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Chọn D

27 tháng 9 2021

Giải ra giúp e vs ạ😭😭

5: ĐKXĐ: \(\left\{{}\begin{matrix}x^2+3x-4>=0\\2x^2-2x>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+4\right)\left(x-1\right)>=0\\2x\left(x-1\right)>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=1\\x< =-4\end{matrix}\right.\\\left[{}\begin{matrix}x>=1\\x< =0\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x>=1\\x< =-4\end{matrix}\right.\)

\(\sqrt{x^2+3x-4}< \sqrt{2x^2-2x}\)

=>\(x^2+3x-4< 2x^2-2x\)

=>\(2x^2-2x-x^2-3x+4>0\)

=>\(x^2-5x+4>0\)

=>(x-1)(x-4)>0

=>\(\left[{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\)

Kết hợp ĐKXĐ, ta được:

\(\left[{}\begin{matrix}x>4\\x< =-4\end{matrix}\right.\)

7: ĐKXĐ: x>=-1

\(2\sqrt{x+2+2\sqrt{x+1}}-\sqrt{x+1}=4\)

=>\(2\cdot\sqrt{x+1+2\sqrt{x+1}+1}-\sqrt{x+1}=4\)

=>\(2\cdot\sqrt{\left(\sqrt{x+1}+1\right)^2}-\sqrt{x+1}=4\)

=>\(2\left(\sqrt{x+1}+1\right)-\sqrt{x+1}=4\)

=>\(\sqrt{x+1}+2=4\)

=>\(\sqrt{x+1}=2\)

=>x+1=4

=>x=3(nhận)

3: \(\Leftrightarrow\dfrac{x-1}{2x-3}< 0\)

hay 1<x<3/2

17 tháng 12 2021

Câu 58: B

Câu 59: C

7 tháng 3 2022

\(5;;\sqrt{\left(x+5\right)\left(3x+4\right)}>4\left(x-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4\left(x-1\right)\le0\\\left(x+5\right)\left(3x+4\right)\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}4\left(x-1\right)\ge0\\\left(x+5\right)\left(3x+4\right)\ge0\\\left(x+5\right)\left(3x+4\right)>16\left(x-1\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(TH:\left\{{}\begin{matrix}4\left(x-1\right)\le0\\\left(x+5\right)\left(3x+4\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x\le-5\\x\ge-\dfrac{4}{3}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x\in(-\infty;-5]\cup\left[-\dfrac{4}{3};1\right]\left(1\right)\)

\(TH:\left\{{}\begin{matrix}4\left(x-1\right)\ge0\\\left(x+5\right)\left(3x+4\right)\ge0\\\left(x+5\right)\left(3x+4\right)>16\left(x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x\le-5\\x\ge-\dfrac{4}{3}\end{matrix}\right.\\-\dfrac{1}{13}< x< 4\\\end{matrix}\right.\)\(\Rightarrow x\in[1;4)\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow x\in(-\infty;5]\cup[\dfrac{-4}{3};4)\)

 

7 tháng 3 2022

\(6;;;;\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}< 181-14x\)

(đoạn 49x^2+7x+42 chắc bạn viết sai đề dấu"-" thành "+")

\(đk:\left\{{}\begin{matrix}7x+7\ge0\\7x-6\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge\dfrac{6}{7}\)

\(bpt\Leftrightarrow\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{\left(7x+7\right)\left(7x-6\right)}+14x+1< 182\left(1\right)\)

\(đặt:\sqrt{7x+7}+\sqrt{7x-6}=t>0\)

\(\Rightarrow t^2=14x+1+2\sqrt{\left(7x+7\right)\left(7x-6\right)}\)

\(\Rightarrow\left(1\right)\Leftrightarrow t^2+t< 182\Leftrightarrow-14< t< 13\)

\(\Rightarrow\sqrt{7x+7}+\sqrt{7x-6}< 13\Leftrightarrow14x+1+2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 169\)

\(\Leftrightarrow2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 168-14x\)

\(\Leftrightarrow\left\{{}\begin{matrix}168-14x\ge0\\\left(7x+7\right)\left(7x-6\right)\ge0\\4\left(7x+7\right)\left(7x-6\right)< \left(168-14x\right)^2\\\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le12\\\left[{}\begin{matrix}x\le-1\\x\ge\dfrac{6}{7}\end{matrix}\right.\\x< 6\\\end{matrix}\right.\)\(\Rightarrow\dfrac{6}{7}\le x< 6\)

 

 

Chọn D