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 a, b,c là 3 số thực dương. Chứng minh rằng
\(\sqrt{\dfrac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\dfrac{b^3}{b^3+\left(a+c\right)^3}}+\sqrt{\dfrac{c^3}{c^3+\left(b+a\right)^3}}\ge1\)
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)
Tìm GTNN của biểu thức: \(\dfrac{\sqrt{x}}{2x+1}\)
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 $x=0$
Giá trị của m để đường thẳng y = m - 1 x + 2 song song với đường thẳng y = 5 x-2m là
Lời giải:
Để $y=(m-1)x+2$ song song với $y=5x-2m$ thì:
$m-1=5$ và $2\neq -2m$
$\Leftrightarrow m=6$
tìm x biết:
a \(\sqrt{\left(x+1\right)^2}\) = 5
b, 5\(\sqrt{x-9}\) - \(\sqrt{4\left(x-1\right)}\) + \(\sqrt{36\left(x-1\right)}\) -18 = 0
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)
tính
a, \(\sqrt{169}\) - \(\sqrt{225}\)
b \(\dfrac{\sqrt{144}}{9}\)
c \(\sqrt{18}\) \(\div\) \(\sqrt{2}\)
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: \(\sqrt{18}:\sqrt{2}=\sqrt{\dfrac{18}{2}}=\sqrt{9}=3\)
tính
a, \(\sqrt{\left(3-\sqrt{5}\right)^2}\) - \(\sqrt{5}\)
b \(\sqrt{\left(4-2\sqrt{3}\right)^2}\)
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: \(\sqrt{\left(4-2\sqrt{3}\right)^2}\)
\(=\left|4-2\sqrt{3}\right|\)
\(=4-2\sqrt{3}\)
khử mẫu
a \(\dfrac{3}{2-\sqrt{5}}\) b \(\dfrac{4}{\sqrt{3}-\sqrt{5}}\)
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: \(\dfrac{4}{\sqrt{3}-\sqrt{5}}\)
\(=\dfrac{4\left(\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{4\left(3+\sqrt{5}\right)}{3-5}=-2\left(3+\sqrt{5}\right)\)