K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

NV
13 tháng 2 2022

\(\sqrt{2x-1}< 8-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\8-x\ge0\\2x-1< \left(8-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le8\\x^2-18x+65>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le8\\\left[{}\begin{matrix}x>13\\x< 5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{2}\le x< 5\)

11 tháng 3 2021

undefined

11 tháng 3 2021

undefined

NV
16 tháng 4 2022

a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

NV
16 tháng 4 2022

b.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

11 tháng 1 2022
Not biếtmdnhdhd
11 tháng 1 2022

Hummmm

NV
16 tháng 3 2022

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-3\ge0\\2x^2-3x+1\ge0\\x^2+2x-3\le2x^2-3x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\\x^2-5x+4\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\le-3\\x\ge4\end{matrix}\right.\)

24 tháng 1 2021

\(\sqrt[3]{2x^2+1}>\sqrt[3]{3x^2-1}\)

\(\Leftrightarrow2x^2+1>3x^2-1\)

\(\Leftrightarrow x^2< 2\)

\(\Leftrightarrow-\sqrt{2}< x< \sqrt{2}\)

\(\Leftrightarrow\left(\sqrt[3]{x+1}-1\right)+\left(\sqrt{2x+4}-2\right)< -x\sqrt{2}\)

=>\(\dfrac{x+1-1}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{2x+4-4}{\sqrt{2x+4}+2}+x\sqrt{2}< 0\)

=>x<0

=>-1<x<0

2 tháng 1 2022

ĐKXĐ : \(1\le x\le3\)

Ta có \(\sqrt{x-1}+\sqrt{3-x}+4x\sqrt{2x}\ge x^3+10\)

<=> \(-2\sqrt{x-1}-2\sqrt{3-x}-8x\sqrt{2x}\le-2x^3-20\)

<=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+2x^3-8x\sqrt{2x}+16\le0\)(1)

Đặt \(\sqrt{2x}=y\) => \(x=\dfrac{y^2}{2}\)

Khi đó \(2x^3-8x\sqrt{2x}+16=\dfrac{y^6}{4}-4y^3+16=\left(\dfrac{y^3-8}{2}\right)^2\)

Khi đó (1) <=> \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2\le0\)(1)

mà \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2\ge0\forall x;y\)(2) 

Từ (2)(1) => \(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{3-x}-1\right)^2+\left(\dfrac{y^3-8}{2}\right)^2=0\)

<=> \(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{3-x}-1=0\\\dfrac{y^3-8}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\3-x=1\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=2\\\sqrt{2x}=2\end{matrix}\right.\Leftrightarrow x=2\)

Vậy x = 2 là nghiệm bất phương trình