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\(9x^2+\sqrt{4x-5}>\sqrt{x}+25\)
ĐK: \(x\ge\frac{5}{4}\)
\(9x^2+\sqrt{4x-5}>\sqrt{x}+25\)
<=> \(9x^2-25+\sqrt{4x-5}-\sqrt{x}>0\)
<=> \(\left(3x-5\right)\left(3x+5\right)+\frac{3x-5}{\sqrt{4x-5}+\sqrt{x}}>0\)
<=> \(\left(3x-5\right)\left(3x+5+\frac{1}{\sqrt{4x-5}+\sqrt{x}}\right)>0\)
<=> 3x - 5 > 0 vì \(3x+5+\frac{1}{\sqrt{4x-5}+\sqrt{x}}>0\) với mọi \(x\ge\frac{5}{4}\)
<=> x > 5/3 thỏa mãn đk
Em trục căn thức:
\(\sqrt{x+3}-2\sqrt{x}=\sqrt{2x+2}-\sqrt{3x+1}\)
<=> \(\frac{-3x+3}{\sqrt{x+3}+2\sqrt{x}}=\frac{-x+1}{\sqrt{2x+2}+\sqrt{3x+1}}\)
=> nhân tử chung là -x + 1 . Tự làm tiếp nhé!
làm như cô thì vẫn cần phải đánh giá rất khó chịu nhé
\(\sqrt{x+3}-2\sqrt{x}=\sqrt{2x+2}-\sqrt{3x+1}\left(ĐKXĐ:x\ge0\right)\)
\(< =>\sqrt{x+3}-\sqrt{2x+2}+\sqrt{3x+1}-2\sqrt{x}=0\)
\(< =>\frac{\sqrt{x+3}^2-\sqrt{2x+2}^2}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{\sqrt{3x+1}^2-4\sqrt{x}^2}{\sqrt{3x+1}+2\sqrt{x}}=0\)
\(< =>\frac{x+3-2x-2}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{3x+1-4x}{\sqrt{3x+1}+2\sqrt{x}}=0\)
\(< =>\frac{1-x}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{1-x}{\sqrt{3x+1}+2\sqrt{x}}=0\)
\(< =>\left(1-x\right)\left(\frac{1}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{1}{\sqrt{3x+1}+2\sqrt{x}}\right)=0< =>x=1\)
a: ĐKXĐ: \(\begin{cases}5x^2+14x+9\ge0\\ x^2-x-20\ge0\\ x+1\ge0\end{cases}\Rightarrow\begin{cases}\left(x+1\right)\left(5x+9\right)\ge0\\ \left(x-5\right)\left(x+4\right)\ge0\\ x\ge-1\end{cases}\)
=>x>=5
TA có: \(\sqrt{5x^2+14x+9} \le 5\sqrt{x+1} + \sqrt{x^2-x-20}\)
=>\(5x^2+14x+9 \le 25(x+1) + x^2-x-20 + 10\sqrt{(x+1)(x^2-x-20)}\)
=>\(5x^2+14x+9 \le x^2 + 24x + 5 + 10\sqrt{(x+1)^2(x-5)}\)
=>\(4x^2 - 10x + 4 \le 10(x+1)\sqrt{x-5}\)
=>\(2x^2 - 5x + 2 \le 5(x+1)\sqrt{x-5}\)
=>\((2x-1)(x-2) \le 5(x+1)\sqrt{x-5}\) (1)
Đặt \(t=\sqrt{x-5}\ge0\implies x=t^2+5\)
(1) sẽ trở thành: \(2(t^2+5)^2 - 5(t^2+5) + 2 \le 5(t^2+6)t\)
=>\(2(t^4 + 10t^2 + 25) - 5t^2 - 25 + 2 \le 5t^3 + 30t\)
=>\(2t^4 + 20t^2 + 50 - 5t^2 - 23 \le 5t^3 + 30t\)
=>\(2t^4 - 5t^3 + 15t^2 - 30t + 27 \le 0\)
=>\((t-1)(2t-3)(t^2 + 6) \le 0\)
=>(t-1)(2t-3)<=0
=>1<=t<=3/2
=>\(1\le\sqrt{x-5}\le\frac{3}{2}\)
=>\(1\le x-5\le\frac{9}{4}\)
\(\iff6\le x\le\frac{29}{4}\)
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