Chương I - Căn bậc hai. Căn bậc ba

\(M=\sqrt{x^2-4x+5}=\sqrt{x^2-2.x.2+2^2+1}=\sqrt{\left(x-2\right)^2+1}\)Ta có (x-2)²≥0⇒(x-2)²+1≥1⇒\(\sqrt{\left(x-2\right)^2+1}\ge1\)

Vậy GTNN của M là 1

Đẳng thức xảy ra khi (x-2)²=0⇒x-2=0⇒x=2

Lời giải:

a)

$\sqrt{98}-\sqrt{72}+0.5\sqrt{8}=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}$

$=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}$

b)

$\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}$

$=4\sqrt{a}+4\sqrt{10}.\sqrt{a}-9\sqrt{10}.\sqrt{a}$

$=(4+4\sqrt{10}-9\sqrt{10})\sqrt{a}=(4-5\sqrt{10}).\sqrt{a}$

c)

$(2\sqrt{3}+\sqrt{5})\sqrt{3}-\sqrt{60}=2.3+\sqrt{15}-2\sqrt{15}$

$=6-\sqrt{15}$

d)

$(\sqrt{99}-\sqrt{18}-\sqrt{11})\sqrt{11}+3\sqrt{32}$

$=\sqrt{99}.\sqrt{11}-\sqrt{18}.\sqrt{11}-11+3\sqrt{32}$

$=\sqrt{9}.\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}-11+12\sqrt{2}$

$=3.11+\sqrt{2}(12-3\sqrt{11})-11$

$=22+\sqrt{2}(12-3\sqrt{11})$

a) Ta có: \(4\sqrt{7}=\sqrt{4^2.7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)

Vì 112 < 117 => \(\sqrt{112}< \sqrt{117}\) => \(4\sqrt{7}< 3\sqrt{13}\)

b) Ta có: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{21}{4}}=\sqrt{\dfrac{147}{28}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}=\sqrt{\dfrac{114}{28}}\)

Vì: \(\dfrac{147}{28}>\dfrac{114}{28}\Rightarrow\sqrt{\dfrac{147}{28}}>\sqrt{\dfrac{114}{28}}\Rightarrow\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

ĐKXĐ: \(x^2+9>=0\)

=>\(x\in R\)

Ta có: \(42\sqrt{\dfrac{25}{6}}-10\sqrt{\dfrac{3}{2}}-12\sqrt{\dfrac{98}{3}}\)

= \(35\sqrt{6}-5\sqrt{6}-28\sqrt{6}\)

=\(2\sqrt{6}\)

Ta có: \(\sqrt{1+\dfrac{\sqrt{3}}{2}}-\) \(\sqrt{1-\dfrac{\sqrt{3}}{2}}\)

= 1

điều kiện xác định : \(a\ge0;a\ne1\)

\(A=\dfrac{2a^2+4}{1+a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}=\dfrac{2a^2+4}{1+a^3}-\left(\dfrac{1}{1+\sqrt{a}}+\dfrac{1}{1-\sqrt{a}}\right)\)

\(\Leftrightarrow A=\dfrac{2a^2+4}{\left(a+1\right)\left(a^2-a+1\right)}+\dfrac{2}{a-1}=\dfrac{\left(2a^2+4\right)\left(a-1\right)+2\left(a^3+1\right)}{\left(a^2-1\right)\left(a^2-a+1\right)}\)

\(\Leftrightarrow A=\dfrac{4a^3-2a^2+4a-2}{\left(a^2-1\right)\left(a^2-a+1\right)}=\dfrac{\left(a^2+1\right)\left(4a-2\right)}{\left(a^2-1\right)\left(a^2-a+1\right)}\)

tới đây mk nghỉ đề sai rồi bn à