Tính
a,\(\sqrt{4+\sqrt{15}}\)
b,\(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}\)
c,\(\left(\sqrt{5}+\sqrt{7}\right)\sqrt{12-2\sqrt{35}}\)
Tính
A= \(\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
B=\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
C=\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
a: Ta có: \(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1-2-\sqrt{3}\)
=-3
b: Ta có: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
=1
c: Ta có: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
Rút gọn :
\(A=\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
\(B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{5}}}}\)
\(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(D=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(E=\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
A = \(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
A = \(\dfrac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
A = \(\dfrac{\sqrt{3}+\sqrt{2}+3-\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{5}+1-\left(\sqrt{5}+\sqrt{2}\right)}\)
A = \(\dfrac{\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}-\sqrt{2}}\) = \(\dfrac{3}{1}\) = \(3\)
C = \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
C = \(\left(4+\sqrt{15}\right).\left(\sqrt{40-10\sqrt{15}}-\sqrt{24-6\sqrt{15}}\right)\)
C = \(\left(4+\sqrt{15}\right)\left(\sqrt{\left(5-\sqrt{15}\right)^2}-\sqrt{\left(\sqrt{15}-3\right)^2}\right)\)
C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\left(\sqrt{15}-3\right)\right)\)
C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\sqrt{15}+3\right)\)
C = \(\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
C = \(32-8\sqrt{15}+8\sqrt{15}-30=2\)
D = \(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
D = \(\left(\sqrt{30-10\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)\left(3+\sqrt{5}\right)\)
D = \(\left(\sqrt{\left(5-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\right)\left(3+\sqrt{5}\right)\)
D = \(\left(5-\sqrt{5}-\left(\sqrt{5}-1\right)\right)\left(3+\sqrt{5}\right)\)
D = \(\left(5-\sqrt{5}-\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\)
D = \(\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
D = \(18+6\sqrt{5}-6\sqrt{5}-10=8\)
E = \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{5}}\)
E = \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
E = \(3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
1)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
2)\(\sqrt{35+12\sqrt{6}}-\sqrt{35-12\sqrt{6}}\)
3)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
`1)A=sqrt{4+sqrt{10+2sqrt5}}+sqrt{4-sqrt{10+2sqrt5}}`
`<=>A^2=4+sqrt{10+2sqrt5}+4-sqrt{10+2sqrt5}+2sqrt{16-10-2sqrt5}`
`<=>A^2=8+2sqrt{6-2sqrt5}`
`<=>A^2=8+2sqrt{(sqrt5-1)^2}`
`<=>A^2=8+2(sqrt5-1)`
`<=>A^2=6+2sqrt5=(sqrt5+1)^2`
`<=>A=sqrt5+1(do \ A>0)`
`b)B=sqrt{35+12sqrt6}-sqrt{35-12sqrt6}`
Vì `35+12sqrt6>35-12sqrt6`
`=>B>0`
`B^2=35+12sqrt6+35-12sqrt6-2sqrt{35^2-(12sqrt6)^2}`
`<=>B^2=70-2sqrt{361}`
`<=>B^2=70-2sqrt{19^2}=70-38=32`
`<=>B=sqrt{32}=4sqrt2(do \ B>0)`
`3)(4+sqrt{15})(sqrt{10}-sqrt6)sqrt{4-sqrt{15}}`
`=sqrt{4+sqrt{15}}.sqrt{4-sqrt{15}}.sqrt{4+sqrt{15}}(sqrt{10}-sqrt6)`
`=sqrt{16-15}.sqrt2(sqrt5-sqrt3).sqrt{4+sqrt{15}}`
`=sqrt{8+2sqrt{15}}(sqrt5-sqrt3)`
`=sqrt{5+2sqrt{5.3}+3}(sqrt5-sqrt3)`
`=sqrt{(sqrt5+sqrt3)^2}(sqrt5-sqrt3)`
`=(sqrt5+sqrt3)(sqrt5-sqrt3)`
`=5-3=2`
* Tính
a. A=\(\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
b. B=\(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)
\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)
=1
b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+1\)
=1
thực hiện phép tính
a)\(\dfrac{3}{5}\)-\(\dfrac{1}{2}\)\(\sqrt{1\dfrac{11}{25}}\)
b)(5+2\(\sqrt{6}\))(5-2\(\sqrt{6}\))
c)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
d)\(\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)(với x,y>0)
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
Bài 1: Tính
a) \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
b) \(1\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
c) \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
d) \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
Bài 2: Cho (d₁): y = \(\dfrac{1}{2}x-4\) và (d₂): y = \(-3x+3\) . Vẽ (d₁) và (d₂) trên cùng một hệ trục tọa độ. Tìm tọa độ giao điểm A của 2 đường thẳng trên.
