Giải pt:$\sqrt{5-3x}+\sqrt{x+1}=\sqrt{3x^2-4x+4}$
Giải pt:$\sqrt{5-3x}+\sqrt{x+1}=\sqrt{3x^2-4x+4}$
#1
Đã gửi 22-11-2017 - 11:41
#2
Đã gửi 22-11-2017 - 17:29
PT <=> $(5-3x)+(x+1)+2\sqrt{(5-3x)(x+1)}=3x^{2}-4x+4<=> 2\sqrt{-3x^{2}+2x+5}=3x^{2}-2x-2<=>2\sqrt{-3x^{2}+2x+5}=-(-3x^{2}+2x+5)+3$
Đặt $t=\sqrt{-3x^{2}+2x+5}...$
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#3
Đã gửi 22-11-2017 - 18:15
ĐK: $-1\leqslant x\leqslant \frac{3}{5}$
Đặt: $\left\{\begin{matrix}\sqrt{5-3x}=a \\ \sqrt{x+1}=b \end{matrix}\right.$ (a,b>=0)
$\Rightarrow \left\{\begin{matrix}5-3x=a^2 \\ x+1=b^2 \end{matrix}\right. \Rightarrow a^2+3b^2=8$
Ta có: $3x^2-4x+4=a^2-a^2b^2+b^2+3$
$\Rightarrow a+b=\sqrt{a^2-a^2b^2+b^2+3}$
$\Leftrightarrow a^2+2ab+b^2=a^2-a^2b^2+b^2+3\Leftrightarrow a^2b^2+2ab-3=0\Leftrightarrow (ab-1)(ab+3)=0$
$\Leftrightarrow ab=1(thỏa mãn) hoặc ab=-3(loại)$
Từ $\left\{\begin{matrix}a^2+3b^2=8 \\ ab=1 \end{matrix}\right. \Rightarrow b^2=\frac{4+\sqrt{13}}{3} hoặc b^2=\frac{4-\sqrt{13}}{3}$
$\Rightarrow x=b^2-1=\frac{1+\sqrt{13}}{3} hoặc \frac{1-\sqrt{13}}{3}$
Bài viết đã được chỉnh sửa nội dung bởi hangnguyen2003: 22-11-2017 - 18:17
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