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a) ĐK: \(x\ge2\)
\(\sqrt{x-1}=1+\sqrt{x-2}\)
<=>\(x-1=1+x-2+2\sqrt{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(loại\right)\\x=2\left(tm\right)\end{cases}}}\)
b) ĐK: x>=10/3
Đặt:\(\sqrt{3x-10}=t\left(t\ge0\right)\Rightarrow3x=t^2+10\)
\(x^2+3\left(t^2+10\right)+20=2t\)
\(\Leftrightarrow x^2+3t^2-2t+50=0\)
\(\Leftrightarrow x^2+3\left(t^2-2.t.\frac{1}{3}+\frac{1}{9}\right)-\frac{1}{3}+50=0\)
<=>\(x^2+3\left(t-\frac{1}{3}\right)^2+\frac{149}{3}=0\)phương trình voo ngiệm
vào trong câu hỏi khác của mình rồi trả lời với mình xin các cậu đúng cho 3 k
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ĐKXĐ: \(x\ge-\dfrac{10}{3}\)
\(\left(x^2+6x+9\right)+\left(3x+10-2\sqrt{3x+10}+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\\sqrt{3x+10}-1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-3\)
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\(2\sqrt{2\left(x+2\right)}\)+4\(\sqrt{2-x}=\sqrt{9x^2+16}\)
=>\(x=\frac{4\sqrt{2}}{3}\)
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\(dk:x\ge\frac{-10}{3}\)
\(x^2+9x+20=2\sqrt{3x+10}\Leftrightarrow x^2+6x+10+\left(3x+10\right)-2\sqrt{3x+10}=0\Leftrightarrow\left(x^2+6x+9\right)+\left(3x+10-2\sqrt{3x+10}+1\right)=\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\Rightarrow\left\{{}\begin{matrix}x+3=0\\\sqrt{3x+10}=1\end{matrix}\right.\Leftrightarrow x=-3\left(tmdk\right)\)
Điều kiện 3x + 10 ≥ 0 =>x ≥ -10 /3
Pt <=> (3x + 10)² + 7(3x + 10) + 10 = 18\(\sqrt{\left(3x+10\right)}\)
Đặt y = \(\sqrt{\left(3x+10\right)}\) ≥ 0 pt trở thành
y⁴ + 7y² - 18y + 10 = 0
<=> (y - 1)²(y² + 2y + 10) = 0
<=> (y-1)^2 [(y+1)^2 +9] =0
mà (y+1)^2 +9 > 0 =>y=1 => x= -3
Đk:\(x\ge-\frac{10}{3}\)
\(pt\Leftrightarrow\left(x^2+6x+9\right)+\left(3x+9\right)-\left(2\sqrt{3x+10}-2\right)=0\)
\(\Leftrightarrow\left(x+3\right)^2+3\left(x+3\right)-2\frac{\left(3x+10\right)-1}{\sqrt{3x+10}+2}=0\)(do \(\sqrt{3x+10}+2>0\) )
\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)+3-2\frac{3}{\sqrt{3x+10}+2}\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)+3-\frac{6}{\sqrt{3x+10}+2}\right]=0\)
Do \(\sqrt{3x+10}+2\ge0\) với mọi x
\(\Rightarrow\frac{6}{\sqrt{3x+10}}+2\le3\)
\(\Rightarrow\left(x+3\right)+3-\frac{6}{\sqrt{3x+10}+2}>0\)(loại)
\(\Rightarrow x+3=0\Leftrightarrow x=-3\)(thỏa mãn)
Vậy pt có nghiệm duy nhất x=-3.