - √2 2 - √3
b) B = 6 √8/9 - 5 √ 32/25 + 14 √ 18/49
c) C = √15 - √5 5 - 2√5
-------------------- -- ----------------
√3 - 1 2√5 - 4
d) E = ( 4 + √15 ) ( √10 - √6 ) √4- √15 ( Đoạn chỗ căn 4 kéo dài hết căn 15 , là đoạn ngoặc cuối )
Tính:
a) \(\left(2\sqrt{5}-\sqrt{7}\right).\left(2\sqrt{5}+\sqrt{7}\right)\)
b)\(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right).\sqrt{3}\)
c)\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
d)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
e)\(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
g)\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
Bài 1: Rút gọn
a)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\) ,
b)\(\sqrt{4+\sqrt{15}}\)+\(\sqrt{4-\sqrt{15}}\)-\(2\sqrt{3-\sqrt{5}}\)
c)A=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
d)B=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
e)C=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f)D= \(\dfrac{\left(5+4\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Tính
a) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}\)
b) \(\left(2\sqrt{3}+4\right)\left(\sqrt{3}-2\right)\)
c) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}\)
d) \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}\)
e) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)
f) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)
D=√147 +√54 -4√27
A =√112 - 7√(1/7) - 14√(1/28)-21/√7
B = 3√2(4 -√2) + 3(1 - 2√2)^2
C = 2√27 + 5√12 - 3√48
E = (√15 - 2√3)^2 + 12√5
F = 3√50 - 7√8 + 12√18
G = 2√80 - 2√245 + 2√180
H =√28 - 4√63 + 7√112
I =√44 -√176 + 2√275
Bài 1: Rút gọn
a)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\) ,
b)\(\sqrt{4+\sqrt{15}}\)+\(\sqrt{4-\sqrt{15}}\)-\(2\sqrt{3-\sqrt{5}}\)
c)A=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
d)B=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
e)C=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f)D=\(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Mình sửa lại để m.n dễ nhìn hơn!
1, \(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6+\sqrt{6}}{\sqrt{6}}\)
2, \(\dfrac{6-6\sqrt{3}}{1-\sqrt{3}}+\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)
3, \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
4, \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
5, \(\left(\dfrac{3\sqrt{125}}{15}-\dfrac{10-4\sqrt{5}}{\sqrt{5}-2}\right)\cdot\dfrac{1}{\sqrt{5}}\)
