ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

- Với \(x=-\dfrac{3}{2}\) là nghiệm của BPT

- Với \(x>-\dfrac{3}{2}\Rightarrow2x+3>0\)

\(\Rightarrow\dfrac{3\left(2x-3\right)\left(2x+3\right)}{\sqrt{3x^2-3}}\le2x+3\)

\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{\sqrt{3x^2-3}}\le1\)

\(\Rightarrow3\left(2x-3\right)\le\sqrt{3x^2-3}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3< 0\\\left\{{}\begin{matrix}2x-3\ge0\\9\left(2x-3\right)^2\le3x^2-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{2}< x< \dfrac{3}{2}\\\left[{}\begin{matrix}x\ge\dfrac{3}{2}\\11x^2-36x+28\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{3}{2}< x< \dfrac{3}{2}\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{14}{11}\le x\le2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-\dfrac{3}{2}< x< \dfrac{3}{2}\\\dfrac{3}{2}\le x\le2\end{matrix}\right.\) \(\Rightarrow-\dfrac{3}{2}< x\le2\)

Kết hợp ĐKXĐ \(\Rightarrow\left[{}\begin{matrix}-\dfrac{3}{2}< x< -1\\1< x\le2\end{matrix}\right.\)

- Với \(x< -\dfrac{3}{2}\Rightarrow2x+3< 0\)

\(\dfrac{3\left(2x-3\right)\left(2x+3\right)}{\sqrt{3x^2-3}}\le2x+3\Leftrightarrow\dfrac{3\left(2x-3\right)}{\sqrt{3x^2-3}}\ge1\)

\(\Rightarrow3\left(2x-3\right)\ge\sqrt{3x^2-3}\)

Do \(x< -\dfrac{3}{2}\Rightarrow3\left(2x-3\right)< 0\Rightarrow\) BPT vô nghiệm

Vậy nghiệm của BPT là \(\left[{}\begin{matrix}-\dfrac{3}{2}\le x< -1\\1< x\le2\end{matrix}\right.\)

ĐK: \(x\ge1;x\le-2\)

\(\sqrt{x^2-1}+\sqrt{x^2-x}\le\sqrt{x^2+x-2}\)

\(\Leftrightarrow2x^2-x-1+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le x^2+x-2\)

\(\Leftrightarrow x^2-2x+1+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le0\)

\(\Leftrightarrow\left(x-1\right)^2+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\left(x^2-1\right)\left(x^2-x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy bất phương trình có nghiệm \(x=1\)

Bạn cần giúp nhanh nhưng lại không ghi đầy đủ đề bài?

\(\sqrt{\dfrac{72x}{128}}=\dfrac{3}{4}\)

\(\Leftrightarrow x\cdot\dfrac{9}{16}=\dfrac{9}{16}\)

hay x=1

ĐK:x\(\ge-1\)(*)

bpt\(\Leftrightarrow3\left(x^2-x+1\right)+2\left(x+1\right)< 5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Leftrightarrow\left(\sqrt{x^2-x+1}-\sqrt{x+1}\right)\left(3\sqrt{x^2-x+1}-2\sqrt{x+1}\right)< 0\)

Đến đây bn chia 2 TH rồi giải bình thường nhá:D

\(\sqrt{2023-\sqrt{x}}=2023-x\left(ĐK:x\ge0\right)\)

Đặt \(t=\sqrt{x}\left(t\le2023\right)\)

Pt trở thành : \(\sqrt{2023-t}=2023-t^2\)

\(\Leftrightarrow2023-t=\left(2023-t^2\right)^2\)

\(\Leftrightarrow t^4-4046t+4092529=2023-t\)

\(\Leftrightarrow t^4-4045+4090506=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2023\left(n\right)\\t=2022\left(n\right)\end{matrix}\right.\)

+) Với \(t=2023\Rightarrow x^2=2023\Rightarrow x=\pm17\sqrt{7}\)

+) Với \(t=2022\Rightarrow x^2=2022\Leftrightarrow x=\pm\sqrt{2022}\)

Vì \(x\ge0\) \(\Rightarrow x\in\left\{17\sqrt{7};\sqrt{2022}\right\}\)

Vậy \(S=\left\{17\sqrt{7};\sqrt{2022}\right\}\)