Rut gon bieu thuc
1)\(\frac{\sqrt{165^2-124^2}}{164}\)
2)\(\frac{\sqrt{149^2-76^2}}{\sqrt{457^2-384^2}}\)
Giải phương trình
a) \(\sqrt{\frac{165^2-124^2}{164}}\)
b) \(\sqrt{\frac{149^2-76^2}{457^2-384^2}}\)
c) \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
Giúp mình với nhé! Cảm ơn !!!
a) \(\sqrt{\frac{165^2-124^2}{164}}=\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}\)
\(=\sqrt{\frac{11849}{164}}=\sqrt{72,25}=8,5\)
b)\(\sqrt{\frac{149^2-76^2}{457^2-384^2}}=\sqrt{\frac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}\) \(=\sqrt{\frac{73.225}{73.841}}=\sqrt{\frac{225}{841}}=\sqrt{\frac{15^2}{29^2}}=\frac{15}{29}\)
c)\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\) \(=\sqrt{2^2+3+2.2.\sqrt{3}}-\sqrt{2^2+3-2.2.\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\sqrt{3}^2}-\sqrt{2^2-2.2.\sqrt{3}+\sqrt{3}^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}=\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
Tính
a) \(\sqrt{\frac{165^2-124^2}{164}}\)
b) \(\sqrt{\frac{149^2-76^2}{457^2-384^2}}\)
c) \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
Mấy bạn giúp My mấy câu này với nhé! My cám ơn !!!
a) \(\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}=\sqrt{\frac{289}{4}}=\frac{17}{2}\)
b) tương tự ý a
c) \(\left(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\right)^2=7+4\sqrt{3}+7-4\sqrt{3}-2.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}\)
\(=14-2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=14-2\sqrt{49-48}\)
\(=14-2.1=12\)
\(\Rightarrow\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{12}=2\sqrt{3}\)
a.\(\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}\)
b.\(\sqrt{1,44.1,21-1,44.0,4}\)
c.\(\sqrt{\dfrac{165^2-124^2}{164}}\)
d.\(\sqrt{\dfrac{149^2-76^2}{457^2-384^2}}\)
a: \(=\sqrt{\dfrac{25}{16}\cdot\dfrac{49}{9}\cdot\dfrac{1}{100}}=\dfrac{5}{4}\cdot\dfrac{7}{3}\cdot\dfrac{1}{10}=\dfrac{35}{120}=\dfrac{7}{24}\)
b: \(=\sqrt{1.44\cdot0.81}=1.2\cdot0.9=1.08\)
c: \(=\sqrt{\dfrac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\dfrac{1}{4}\cdot289}=\dfrac{17}{2}\)
d: \(=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}=\sqrt{\dfrac{225}{841}}=\dfrac{15}{29}\)
Tính:
a. \(\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01};\) b. \(\sqrt{1,44.1,21-1,44.0,4};\)
c. \(\sqrt{\dfrac{165^2-124^2}{164}};\) d. \(\sqrt{\dfrac{149^2-76^2}{457^2-384^2}}.\)
a) HD: Đổi hỗn số và số thập phân thành phân số.
ĐS: .
b) =
= = =
= .
d) ĐS: .
\(\sqrt{\frac{149^2-76^2}{457^2-384^2}}\)
\(\sqrt{\frac{149^2-76^2}{457^2-384^2}}=\sqrt{\frac{73.225}{73.841}}=\sqrt{\frac{225}{841}}=\frac{\sqrt{225}}{\sqrt{841}}=\frac{15}{29}\)
rut gon bieu thuc:\(\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\frac{a\sqrt{a^2-b}}{2}\)
\(\sqrt{\dfrac{149^2-76^2}{457^2-384^2}}\)
\(\sqrt{\dfrac{149^2-76^2}{457^2-384^2}}=\sqrt{\dfrac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}\)
\(=\sqrt{\dfrac{73\cdot225}{73\cdot841}}=\sqrt{\dfrac{225}{841}}=\dfrac{15}{29}\)
Cho bieu thuc:
P=\(\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
a. Rut gon bieu thuc P
b.Tim GTLN cua P sau khi rut gon
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
RUT GON BIEU THUC \(\frac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}\)
nhan ca tu va mau voi\(\sqrt{2}\) ta dc
\(\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{2x-4-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}\)(dkx>=2)
=\(\frac{\left|\sqrt{2x-4}-2\right|}{2}\)