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\(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(ĐK:x\ge3\right)\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x-3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(c\right)\\x=6\left(c\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình là \(S=\left\{3;6\right\}\)
\(\sqrt{3x-2}+\sqrt{3+x}=\sqrt{5x+4}\)
→ \(\left(\sqrt{3x-2}+\sqrt{3+x}\right)^2=\left(\sqrt{5x+4}\right)^2\)
→ \(3x-2+3+x+2\sqrt{\left(2x-2\right)\left(3+x\right)}=5x+4\)
➝ \(4x+3+2\sqrt{6x+2x^2-6-2x}=5x+4\)
→ \(2\sqrt{2x^2+4x-6}=5x+4-4x-3\)
→ \(2\sqrt{2x^2+4x-6}=x+1\)
→ \(\left(2\sqrt{2x^2+4x-6}\right)^2=\left(x+1\right)^2\)
→ \(4\left(2x^2+4x-6\right)=x^2+2x+1\)
→ \(8x^2+16x-24=x^2+2x+1\)
→ \(8x^2+16x-24-x^2-2x-1=0\)
→ \(7x^2+14x-25=0\)
→ \(x_1=\frac{-7+4\sqrt{14}}{7}\)
\(x_2=\frac{-7-4\sqrt{14}}{7}\)
ĐKXĐ : \(\left\{{}\begin{matrix}3x-2\ge0\\3+x\ge0\\5x+4\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge\frac{2}{3}\\x\ge-3\\x\ge-\frac{4}{5}\end{matrix}\right.\)
=> \(x\ge\frac{2}{3}\) (1)
Ta có : \(\sqrt{3x-2}+\sqrt{3+x}=\sqrt{5x+4}\)
<=> \(\left(\sqrt{3x-2}+\sqrt{3+x}\right)^2=\left(\sqrt{5x+4}\right)^2\)
<=> \(\left(3x-2\right)+2\sqrt{\left(3x-2\right)\left(3+x\right)}+\left(3+x\right)=5x+4\)
<=> \(3x-2+2\sqrt{\left(3x-2\right)\left(3+x\right)}+3+x=5x+4\)
<=> \(2\sqrt{\left(3x-2\right)\left(3+x\right)}=5x+4+2-3-x-3x\)
<=> \(2\sqrt{\left(3x-2\right)\left(3+x\right)}=x+3\)
<=> \(\sqrt{\left(3x-2\right)\left(3+x\right)}=\frac{x+3}{2}\)
ĐKXĐ : \(\frac{x+3}{2}\ge0\)
=> \(x+3\ge0\)
=> \(x\ge-3\) (2)
Từ (1) và (2)
=> \(x\ge\frac{2}{3}\)
<=> \(\left(\sqrt{\left(3x-2\right)\left(3+x\right)}\right)^2=\left(\frac{x+3}{2}\right)^2\)
<=> \(\left(3x-2\right)\left(3+x\right)=\frac{\left(x+3\right)^2}{4}\)
<=> \(9x-6+3x^2-2x=\frac{x^2+6x+9}{4}\)
<=> \(\frac{4\left(9x-6+3x^2-2x\right)}{4}=\frac{x^2+6x+9}{4}\)
<=> \(4\left(9x-6+3x^2-2x\right)=x^2+6x+9\)
<=> \(36x-24+12x^2-8x=x^2+6x+9\)
<=> \(36x-24+12x^2-8x-x^2-6x-9=0\)
<=> \(22x-33+11x^2=0\)
<=> \(11x^2+33x-11x-33=0\)
<=> \(11x\left(x-1\right)+33\left(x-1\right)=0\)
<=> \(\left(11x+33\right)\left(x-1\right)=0\)
<=> \(\left\{{}\begin{matrix}11x+33=0\\x-1=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=-3\left(L\right)\\x=1\left(TM\right)\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là x = 1 .
ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ge0\\2x+3\ge0\\x+20\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-\frac{3}{2}\\x\ge-20\end{matrix}\right.\)
\(\sqrt{x+1}+\sqrt{2x+3}=\sqrt{x+20}\)
\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=\left(\sqrt{x+20}\right)^2\)
\(\Leftrightarrow x+1+2\sqrt{\left(x+1\right)\left(2x+3\right)}+2x+3=x+20\)
\(\Leftrightarrow3x+4+2\sqrt{\left(x+1\right)\left(2x+3\right)}=x+20\)
\(\Leftrightarrow2\sqrt{\left(x+1\right)\left(2x+3\right)}=-2x+16\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=16-2x\)
\(\Leftrightarrow\left\{{}\begin{matrix}16-2x\ge0\\4\left(2x^2+5x+3\right)=\left(16-2x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\8x^2+20x+12=256-64x+4x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\4x^2+84x-244=0\end{matrix}\right.\)
còn lại bn tự làm nha
ĐKXĐ : \(x-1\ge0\)
=> \(x\ge1\)
Ta có : \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=5\)
<=> \(\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=5\)
<=> \(\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=5\)
<=> \(\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}=5\)
<=> \(|\sqrt{x-1}-1|+|\sqrt{x-1}+1|=5\)
<=> \(|\sqrt{x-1}-1|+\sqrt{x-1}+1=5\) ( 1 )
+, TH 1 : \(\sqrt{x-1}-1\ge0\) <=> \(x\ge2\) . Khi đó phương trình (1) được :
\(\sqrt{x-1}-1+\sqrt{x-1}+1=5\)
<=> \(2\sqrt{x-1}=5\)
<=> \(\sqrt{x-1}=2,5\)
<=> \(x-1=6,25\)
<=> \(x=7,25\) ( TM )
TH 2 : \(\sqrt{x-1}-1\le0\) <=> \(x\le2\) . Khi đó phương trình (1) được :
\(1-\sqrt{x-1}+\sqrt{x-1}+1=5\)
<=> \(2=5\) ( Vô lý )
Vậy phương trình trên có nghiệm duy nhất là x = 7,25 .
Bp cả hai vế ta được:X - 2 ≤9
=>X ≤11
vậy 0<X=<11
ĐK: \(x\ge2\) nên nghiệm là \(2\le x\le11\) nha Nguyễn Hoàng Anh