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1: =>3x^2+5x-7=3x+14
=>2x=21
=>x=21/2
2;=>x+4=4
=>x=0
3: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\4x^2-20x+25-4x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\4x^2-24x+32=0\end{matrix}\right.\)
=>x>=5/2 và x^2-6x+8=0
=>x=4
4: \(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2+2x-1=x^2-2x+1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
5: \(\Leftrightarrow\sqrt{2x+16}=x-4\)
=>x>=4 và x^2-8x+16=2x+16
=>x>=4 và x^2-10x=0
=>x=10
1) ĐK: \(x\ge-1\)
\(\sqrt{9x^2+9x+4}>9x+3-\sqrt{x+1}\)
<=> \(\sqrt{9x^2+9x+4}+\sqrt{x+1}>9x+3\)(1)
TH1: 9x + 3 \(\le\)0 <=> x\(\le-\frac{1}{3}\)
(1) luôn đúng
Th2: x\(>-\frac{1}{3}\)
<=> \(\left(\frac{1}{2}x+1-\sqrt{x+1}\right)+\left(\frac{17}{2}x+2-\sqrt{9x^2+9x+4}\right)< 0\)
<=> \(\frac{\frac{1}{4}x^2}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{\frac{253}{4}x^2}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}< 0\)
<=> \(\frac{x^2}{4}\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)< 0\)vô nghiệm
Vì với x \(>-\frac{1}{3}\):
ta có: \(\frac{1}{2}x+1+\sqrt{x+1}>0\)
\(\frac{17}{2}x+2+\sqrt{9x^2+9x+4}=\frac{17}{2}x+2+\sqrt{3\left(x+\frac{1}{2}\right)^2+\frac{7}{4}}>\frac{17}{2}x+2+1>0\)
=> \(\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)>0\)với x \(>-\frac{1}{3}\) và \(x^2\ge0\)với mọi x
=> \(\frac{x^2}{4}\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)\ge0\)với x\(>-\frac{1}{3}\)
Vậy \(x< -\frac{1}{3}\)
Xin lỗi bạn kết luận bài 1 là:
\(-1\le x\le-\frac{1}{3}\)
Bài 2) \(2+\sqrt{x+2}-x\sqrt{x+2}=x\left(\sqrt{x+2}-x\right)\)(2)
ĐK: \(x\ge-2\)
(2) <=> \(2+\sqrt{x+2}+x^2-2x\sqrt{x+2}=0\)
<=> \(8+4\sqrt{x+2}+4x^2-8x\sqrt{x+2}=0\)
<=> \(\left(2x-1\right)^2-4\left(2x-1\right)\sqrt{x+2}+4\left(x+2\right)-1=0\)
<=> \(\left(2x-1-2\sqrt{x+2}\right)^2-1=0\)
<=> \(\left(x-1-\sqrt{x+2}\right)\left(x-\sqrt{x+2}\right)=0\)
<=> \(\orbr{\begin{cases}x-1=\sqrt{x+2}\left(3\right)\\x=\sqrt{x+2}\left(4\right)\end{cases}}\)
(3) <=> \(\hept{\begin{cases}x\ge1\\x^2-3x-1=0\end{cases}}\Leftrightarrow x=\frac{3+\sqrt{13}}{2}\left(tm\right)\)
(4) <=> \(\hept{\begin{cases}x\ge0\\x^2-x-2=0\end{cases}\Leftrightarrow}x=2\left(tm\right)\)
Kết luận:...
đề đungs \(\sqrt{10x+1}+\sqrt{3x-5}=\sqrt{9x+4}+\sqrt{2x-2}\). ĐK: \(x\ge\frac{5}{3}\)
\(\Leftrightarrow\)\(\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)
\(\Leftrightarrow\)\(\frac{10x+1-9x-4}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{3x-5-2x+2}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(\frac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)
\(\Leftrightarrow\)\(x=3\) ( nhan )
\(\sqrt{3x^2+6x+16}+\sqrt{x^2+2x}=2\sqrt{x^2+2x+4}\)
\(\Leftrightarrow\sqrt{3\left(x^2+2x\right)+16}+\sqrt{x^2+2x}=2\sqrt{x^2+2x+4}\)
Đặt \(t=x^2+2x\), \(\left(t\ge0\right)\) phương trình trở thành:
\(\sqrt{3t+16}+\sqrt{t}=2\sqrt{t+4}\)
\(3t+16+t+2\sqrt{t\left(3t+16\right)}=4\left(t+4\right)\)
\(\Leftrightarrow2\sqrt{t\left(3t+16\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\3t+16=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\) \(\Leftrightarrow t=0\).
Với \(t=0\Leftrightarrow x^2+2x=0\) \(\Leftrightarrow x\left(x+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\).
Vậy x = 0 hoặc x = -2.
\(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2+16}\)
\(\Leftrightarrow\left(2\sqrt{2x+4}+4\sqrt{2-x}\right)^2=\left(\sqrt{9x^2+16}\right)^2\)
\(\Leftrightarrow4\left(2x+4\right)+16\left(2-x\right)+16\sqrt{2x+4}\sqrt{2-x}=9x^2+16\)
\(\Leftrightarrow4.2\left(4-x^2\right)+16\sqrt{2\left(4-x^2\right)}=x^2+8x\)
Đặt \(\sqrt{2\left(4-x^2\right)}=a\)
\(\Rightarrow4a^2+16a=x^2+8x\)
\(\Leftrightarrow\left(2a-x\right)\left(2a+x+8\right)=0\)
Làm nốt