\(\sqrt{x^2-2x+1}=2x\)
giai phuong trinh
giai phuong trinh: \(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x-1}\)
Giai phuong trinh \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
giai phuong trinh \(\sqrt{\frac{x+7}{x+1}}+8=2x^2+\sqrt{2x-1}\)
ĐKXĐ: \(x\ge\frac{1}{2}\)
Đề \(\Rightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}+8-2x^2-\left(\sqrt{2x-1}-\sqrt{3}\right)=0\)
Nhân liên hợp ta được:
\(\frac{\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{\left(\sqrt{2x-1}-\sqrt{3}\right)\left(\sqrt{2x+1}+\sqrt{3}\right)}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{2x-1-3}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(2-x\right)\left(2+x\right)-\frac{2x-4}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\left(x-2\right)\left[\frac{-2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}\right]=0\)
mà \(-\frac{2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}< 0\)
=> x - 2 = 0 => x = 2
Vậy x = 2
Giai phuong trinh ; 2\(\sqrt{x^2-x}-2\sqrt{x}\sqrt{2x-1}+3x=1\)
1) giai phuong trinh:
a) \(x+\sqrt{2x+3}=2x\left(x-2\right)\)
Lời giải:
ĐK: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow \sqrt{2x+3}=2x^2-5x$
$\Leftrightarrow \sqrt{2x+3}-3=2x^2-5x-3$
$\Leftrightarrow \frac{2(x-3)}{\sqrt{2x+3}+3}=(2x+1)(x-3)$
$\Leftrightarrow (x-3)\left[\frac{2}{\sqrt{2x+3}+3}-(2x+1)\right]=0$
Xảy ra 2 TH:
TH1: $x-3=0\Rightarrow x=3$ (thỏa mãn)
TH2: $\frac{2}{\sqrt{2x+3}+3}=2x+1$
Đặt $\sqrt{2x+3}=t(t\geq 0)$ thì pt trở thành: \frac{2}{t+3}=t^2-2$
$\Leftrightarrow 2=(t^2-2)(t+3)\Leftrightarrow t^3+3t^2-2t-8=0$
$\Leftrightarrow (t+2)(t^2+t-4)=0$
Do $t\geq 0$ nên $t=\frac{-1+\sqrt{17}}{2}$
$\Leftrightarrow \sqrt{2x+3}=\frac{-1+\sqrt{17}}{2}\Leftrightarrow x=\frac{3-\sqrt{17}}{4}$ (thỏa mãn)
Vậy........
Giai phuong trinh :\(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x+8}=1+\sqrt{3}\)
Giai phuong trinh:
\(\sqrt{x^2+2x-2}+2\sqrt{x^2+2x-2}=x+2\)
Bình phương hai vế lên ta được:
x2+2x-2+4(x2+2x-2)=(x+2)2
<=>x2+2x-2+4x2+8x-8=x2+4x+4
<=>x2+4x2-x2+2x+8x-4x-2-8-4=0
<=>4x2+6x-14=0
<=>2x2+3x-7=0
Đến đây bạn tự làm tiếp nha. Nhớ k cho mk đấy
Giai phuong trinh:
\(\sqrt{x^2+2x-2}+2\sqrt{x^2+2x-2}=x+2\)
Giai phuong trinh
\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-2\sqrt{2x-5}}=2\sqrt{2}\)