Helpp!!
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
B1:Tính
a,\(\sqrt{\left(4-\sqrt{17}\right)^2}-\left(\sqrt{17}+2\right)\) b,\(\dfrac{7}{\sqrt{3}-\sqrt{2}}-\sqrt{147}-2\sqrt{18}\)
c,\(\dfrac{6}{\sqrt{5}-2}-\dfrac{6}{\sqrt{5}+2}+\sqrt{8}-4\sqrt{\dfrac{1}{7}}\) ; \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
\(a,=\sqrt{17}-4-\sqrt{17}-2=-6\\ b,=7\left(\sqrt{3}+\sqrt{2}\right)-7\sqrt{3}-6\sqrt{2}\\ =7\sqrt{3}+7\sqrt{2}-7\sqrt{3}-6\sqrt{2}=\sqrt{2}\\ c,=\dfrac{6\sqrt{5}+12-6\sqrt{5}+12}{3}+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}\\ =8+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}=\dfrac{56+14\sqrt{2}-4\sqrt{7}}{7}\\ d,=\left(\dfrac{\sqrt{2}}{4}-\dfrac{6\sqrt{2}}{4}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{-5\sqrt{2}+32\sqrt{2}}{4}\cdot8=54\sqrt{2}\)
Rút gọn:
a) \(\dfrac{\left(5\sqrt{2}+2\sqrt{5}\right)\left(\sqrt{3}-3\sqrt{2}\right)}{\sqrt{30}}\)
b) \(\dfrac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
c) \(\dfrac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
d) \(\dfrac{10\sqrt{18}+5\sqrt{3}-15\sqrt{27}}{\sqrt{3}\left(\sqrt{6}-4\right)}\)
a) A=\(\left(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{\sqrt{15}-\sqrt{35}}{\sqrt{3}-\sqrt{7}}\right).\left(\sqrt{2}+\sqrt{5}\right)\)
b) B=\(\dfrac{12}{3+\sqrt{3}}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{27}-3\sqrt{2}}{\sqrt{3}.\sqrt{2}}\)
c)C=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)(x>0,x≠1,x≠4)
\(A=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{\sqrt{5}\left(\sqrt{3}-\sqrt{7}\right)}{\sqrt{3}-\sqrt{7}}\right).\left(\sqrt{2}+\sqrt{5}\right)\)
\(=\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)=2-5=-3\)
\(B=\dfrac{12\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{2\sqrt{3}.\sqrt{3}}{\sqrt{3}}+\dfrac{3}{\sqrt{2}}-\dfrac{3}{\sqrt{3}}\)
\(=\dfrac{12\left(3-\sqrt{3}\right)}{6}-2\sqrt{3}+\dfrac{3\sqrt{2}}{2}-\sqrt{3}\)
\(=2\left(3-\sqrt{3}\right)-3\sqrt{3}+\dfrac{3\sqrt{2}}{2}=6-5\sqrt{3}+\dfrac{3\sqrt{2}}{2}\) (câu này khả năng đề sai, dấu \(\sqrt{3}.\sqrt{2}\) ở mẫu cuối cùng là dấu trừ mới hợp lý)
\(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
Dấu giữa 2 dấu ngoặc là dấu chia sẽ hợp lý hơn