Đặt \(\sqrt{2x^2-1}=t\) ta có: \(x^2+4x+4=-t^2+2t\sqrt{7}-7\Leftrightarrow\left(x+2\right)^2=-\left(t-\sqrt{7}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}x+2=0\\t-\sqrt{7}=0\end{matrix}\right.\Rightarrow x=-2\)

Bài này dùng bất đẳng thức cho nhanh bạn ơi

mình gửi bài này từ hồi học kì 1 lớp 9, giờ lớp 10 lại có người trả lời :v

\(3x^2+4x+10=2\sqrt{14x^2-7}\)

\(\Leftrightarrow2\sqrt{14x^2-7}=3x^2+4x+10\)

\(\Leftrightarrow\left(2\sqrt{14x^2-7}\right)^2=\left(3x^2+4x+10\right)^2\)

\(\Leftrightarrow56x^2-28=9x^4+76x^2+10+24x^3+80x\)\(\Leftrightarrow56x^2-28-9x^4-76x^2-100-24x^3-80x=0\)\(\Leftrightarrow-20x^2-128-9x^4-24x^3-80x=0\)

\(\Leftrightarrow-9x^4-18x^3-6x^3-12x^2-8x^2-16x-64x-128=0\)\(\Leftrightarrow-9x^3\cdot\left(x+2\right)-6x^2\cdot\left(x+2\right)-8x\cdot\left(x+2\right)-64\left(x+2\right)=0\)\(\Leftrightarrow-\left(x+2\right)\cdot\left(9x^3+18x^2+12x^2-24x+32x+64\right)=0\)\(\Leftrightarrow-\left(x+2\right)\cdot\left(9x^2\left(x+2\right)-12x\cdot\left(x+2\right)+32\left(x+2\right)\right)=0\)

\(\Leftrightarrow-\left(x+2\right)^2\cdot\left(9x^2-12x+32\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-\left(x+2\right)^2=0\\9x^2-12x+32=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x\notin R\end{matrix}\right.\)

\(\Leftrightarrow x=-2\)

Lời giải:

a) \(3x^2+4x+10=2\sqrt{14x^2-7}=2\sqrt{7(2x^2-1)}\)

Áp dụng BĐT AM-GM:

\(3x^2+4x+10\leq 7+(2x^2-1)\)

\(\Leftrightarrow x^2+4x+4\leq 0\)

\(\Leftrightarrow (x+2)^2\leq 0\)

Mà \((x+2)^2\geq 0\forall x\in\mathbb{R}\Rightarrow (x+2)^2=0\)

\(\Leftrightarrow x=-2\) (thử lại thấy thỏa mãn)

b) Có:

\(\sqrt{4x^2+5x+1}+3=2\sqrt{x^2-x+1}+9x\)

\(\Leftrightarrow \sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=9x-3\)

\(\Leftrightarrow \frac{9x-3}{\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}}-(9x-3)=0\)

\(\Leftrightarrow (9x-3)\left(\frac{1}{\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9x-3=0\Leftrightarrow x=\dfrac{1}{3}\\\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}=1\left(2\right)\end{matrix}\right.\)

Xét (2):

Ta thấy:

\(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}\geq \sqrt{4x^2-4x+4}=\sqrt{(2x-1)^2+3}\geq \sqrt{3}>1\)

Do đó \((2)\) vô lý

Vậy PT có nghiệm \(x=\frac{1}{3}\)

giỏi quá ><

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

=>\(1\le\sqrt{x-5}\le\frac{3}{2}\)

=>\(1\le x-5\le\frac{9}{4}\)

\(\iff6\le x\le\frac{29}{4}\)

\(Đk:14x^2-7\ge0\Leftrightarrow x\ge\frac{1}{\sqrt{2}};x\le\frac{-1}{\sqrt{2}}\)

Nhận thấy x = –2 là nghiệm của pt , ta phân tích: 

\(3x^2+4x+10-2\sqrt{14x^2-7}=0\)

\(\Leftrightarrow3.x^2+4x+10-2\sqrt{14x^2-7}=0\)

\(\Leftrightarrow3.x+2^2-2.1+4x-\sqrt{14x^2-7}=0\)

\(\Leftrightarrow\frac{3.x+2^2-2.16^2+8x+1-14x^2+7}{1+4x+\sqrt{14x^2-7}}=0\) nhân liên hợp

\(\Leftrightarrow\frac{3.x+2^2-4.x+2^2}{1+4x+\sqrt{14x^2-7}}=0\)

\(\Leftrightarrow x+2^2.\frac{3-4}{1+4x+\sqrt{14x^2-7}}=0\)

+, \(x+2=0\Leftrightarrow x=-2TM\)

+, \(\frac{3-4}{1+4x+\sqrt{14x^2+7}}=0\)

\(\Leftrightarrow1+4x+\sqrt{14x^2-7}=\frac{4}{3}\)

\(\Leftrightarrow\sqrt{14x^2-7}=\frac{1}{3}-4x;Đk\frac{1}{3}-4x\ge0\)

\(\Leftrightarrow14x^2-7=16x^2-\frac{2}{3x}+\frac{1}{9}\)

\(\Leftrightarrow2x^2-\frac{2}{3x}+\frac{64}{9}=0VN\)

Vậy: \(x=2\)

P/s: Tôi mớp lớp 6, sai chỗ nào thì sửa hộ nhé. Thanks

 vn la cai gi

a,đk -1<x<7

x+1+2 căn 7-x-2 căn x+1=căn (x+1)(7-x)