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WR
1
14 tháng 7 2019
ĐK \(x\ge-\frac{2}{3}\)
Pt
<=> \(x^3+2x^2-4x-3+3\left(x+1\right)\left(x+1-\sqrt{3x+2}\right)=0\)
<=> \(\left(x+3\right)\left(x^2-x-1\right)+3\left(x+1\right).\frac{\left(x+1\right)^2-3x-2}{x+1+\sqrt{3x+2}}=0\)
<=> \(\left(x+3\right)\left(x^2-x-1\right)+3\left(x+1\right).\frac{x^2-x-1}{x+1+\sqrt{3x+2}}=0\)
<=> \(\orbr{\begin{cases}x^2-x-1=0\\x+3+\frac{3\left(x+1\right)}{x+1+\sqrt{3x+2}}=0\left(2\right)\end{cases}}\)
Pt (2) vô nghiệm do VT>0 với mọi \(x\ge-\frac{2}{3}\)
=> \(x=\frac{1\pm\sqrt{5}}{2}\)(tmĐKXĐ)
Vậy \(x=\frac{1\pm\sqrt{5}}{2}\)
DN
1
11 tháng 5 2015
ĐK: x \(\ge\)\(\frac{8}{3}\)
pt <=> \(4.\left(x-3\right)+9-3.\sqrt{5x-6}=\sqrt{3x-8}-1\)
<=> \(4.\left(x-3\right)+3.\left(3-\sqrt{5x-6}\right)=\sqrt{3x-8}-1\)
<=> \(4.\left(x-3\right)+3.\frac{\left(3-\sqrt{5x-6}\right)\left(3+\sqrt{5x-6}\right)}{3+\sqrt{5x-6}}=\frac{\left(\sqrt{3x-8}-1\right)\left(\sqrt{3x-8}+1\right)}{\sqrt{3x-8}+1}\)
<=> \(4.\left(x-3\right)+3.\frac{9-5x+6}{3+\sqrt{5x-6}}=\frac{3x-8-1}{\sqrt{3x-8}+1}\)
<=> \(4.\left(x-3\right)+15.\frac{3-x}{3+\sqrt{5x-6}}-3.\frac{x-3}{\sqrt{3x-8}+1}=0\)
<=> \(\left(x-3\right)\left(4-\frac{15}{3+\sqrt{5x-6}}-\frac{3}{\sqrt{3x-8}+1}\right)=0\)
<=> x = 3 (thoả mãn) hoặc \(4-\frac{15}{3+\sqrt{5x-6}}-\frac{3}{\sqrt{3x-8}+1}=0\) (2)
Giải (2): (2) <=> \(\frac{15}{6}-\frac{15}{3+\sqrt{5x-6}}+\frac{3}{2}-\frac{3}{\sqrt{3x-8}+1}=0\)
<=> \(15\left(\frac{1}{6}-\frac{1}{3+\sqrt{5x-6}}\right)+3.\left(\frac{1}{2}-\frac{1}{\sqrt{3x-8}+1}\right)=0\)
<=> \(15.\frac{\sqrt{5x-6}-3}{6.\left(3+\sqrt{5x-6}\right)}+3.\frac{\sqrt{3x-8}-1}{2.\left(\sqrt{3x-8}+1\right)}=0\)
<=> \(15.\frac{5.\left(x-3\right)}{6.\left(3+\sqrt{5x-6}\right)^2}+3.\frac{3.\left(x-3\right)}{2.\left(\sqrt{3x-8}+1\right)^2}=0\)
<=> \(\left(x-3\right).\left(\frac{75}{6.\left(3+\sqrt{5x-6}\right)^2}+\frac{9}{2.\left(\sqrt{3x-8}+1\right)^2}\right)=0\)
<=> x = 3 Vì \(\frac{75}{6.\left(3+\sqrt{5x-6}\right)^2}+\frac{9}{2.\left(\sqrt{3x-8}+1\right)^2}>0\) với mọi x \(\ge\frac{8}{3}\)
Vậy pt có 1 nghiệm duy nhất x = 3
23 tháng 9 2021
\(ĐK:x\in R\)
Đặt \(\sqrt{x^2+3}=t\left(t\ge0\right)\)
\(PT\Leftrightarrow2t^2-\left(7x+1\right)t+3x^2+3x=0\\ \Delta=\left(7x+1\right)^2-4\cdot2\left(3x^2+3x\right)=25x^2-10x+1=\left(5x-1\right)^2\ge0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{7x+1-5x+1}{4}\\t=\dfrac{7x+1+5x-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{2x+2}{4}=\dfrac{x+1}{2}\\t=\dfrac{12x}{4}=3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=\dfrac{x+1}{2}\\\sqrt{x^2+3}=3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3=\dfrac{x^2+2x+1}{4}\\x^2+3=9x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x^2-2x+11=0\\x^2=\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\Delta=4-132< 0\\\left[{}\begin{matrix}x=\dfrac{\sqrt{6}}{4}\\x=-\dfrac{\sqrt{6}}{4}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{\sqrt{6}}{4};\dfrac{\sqrt{6}}{4}\right\}\)
NG
0
NN
3
25 tháng 12 2015
\(\Leftrightarrow\left(6x+2\right)\sqrt{2x^2-1}=10x^2+3x-6\)
\(\Leftrightarrow\left(2x^2-1\right)\left(36x^2+24x+4\right)=100x^4+9x^2+36+60x^3-36x-120x^2\)
23 tháng 3 2017
Câu 1:
Đặt \(3x-16y-24=k\left(k\in N\right)\) khi đó:
\(\sqrt{9x^2+16x+32}=k\Rightarrow9x^2+16x+32=k^2\)
\(\Rightarrow9\left(x+\dfrac{8}{9}\right)^2+\dfrac{224}{9}=k^2\)
\(\Rightarrow\dfrac{1}{9}\left(\left(9x+8\right)^2-9k^2\right)=-\dfrac{224}{9}\)
\(\Rightarrow\left(9x+8+3k\right)\left(9x+8-3k\right)=-224\)
tự giải nốt
23 tháng 3 2017
Câu 2:
\(4x^3+5x^2+1=\sqrt{3x+1}-3x\)
\(\Leftrightarrow4x^3+5x^2+3x+1=\sqrt{3x+1}\)
\(\Leftrightarrow 16x^6+40x^5+49x^4+38x^3+19x^2+6x+1=3x+1\)
\(\Leftrightarow x(4x+1)(4x^4+9x^3+10x^2+7x+3)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{4}\end{matrix}\right.\)