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\(\sqrt{3x-2}+\sqrt{3+x}=\sqrt{5x+4}\)
→ \(\left(\sqrt{3x-2}+\sqrt{3+x}\right)^2=\left(\sqrt{5x+4}\right)^2\)
→ \(3x-2+3+x+2\sqrt{\left(2x-2\right)\left(3+x\right)}=5x+4\)
➝ \(4x+3+2\sqrt{6x+2x^2-6-2x}=5x+4\)
→ \(2\sqrt{2x^2+4x-6}=5x+4-4x-3\)
→ \(2\sqrt{2x^2+4x-6}=x+1\)
→ \(\left(2\sqrt{2x^2+4x-6}\right)^2=\left(x+1\right)^2\)
→ \(4\left(2x^2+4x-6\right)=x^2+2x+1\)
→ \(8x^2+16x-24=x^2+2x+1\)
→ \(8x^2+16x-24-x^2-2x-1=0\)
→ \(7x^2+14x-25=0\)
→ \(x_1=\frac{-7+4\sqrt{14}}{7}\)
\(x_2=\frac{-7-4\sqrt{14}}{7}\)
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 : \(\left\{{}\begin{matrix}2x+9\ge0\\4-x\ge0\\3x+1\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}2x\ge-9\\-x\ge-4\\3x\ge-1\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x\ge-\frac{9}{2}\\x\le4\\x\ge-\frac{1}{3}\end{matrix}\right.\)
<=> \(4\ge x\ge-\frac{1}{3}\)
Ta có : \(\sqrt{2x+9}=\sqrt{4-x}+\sqrt{3x+1}\)
<=> \(\left(\sqrt{2x+9}\right)^2=\left(\sqrt{4-x}+\sqrt{3x+1}\right)^2\)
<=> \(2x+9=\left(4-x\right)+2\sqrt{\left(4-x\right)\left(3x+1\right)}+\left(3x+1\right)\)
<=> \(2x+9=4-x+2\sqrt{12x-3x^2+4-x}+3x+1\)
<=> \(2x+9-4+x-3x-1=2\sqrt{12x-3x^2+4-x}\)
<=> \(4=2\sqrt{12x-3x^2+4-x}\)
<=> \(4^2=\left(2\sqrt{12x-3x^2+4-x}\right)^2\)
<=> \(16=4\left(12x-3x^2+4-x\right)\)
<=> \(4=12x-3x^2+4-x\)
<=> \(0=12x-3x^2-x\)
<=> \(0=11x-3x^2\)
<=> \(0=x\left(11-3x\right)\)
<=> \(\left\{{}\begin{matrix}x=0\\11-3x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\\-3x=-11\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\\x=\frac{11}{3}\end{matrix}\right.\) ( TM )
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)
\(\Leftrightarrow\sqrt{x-1-2.\sqrt{x-1}.2+4}+\sqrt{x-1-2.\sqrt{x-1}.3+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)=5
bạn giải tiếp nhé
Xin lỗi bạn nha mình làm sai 
Nhờ bạn sửa lại \(x\ge3\) và x<3 và nghiệm là \(1\le x\le5\) nha Trần Ngọc Thảo
Ta có:\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}\)(ĐK: \(x\ge1\))
\(=\sqrt{\left(x-1\right)-2\sqrt{x-1}.2+4}+\sqrt{\left(x-1\right)+2\sqrt{x-1}.3+9}\)
\(=\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}\)
\(=\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}+3\right|\)
Thay vào phương trình ta được:
\(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}+3\right|=5\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\sqrt{x-1}+3=5\)(vì \(\sqrt{x-1}\ge0\Rightarrow\sqrt{x-1}+3>0\))
-TH: \(\sqrt{x-1}-2\ge0\Leftrightarrow\sqrt{x-1}\ge2\Leftrightarrow x-1\ge4\Leftrightarrow x\ge3\)thì ta có:
\(\sqrt{x-1}-2+\sqrt{x-1}+3=5\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
\(\Leftrightarrow x=5\)
-TH:\(\sqrt{x-1}-2< 0\Leftrightarrow x< 3\) thì ta có:
\(2-\sqrt{x-1}+\sqrt{x-1}+3=5\)
\(\Leftrightarrow5=5\)(luôn đúng \(\forall1\le x< 3\))
Vậy nghiệm của phương trình là \(1\le x< 3\) và \(x=5\)
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 .
\(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
\(\Leftrightarrow10+\sqrt{3x}=\left(2+\sqrt{6}\right)^2\)
\(\Leftrightarrow\sqrt{3x}=4+6+4\sqrt{6}-10\)
\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)
\(\Leftrightarrow3x=96\)
\(\Leftrightarrow x=32\)
Vậy ...