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Do vế trái dương nên pt chỉ có nghiệm khi \(x\ge\dfrac{3}{4}\), kết hợp điều kiện \(2x^4-3x^2+1\ge0\Rightarrow x\ge1\)
Khi đó:
\(4x-3=\sqrt{2x^4-3x^2+1}+\sqrt{2x^4-x^2}\ge\sqrt{2x^4-3x^2+1+2x^4-x^2}\)
\(\Rightarrow4x-3\ge\sqrt{4x^4-4x^2+1}\)
\(\Rightarrow4x-3\ge\left|2x^2-1\right|=2x^2-1\)
\(\Rightarrow2x^2-4x+2\le0\)
\(\Rightarrow2\left(x-1\right)^2\le0\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ge\frac{1}{2}\)
Bình phương 2 vế ta dc:
\(x^2+2x+2x-1+2\sqrt{\left(x^2+2x\right)\left(2x-1\right)}=3x^2+4x+1\)
\(\Leftrightarrow3x^2+4x+1-x^2-2x-2x+1=2\sqrt{\left(x^2+2x\right)\left(2x-1\right)}\)
\(\Leftrightarrow2x^2+2=2\sqrt{\left(x^2+2x\right)\left(2x-1\right)}\)
\(\Leftrightarrow x^2+1=\sqrt{\left(x^2+2x\right)\left(2x-1\right)}\)
\(\Rightarrow x^4+2x^2+1=2x^3+3x^2-2x\)
\(\Leftrightarrow x^4+2x^2+1-2x^3-3x^2+2x=0\)
\(\Leftrightarrow\left(x^2-x-1\right)^2=0\Leftrightarrow x^2-x-1=0\)
\(\Delta=\left(-1\right)^2-4.\left(-1\right)=5>0\)
\(\Rightarrow x_1=\frac{1+\sqrt{5}}{2}\left(TM\right);x_2=\frac{1-\sqrt{5}}{2}\left(loai\right)\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow (x^2-2x)+(\sqrt{4x+1}-3)+(\sqrt{x-1}-1)=0$
$\Leftrightarrow x(x-2)+\frac{4(x-2)}{\sqrt{4x+1}+3}+\frac{x-2}{\sqrt{x-1}+1}=0$
$\Leftrightarrow (x-2)\left[x+\frac{4}{\sqrt{4x+1}+3}+\frac{1}{\sqrt{x-1}+1}\right]=0$
Dễ thấy với mọi $x\geq 1$ thì biểu thức trong ngoặc vuông luôn dương.
$\Rightarrow x-2=0$
$\Leftrightarrow x=2$ (tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\sqrt{x^2+1}=t>0\)
\(\Rightarrow\left(4x-1\right)t=2t^2-2x\)
\(\Leftrightarrow2t^2-\left(4x-1\right)t-2x=0\)
\(\Delta=\left(4x-1\right)^2+16x=\left(4x+1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{4x-1-\left(4x+1\right)}{4}=-\dfrac{1}{2}\left(loại\right)\\t=\dfrac{4x-1+4x+1}{4}=2x\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+1}=2x\) (\(x\ge0\))
\(\Leftrightarrow x^2+1=4x^2\)
\(\Rightarrow x=\dfrac{\sqrt{3}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có, bình phương 2 vế lên là chứng minh xong
Dấu "=" xảy ra khi a hoặc b bằng 0
BĐT này thường được sử dụng trong trường hợp ẩn x nằm ở 2 căn là trái dấu nhau
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\leftrightarrow\sqrt{4x^2-1}-\sqrt{2x^2-x}-\sqrt{2x+1}+\sqrt{x}=0\)
\(\leftrightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-\sqrt{x\left(2x-1\right)}-\sqrt{2x+1}+\sqrt{x}=0\)
\(\leftrightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-1\right)-\sqrt{x}\left(\sqrt{2x-1}-1\right)=0\)
\(\leftrightarrow\left(\sqrt{2x+1}-\sqrt{x}\right)\left(\sqrt{2x-1}-1\right)=0\)
\(\leftrightarrow\sqrt{2x+1}-\sqrt{x}=0hoặc\sqrt{2x-1}-1=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{x^2+2x}+\sqrt{2x-1}\right)^2=3x^2+4x+1\)
\(x^2+4x-1+2\sqrt{2x^3+3x^2-2x}=3x^2+4x+1\)
\(2\sqrt{2x^3+3x^2-2x}=2x^2+2\)
\(\sqrt{2x^3+3x^2-2x}=x^2+1\)
\(2x^3+3x^2-2x=x^4+2x^2+1\)
\(x^4-2x^3-x^2+2x+1=0\)
pt đối xứng bậc 4 tự làm được chưa?
\(\left(\sqrt{x^2+2x}+\sqrt{2x+1}\right)^2=3x^2+4x+1\)
\(x^2+4x-1+2\sqrt{2x^3+3x^2-2x}=3x^2+4x+1\)
\(\sqrt{2x^3+3x^2+2x}=x^2+1\)
\(2x^3+3x^2+2x=x^4+2x^2+1\)
\(x^4-2x^3-x^2+2x+1=0\)
\(\left(x^2-x-1\right)^2=0\)
\(x^2-x-1=0\)
\(x^2-x+\frac{1}{4}-\frac{5}{4}=0\)
\(\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)
\(x-\frac{1}{2}=\frac{\sqrt{5}}{2}\)
\(x=\frac{1+\sqrt{5}}{2}\)
lên thánh nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
cộng hai vế cho 2x ta có
\(\left(2x+1\right)^2-2x=\sqrt{4x+1}\)
\(2x+1-\sqrt{2x}=4x+1\)
\(-\sqrt{2x}=2x\)
suy ra x=\(0;\frac{1}{2}\)
\(2x^2+2x+1=\sqrt{4x+1}\) ( ĐK : \(x\ge-\dfrac{1}{4}\) )
\(\Leftrightarrow4x^2+4x+2-2\sqrt{4x+1}=0\)
\(\Leftrightarrow4x^2+\left(4x+1-2\sqrt{4x+1}+1\right)=0\)
\(\Leftrightarrow4x^2+\left(\sqrt{4x+1}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2=0\\\left(\sqrt{4x+1}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=0\)