Ôn tập chương 1: Căn bậc hai. Căn bậc ba

ĐKXĐ: x>=4

\(\dfrac{\sqrt{3x-12}}{\sqrt{3}}+\sqrt{\dfrac{x-4}{16}}=5\)

=>\(\sqrt{x-4}+\dfrac{\sqrt{x-4}}{4}=5\)

=>\(\dfrac{5}{4}\cdot\sqrt{x-4}=5\)

=>\(\sqrt{x-4}=4\)

=>x-4=16

=>x=16+4=20(nhận)

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{A}+45^0=180^0\)

=>\(\widehat{A}=135^0\)

\(cos\left(\dfrac{A}{2}\right)=cos67,5=cos\left(90-22,5\right)=sin22,5\)

\(1+cot^222,5^0=\dfrac{1}{sin^222,5^0}\)

=>\(\dfrac{1}{sin^222,5^0}=1+\left(1+\sqrt{2}\right)^2=4+2\sqrt{2}\)

=>\(sin^222,5^0=\dfrac{1}{4+2\sqrt{2}}=\dfrac{2-\sqrt{2}}{4}\)

=>\(sin22,5^0=\dfrac{\sqrt{2-\sqrt{2}}}{2}\)

=>\(\dfrac{\sqrt{2-\sqrt{2}}}{2}=\dfrac{\sqrt{2-\sqrt{a}}}{b}\)

=>a=2; b=2

=>a+b=4

\(A=\dfrac{4}{3}\sqrt{2\left(9+6a+a^2\right)}+\sqrt{\dfrac{32}{9}}\cdot a+2\sqrt{2}\)

\(A=\dfrac{4}{3}\cdot\sqrt{2\cdot\left(3^2+2\cdot3\cdot a+a^2\right)}+\dfrac{\sqrt{32}}{\sqrt{9}}\cdot a+2\sqrt{2}\)

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

\(A=\dfrac{4\sqrt{2}}{3}\cdot\left|a+3\right|+\dfrac{4\sqrt{2}}{3}\cdot a+2\sqrt{2}\)

\(A=-\dfrac{4\sqrt{2}}{3}\left(a+3\right)+\dfrac{4\sqrt{2}a+6\sqrt{2}}{3}\)

\(A=\dfrac{-4\sqrt{2}a-12\sqrt{2}+4\sqrt{2}a+6\sqrt{2}}{3}\) 

\(A=\dfrac{-6\sqrt{2}}{3}\)

\(A=-2\sqrt{2}\) 

\(\Rightarrow2\sqrt{2}A=2\sqrt{2}\cdot-2\sqrt{2}=-8\)

Với \(x>0\) ta có: \(\dfrac{1}{\sqrt{1+x^3}}=\dfrac{1}{\sqrt{\left(1+x\right)\left(x^2-x+1\right)}}\ge\dfrac{2}{1+x+x^2-x+1}=\dfrac{2}{x^2+2}\)

Do đó:

\(\sqrt{\dfrac{a^3}{a^3+\left(b+c\right)^3}}=\dfrac{1}{\sqrt{1+\left(\dfrac{b+c}{a}\right)^3}}\ge\dfrac{2}{\left(\dfrac{b+c}{a}\right)^2+2}=\dfrac{2a^2}{2a^2+\left(b+c\right)^2}\)

Mà \(\left(b+c\right)^2\le2\left(b^2+c^2\right)\Rightarrow\dfrac{2a^2}{2a^2+\left(b+c\right)^2}\ge\dfrac{2a^2}{2a^2+2\left(b^2+c^2\right)}=\dfrac{a^2}{a^2+b^2+c^2}\)

\(\Rightarrow\sqrt{\dfrac{a^3}{a^3+\left(b+c\right)^3}}\ge\dfrac{a^2}{a^2+b^2+c^2}\)

Chứng minh tương tự ta được:

\(\sqrt{\dfrac{b^3}{b^3+\left(a+c\right)^3}}\ge\dfrac{b^2}{a^2+b^2+c^2}\) ; \(\sqrt{\dfrac{c^3}{c^3+\left(a+b\right)^3}}\ge\dfrac{c^2}{a^2+b^2+c^2}\)

Cộng vế các BĐT nói trên ta sẽ được đpcm (hơi dài nên làm biếng ghi lại)

Lời giải:
ĐKXĐ: $x\geq 0$

Ta thấy: $\sqrt{x}\geq 0; 2x+1>0$ với mọi $x\geq 0$

$\Rightarrow \frac{\sqrt{x}}{2x+1}\geq 0$

Vậy GTNN của biểu thức là $0$. Giá trị này đạt được khi $x=0$

Lời giải:

Để $y=(m-1)x+2$ song song với $y=5x-2m$ thì:

$m-1=5$ và $2\neq -2m$

$\Leftrightarrow m=6$

a: \(\sqrt{\left(x+1\right)^2}=5\)(ĐKXĐ: \(x\in R\))

=>|x+1|=5

=>\(\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)

b: Sửa đề: \(5\sqrt{9x-9}-\sqrt{4\left(x-1\right)}+\sqrt{36\left(x-1\right)}-18=0\)

ĐKXĐ: x>=1

\(PT\Leftrightarrow5\cdot3\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0\)

=>\(15\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}=18\)

=>\(19\sqrt{x-1}=18\)

=>\(\sqrt{x-1}=\dfrac{18}{19}\)

=>\(x-1=\left(\dfrac{18}{19}\right)^2=\dfrac{324}{361}\)

=>\(x=\dfrac{324}{361}+1=\dfrac{324+361}{361}=\dfrac{685}{361}\)

Lời giải:

a. PT $\Leftrightarrow |x+1|=5$

$\Leftrightarrow x+1=\pm 5\Leftrightarrow x=4$ hoặc $x=-6$

b. ** Sửa $x-9$ thành $x-1$

ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow 5\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0$

$\Leftrightarrow (5-2+6)\sqrt{x-1}=18$

$\Leftrightarrow 9\sqrt{x-1}=18$

$\Leftrightarrow \sqrt{x-1}=2$

$\Leftrightarrow x-1=4$

$\Leftrightarrow x=5$ (tm)

a: \(\sqrt{169}-\sqrt{225}\)

\(=\sqrt{13^2}-\sqrt{15^2}\)

=13-15

=-2

b: \(\dfrac{\sqrt{144}}{9}\)

\(=\dfrac{\sqrt{12^2}}{9}\)

\(=\dfrac{12}{9}=\dfrac{4}{3}\)

c: \(\sqrt{18}:\sqrt{2}=\sqrt{\dfrac{18}{2}}=\sqrt{9}=3\)

a: \(\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{5}\)

\(=\left|3-\sqrt{5}\right|-\sqrt{5}\)

\(=3-\sqrt{5}-\sqrt{5}=3-2\sqrt{5}\)

b: \(\sqrt{\left(4-2\sqrt{3}\right)^2}\)

\(=\left|4-2\sqrt{3}\right|\)

\(=4-2\sqrt{3}\)

a: \(\dfrac{3}{2-\sqrt{5}}=\dfrac{3\left(2+\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}\)

\(=\dfrac{3\left(2+\sqrt{5}\right)}{-1}=-6-2\sqrt{5}\)

b: \(\dfrac{4}{\sqrt{3}-\sqrt{5}}\)

\(=\dfrac{4\left(\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}\)

\(=\dfrac{4\left(3+\sqrt{5}\right)}{3-5}=-2\left(3+\sqrt{5}\right)\)