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1) \(\sqrt{x^2+1}=\sqrt{5}\)
\(\Leftrightarrow x^2+1=5\)
\(\Leftrightarrow x^2=5-1\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x^2=2^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\sqrt{2x-1}=\sqrt{3}\) (ĐK: \(x\ge\dfrac{1}{2}\))
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=3+1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=\dfrac{4}{2}\)
\(\Leftrightarrow x=2\left(tm\right)\)
3) \(\sqrt{43-x}=x-1\) (ĐK: \(x\le43\))
\(\Leftrightarrow43-x=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-2x+1=43-x\)
\(\Leftrightarrow x^2-x-42=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)
4) \(x-\sqrt{4x-3}=2\) (ĐK: \(x\ge\dfrac{3}{4}\))
\(\Leftrightarrow\sqrt{4x-3}=x-2\)
\(\Leftrightarrow4x-3=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-4x+4=4x-3\)
\(\Leftrightarrow x^2-8x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
5) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{x}+3=2\sqrt{x}+2\)
\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3-2\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1^2\)
\(\Leftrightarrow x=1\left(tm\right)\)
1)
\(\sqrt{x^2+1}=\sqrt{5}\\ \Leftrightarrow x^2+1=5\\ \Leftrightarrow x^2=5-1=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy PT có nghiệm `x=2` hoặc `x=-2`
2)
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{2x-1}=\sqrt{3}\\ \Leftrightarrow2x-1=3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)
Vậy PT có nghiệm `x=2`
3)
\(ĐKXĐ:x\le43\)
PT trở thành:
\(43-x=\left(x-1\right)^2=x^2-2x+1\\ \Leftrightarrow43-x-x^2+2x-1=0\\ \Leftrightarrow-x^2+x+42=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm `x=-6` hoặc `x=7`
4)
ĐKXĐ: \(x\ge\dfrac{3}{4}\)
PT trở thành:
\(\sqrt{4x-3}=x-2\\ \Leftrightarrow4x-3=\left(x-2\right)^2=x^2-4x+4\\ \Leftrightarrow4x-3-x^2+4x-4=0\\ \Leftrightarrow-x^2+8x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm \(x=1\) hoặc \(x=7\)
5)
ĐKXĐ: \(x\ge0\)
PT trở thành:
\(\sqrt{x+3}=2\sqrt{x}+2\\ \Leftrightarrow x+3=\left(2\sqrt{x}+2\right)^2=4x+8\sqrt{x}+4\\ \Leftrightarrow x+3-4x-8\sqrt{x}-4=0\\ \Leftrightarrow-3x-8\sqrt{x}-1=0\left(1\right)\)
Đặt \(\sqrt{x}=t\left(t\ge0\right)\)
Khi đó:
(1)\(\Leftrightarrow3t^2+8t+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-4+\sqrt{13}}{3}\left(loại\right)\\t=\dfrac{-4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)
Vậy PT vô nghiệm.
\(\Leftrightarrow\left\{{}\begin{matrix}43-x\ge0\\43-x=\left(x-1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le43\\43-x=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le43\\x^2-x-42=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le43\\\left(x+6\right)\left(x-7\right)=42\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le43\\\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\) (t/m)
Vậy phương trình đã cho có tập nghiệm \(S=\left\{-6;7\right\}\)thỏa mãn đề
\(ĐK:x\le43\)
\(\sqrt{43-x}=x-1\)
\(\Leftrightarrow\left(\sqrt{43-x}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow43-x=x^2-2x+1\)
\(\Leftrightarrow x^2-x-42=0\)
\(\Delta=\left(-1\right)^2-4.\left(-42\right)=1+168=169>0\)
\(\rightarrow\left\{{}\begin{matrix}x_1=\dfrac{1+\sqrt{169}}{2}=7\left(tm\right)\\x_2=\dfrac{1-\sqrt{169}}{2}=-6\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{7;-6\right\}\)
\(\sqrt{43-x}=x-1\left(đk:x\le43\right)\)
\(\Leftrightarrow\left|43-x\right|=\left(x-1\right)^2\)
\(\Leftrightarrow43-x=x^2-2x+1\)
\(\Leftrightarrow x^2-x-42=0\)
\(\Delta=\left(-1\right)^2-4.\left(-42\right)=169>0\)
Do \(\Delta\) > 0 nên pt có 2 nghiệm phân biện:
\(x_1=\dfrac{1+\sqrt{169}}{2}=7\left(TM\right)\)
\(x_2=\dfrac{1-\sqrt{169}}{2}=-6\left(TM\right)\)
\(a,\sqrt{x^2-10x+2}=x-2\)
\(\Rightarrow x^2-10x+25=\left(x-2\right)^2\)
\(\Rightarrow x^2-10x+25=x^2-4x+4\)
\(\Rightarrow10x+25=4x+4\)
\(\Rightarrow10x-4x=4-25\)
\(\Rightarrow6x=-21\Leftrightarrow x=-\frac{21}{6}=-\frac{7}{2}\)
\(b,\sqrt{43-x}=x-1\)
\(\Rightarrow43-x=\left(x-1\right)^2\)
\(\Rightarrow43-x=x^2-2x+1\)
\(\Rightarrow42=x^2-3x\)
\(\Rightarrow42=x\left(x-3\right)\)
\(c,\sqrt{x-5}=4\)
\(\Rightarrow x-5=16\)
\(\Rightarrow x=16+5=21\)
a) \(\sqrt{x^2-10x+25}=x-2\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow x-5=x-2\)
\(\Leftrightarrow x-x=-2+5\left(vonghiem\right)\)
a)\(\sqrt{x^2-10x+25}=x+2\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x+2\)
\(\Leftrightarrow\left|x-5\right|=x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x+2\\x-5=-x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-x=5+2\\x+x=5-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0x=7\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=7\left(lo\text{ạ}i\right)\\x=\dfrac{3}{2}\left(nh\text{ậ}n\right)\end{matrix}\right.\)
vậy
\(b.\sqrt{43-x}=x-1\left(ĐK:43\ge x\ge1\right)\)
\(\Leftrightarrow43-x=x^2-2x+1\)
\(\Leftrightarrow43-x^2+x-1=0\)
\(\Leftrightarrow x^2-x-42=0\)
\(\Leftrightarrow x^2-7x+6x-42=0\)
\(\Leftrightarrow x\left(x-7\right)+6\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(n\right)\\x=-6\left(l\right)\end{matrix}\right.\)
\(c.\sqrt{x-5}=4\left(ĐK:x\ge5\right)\)
\(\Leftrightarrow x-5=16\)
\(\Leftrightarrow x=21\left(n\right)\)
Bg (x thuộc Z đc không ?)
\(\sqrt{43-x}=x-1\)
=> 43 - x = (x - 1)2
=> 43 - x = x2 - 2x + 1
=> 43 = x2 - 2x + 1 + x
=> 42 = x2 - 2x + x
=> 42 = x2 - (2x - x)
=> 42 = x2 - x
=> 42 = x.(x - 1)
=> 7.6 = -6.(-7) = x.(x - 1)
Vậy x = 7 hoặc x = -6
Nhầm rồi, em xin lỗi ạ:
Kết quả là 7 thôi ạ,
Vì khi rút gọn x.(x - 1) thì phải dương
Theo mình thì x = 7
\(\sqrt{43-x}\) có nghĩa khi \(\sqrt{43-x}\)>0 => x<43
\(\sqrt{43-x}\)= \(\sqrt{\left[x-1\right]^2}\)=> 43-x=\(x^2\)-2x+1
=> [x-7]x[x+6]=0
=> x=7 và x=-6