Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x\sqrt{x}+4x-12\sqrt{x}-27\)
\(=\left(x\sqrt{x}-27\right)+\left(4x-12\sqrt{x}\right)\)
\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)+4\sqrt{x}\left(\sqrt{x}-3\right)\)
\(=\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9+4\sqrt{x}\right)\)
\(=\left(\sqrt{x}-3\right)\left(x+7\sqrt{x}+9\right)\)
a, \(\sqrt{a^2-b^2}-\sqrt{a^3+b^3}\)
\(=\sqrt{\left(a+b\right)\left(a-b\right)}-\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}\)
\(=\sqrt{a+b}\left(\sqrt{a-b}-\sqrt{a^2-ab+b^2}\right)\)
a/ Có: \(\sqrt{99}< \sqrt{100}=10\) mà 10 < 11
=> \(11>\sqrt{99}\)
b/ có: √11 < √16 =4
=> √11 + 1 < 4 + 1 = 5
hay 5 > √11 + 1
c/ Có: √2 > √1 = 1
=> √2 + 1 > 1 + 1 = 2
hay 2 < 1 + √2
d/ 3√11 = √99 ; 12 = √144
mà √99 < √144
=> 3√11 < 12
e/ - 10 = -√100 ; -2√23 = -√92
Có: √100 > √92 => -√100 < - √92
hay -10 < -2√23
f/ Có: √7 < √9 = 3
=> 1 + √7 < 1 + 3 = 4
hay 4 > 1 + √7
Giải:
a) Ta có:
\(11=\sqrt{121}\)
Vì \(\sqrt{121}>\sqrt{99}\)
\(\Leftrightarrow11>\sqrt{99}\)
Vậy ...
b) Ta có:
\(5=4+1=\sqrt{16}+1\)
Vì \(\sqrt{16}+1>\sqrt{11}+1\)
\(\Leftrightarrow5>\sqrt{11}+1\)
Vậy ...
c) Ta có:
\(2=1+1=\sqrt{1}+1\)
Vì \(\sqrt{1}+1< 1+\sqrt{2}\)
\(\Leftrightarrow2< 1+\sqrt{2}\)
Vậy ...
d) Ta có:
\(3\sqrt{11}=\sqrt{9.11}=\sqrt{99}\)
\(12=\sqrt{144}\)
Vì \(\sqrt{99}< \sqrt{144}\)
\(\Leftrightarrow3\sqrt{11}< 12\)
Vậy ...
e) Ta có:
\(-10=-\sqrt{100}\)
\(-2\sqrt{23}=-\sqrt{92}\)
Vì \(-\sqrt{100}< -\sqrt{92}\)
\(\Leftrightarrow-10< -2\sqrt{23}\)
Vậy ...
f) Ta có:
\(4=1+3=1+\sqrt{9}\)
Vì \(1+\sqrt{9}>1+\sqrt{7}\)
\(\Leftrightarrow4>1+\sqrt{7}\)
Vậy ...
a) \(5+\sqrt{10}-\sqrt{5}=\sqrt{5}.\left(\sqrt{5}+\sqrt{2}-1\right)\)
b) ĐK: \(a\ge0\)
\(a-4\sqrt{a}-5=a+\sqrt{a}-5\sqrt{a}-5=\left(\sqrt{a}+1\right)\left(\sqrt{a}-5\right)\)
c) ĐK: \(a\ge0\)
\(a+12\sqrt{a}+32=a+8\sqrt{a}+4\sqrt{a}+32=\left(\sqrt{a}+8\right)\left(\sqrt{a}+4\right)\)
d) ĐK: \(a\ge0\)
\(a-5\sqrt{a}+6=a-2\sqrt{a}-3\sqrt{a}+6=\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)\)
a: \(=2\cdot3+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
b: \(=5\sqrt{10}+2\cdot5-5\sqrt{10}=10\)
c: \(=2\sqrt{7}\cdot\sqrt{7}-\sqrt{12}\cdot\sqrt{7}-\sqrt{7}\cdot\sqrt{7}+2\sqrt{21}=2\cdot7-7=7\)
d: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}=2\cdot11=22\)
mình không biết bạn ơi
a. \(11+2\sqrt{10}=\left(\sqrt{10}+1\right)^2\)
b. \(12-2\sqrt{11}=\left(\sqrt{11}-1\right)^2\)
c.\(23+2\sqrt{22}=\left(\sqrt{22}+1\right)^2\)