Bài 3: Liên hệ giữa phép nhân và phép khai phương

\(\sqrt{108x^3}.\sqrt{3x}=\sqrt{108x^3.3x}=\sqrt{324x^4}=18x^2\)( vì x >= 0)

\(\sqrt{108x^3}.\sqrt{3x}=\sqrt{108x^3.3x}=\sqrt{324x^4}\)

\(=18x^2\) (do \(x\ge0\))

Chúc bạn học tốt!!!

\(\sqrt{20}\cdot\sqrt{72}\cdot\sqrt{4,9}\)

\(=\sqrt{7056}\)

\(=84\)

\(\sqrt{20}.\sqrt{72}.\sqrt{4,9}=\sqrt{20.72.4,9}=\sqrt{7056}=84\)

a,\(\sqrt{0,16.0,65,225}\)

\(=\sqrt{0,16}.\sqrt{0,65}.\sqrt{225}\)

\(=0,4.25.\sqrt{0,65}=10.\sqrt{0,65}\)

\(=\sqrt{6,5}\)

b, \(\sqrt{250.360}\)

\(=\sqrt{250}.\sqrt{360}=\sqrt{25}.\sqrt{10}.\sqrt{36}.\sqrt{10}\)

\(=5.10.6=300\)

Chúc bạn học tốt!!!

pt(1)=>a=90-b thay vào pt(2) giải

\(1=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+\left(b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{a}{bc}\right)\)

\(\Leftrightarrow\left(b+c\right)\left(\dfrac{bc+ac+ab+a^2}{abc}\right)=0\)

\(\dfrac{\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(a+c\right)}{abc}=0\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)

Xét 3 TH

=> P=0 ( đề bài BT ở giữa có 1 số mũ sai nha )

Đề ??

Ta có: \(P=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)

\(=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{xz}\)

\(=\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2z}{xyz}-\dfrac{2x}{xyz}-\dfrac{2y}{xyz}\)

\(=3-\dfrac{2\left(x+y+z\right)}{xyz}\)

\(=3-\dfrac{2xyz}{xyz}=3-2=1\)

Vậy P = 1

Câu 1 nè

\(\sqrt{11}+6\sqrt{2}-3+\sqrt{2}=\sqrt{11}+7\sqrt{2}-3.\)

Câu 2 nè :

a) đề không rõ.

b) \(\sqrt{75.48}=\sqrt{25.16.3^2}=5.4.3=60\)

Ta có \(a^4+b^4-2ab^3-2a^3b+2a^2b^2\) =(a²-ab)²+(b²-ab)²\(\ge0\forall a;b\) suy ra

\(\dfrac{a^4+b^4}{2}\ge ab^3+a^3b-a^2b^2\)(đpcm)

Đẳng thức xảy ra \(\Leftrightarrow a=b\)

a⁴+b⁴ \(\ge\)ab(a+b) (1)

1/2 (a⁴+b⁴)\(\ge\)a²b². (2)

(1) -(2)

=>dpcm