x2 = \(\sqrt{5-x}\)+\(\sqrt{5+x}\)+12
giải giúp mình với ạ
K
Khách
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11 tháng 10 2021
a: \(f\left(-x\right)=-2\cdot\left(-x\right)^3+3\cdot\left(-x\right)\)
\(=2x^3-3x\)
\(=-\left(-2x^3+3x\right)\)
=-f(x)
Vậy: f(x) là hàm số lẻ
c: TXĐ: D=[-2;2]
Nếu \(x\in D\Leftrightarrow-x\in D\)
\(f\left(-x\right)=\sqrt{6-3\cdot\left(-x\right)}-\sqrt{6+3\cdot\left(-x\right)}\)
\(=\sqrt{6+3x}-\sqrt{6-3x}\)
\(=-f\left(x\right)\)
Vậy: f(x) là hàm số lẻ
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
ĐKXĐ: \(-2\le x\le3\)
\(\dfrac{\sqrt{-x^2+x+6}}{2x+5}-\dfrac{\sqrt{-x^2+x+6}}{x-4}\ge0\)
\(\Leftrightarrow\sqrt{-x^2+x+6}\left(\dfrac{1}{2x+5}-\dfrac{1}{x-4}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(-x-9\right)\sqrt{x^2+x+6}}{\left(2x+5\right)\left(x-4\right)}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+6=0\\\dfrac{-x-9}{\left(2x+5\right)\left(x-4\right)}\ge0\end{matrix}\right.\) \(\Leftrightarrow-2\le x\le3\)
Hoặc có thể biện luận như sau:
Ta có: \(\left\{{}\begin{matrix}2x+5>0;\forall x\in\left[-2;3\right]\\x-4< 0;\forall x\in\left[-2;3\right]\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{-x^2+x+6}}{2x+5}\ge0\\\dfrac{\sqrt{-x^2+x+6}}{x-4}\le0\end{matrix}\right.\) ; \(\forall x\in\left[-2;3\right]\)
Do đó nghiệm của BPT là \(-2\le x\le3\)
G
0
MK
1
NC
30 tháng 8 2020
Bài làm:
Ta có: \(\sqrt{7+\sqrt{2x}}=3+\sqrt{5}\)
\(\Leftrightarrow7+\sqrt{2x}=\left(3+\sqrt{5}\right)^2\)
\(\Leftrightarrow7+\sqrt{2x}=14+6\sqrt{5}\)
\(\Leftrightarrow\sqrt{2x}=7+6\sqrt{5}\)
\(\Leftrightarrow2x=\left(7+6\sqrt{5}\right)^2\)
\(\Leftrightarrow2x=229+84\sqrt{5}\)
\(\Rightarrow x=\frac{229+84\sqrt{5}}{2}\)
8 tháng 10 2021
\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
16 tháng 2 2022
\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)
MN
14 tháng 7 2021
Bài 1 :
\(a.\sqrt{x^2-1}\)
\(ĐK:\)
\(x^2-1\ge0\)
\(\Leftrightarrow x^2\ge1\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)
Bài 2 :
\(2\cdot\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{48}-5\sqrt{50}\)
\(=2\cdot\left|\sqrt{2}-3\right|+4\sqrt{3}-25\sqrt{2}\)
\(=-2\cdot\left(\sqrt{2}-3\right)+4\sqrt{3}-25\sqrt{2}\)
\(=-2\sqrt{2}-6+4\sqrt{3}-25\sqrt{2}\)
\(=-27\sqrt{2}-6+4\sqrt{3}\)
đk: \(5\ge x\ge-5\)
Ta có: \(x^2=\sqrt{5-x}+\sqrt{5+x}+12\)
\(\Leftrightarrow\left(x^2-16\right)-\left(\sqrt{5-x}-3\right)-\left(\sqrt{5+x}-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)-\frac{5-x-9}{\sqrt{5-x}+3}-\frac{5+x-1}{\sqrt{5+x}+1}=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)+\frac{x+4}{\sqrt{5-x}+3}-\frac{x+4}{\sqrt{5+x}+1}=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-4+\frac{1}{\sqrt{5-x}+3}-\frac{1}{\sqrt{5+x}+1}\right)=0\)
Nếu x + 4 = 0 => x = -4 (tm)
Nếu \(x-4+\frac{1}{\sqrt{5-x}+3}-\frac{1}{\sqrt{5+x}+1}=0\)
TH1: \(x=4\left(tm\right)\)
TH2: \(x>4\)
Khi đó: \(\hept{\begin{cases}\sqrt{5-x}+3< 1+3=4\\\sqrt{5+x}+1>3+1=4\end{cases}}\Rightarrow\frac{1}{\sqrt{5-x}+3}>\frac{1}{4}>\frac{1}{\sqrt{5+x}+1}\)
\(\Rightarrow x-4+\frac{1}{\sqrt{5-x}+3}-\frac{1}{\sqrt{5+x}+1}>0\)
TH3: \(x< 4\)
Khi đó: \(\hept{\begin{cases}\sqrt{5-x}+3>4\\\sqrt{5+x}+1< 4\end{cases}}\Rightarrow\frac{1}{\sqrt{5-x}+3}< \frac{1}{4}< \frac{1}{\sqrt{5+x}+1}\)
\(\Rightarrow x-4+\frac{1}{\sqrt{5-x}+3}-\frac{1}{\sqrt{5+x}+1}< 0\)
Vậy tập nghiệm của PT \(S=\left\{-4;4\right\}\)