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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
H
1 tháng 7 2019
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
NG
0
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:
\(4x^3+5x^2+1=\sqrt{3x+1}-3x\)
\(\Leftrightarrow4x^3+5x^2+3x+1=\sqrt{3x+1}\)
\(\Leftrightarrow 16x^6+40x^5+49x^4+38x^3+19x^2+6x+1=3x+1\)
\(\Leftrightarow x(4x+1)(4x^4+9x^3+10x^2+7x+3)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{4}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
a/ ĐXĐK: ...
\(\Leftrightarrow9x^2-1-x-8x\sqrt{x+1}=0\)
\(\Leftrightarrow x^2-x-1+8x\left(x-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow x^2-x-1+\frac{8x\left(x^2-x-1\right)}{x+\sqrt{x+1}}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\Rightarrow x=...\\\frac{-8x}{x+\sqrt{x+1}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow-8x=x+\sqrt{x+1}\)
\(\Leftrightarrow-9x=\sqrt{x+1}\) (\(x\le0\))
\(\Leftrightarrow81x^2-x-1=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1-5\sqrt{13}}{162}\\x=\frac{1+5\sqrt{13}}{162}>0\left(l\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
d/
\(\Leftrightarrow3x^2+2\left(x^2+x+1\right)-5x\sqrt{x^2+x+1}=0\)
Đặt \(\sqrt{x^2+x+1}=a\)
\(\Leftrightarrow3x^2-5ax+2a^2=0\)
\(\Leftrightarrow\left(x-a\right)\left(3x-2a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=a\\3x=2a\end{matrix}\right.\) (\(x\ge0\))
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+1}=x\\2\sqrt{x^2+x+1}=3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=x^2\\2\left(x^2+x+1\right)=9x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\7x^2-2x-2=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1+\sqrt{15}}{7}\)