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ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow x^2-4x+4-2x+1+2\sqrt{2x-1}-1=0\)
\(\Leftrightarrow\left(x-2\right)^2-\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2-\left(\sqrt{2x-1}-1\right)^2=0\)
\(\Leftrightarrow\left(x-3+\sqrt{2x-1}\right)\left(x-1-\sqrt{2x-1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=3-x\left(x\le3\right)\\\sqrt{2x-1}=x-1\left(x\ge1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x^2-6x+9\\2x-1=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+10=0\\x^2-4x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4+\sqrt{6}\left(l\right)\\x=4-\sqrt{6}\\x=2+\sqrt{2}\\x=2-\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(1)\) ĐKXĐ : \(x\ge3\)
\(\sqrt{x^2-4x+3}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x^2-4x+4\right)-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)^2-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2-1\right)\left(x-2+1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-3\right)\left(x-1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x-3}+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x\in\left\{\varnothing\right\}\end{cases}}}\)
Vậy \(x=1\)
\(2)\)\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)
\(\Leftrightarrow\)\(\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)
\(\Leftrightarrow\)\(\left|x-1\right|-\left|x-3\right|=10\)
+) Với \(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow}x\ge3}\) ta có :
\(x-1-x+3=10\)
\(\Leftrightarrow\)\(0=8\) ( loại )
+) Với \(\hept{\begin{cases}x-1< 0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x< 3\end{cases}\Leftrightarrow}x< 1}\) ta có :
\(1-x+x-3=10\)
\(\Leftrightarrow\)\(0=12\) ( loại )
Vậy không có x thỏa mãn đề bài
Chúc bạn học tốt ~
PS : mới lp 8 sai đừng chửi nhé :v
Như thế này @Cold Wind
\(\sqrt{2y-2}+\sqrt{4-x}-x^2+6x-11=0\)
\(\Leftrightarrow\sqrt{2y-2}+\sqrt{4-x}=x^2-6x+11\)
\(\Leftrightarrow\sqrt{2y-2}+\sqrt{4-2y}=4y^2-12y+11\)
Ta có \(VT^2\le\left(1+1\right)\left(2y-2+4-2y\right)=2^2\)
\(\Leftrightarrow VT\le2\)
Mà \(VP=4y^2-12y+11=\left(2y-3\right)^2+2\ge2\)
\(VT\le VP=2\Leftrightarrow VT=VP=2\)
\(\Leftrightarrow\left(2y-3\right)^2+2=2\Leftrightarrow2y-3=0\Leftrightarrow y=\dfrac{3}{2}\Leftrightarrow x=3\)
2/ x2 - 6x + 4 + \(2\sqrt{2x-1}\)= 0
<=> (x2 - 4x + 4) - (2x - 1 - \(2\sqrt{2x-1}\)+1) = 0
<=> (x - 2)2 - (1 - \(\sqrt{2x-1}\))2 = 0
\(\Leftrightarrow\left(x-1-\sqrt{2x-1}\right)\left(x-3+\sqrt{2x-1}\right)=0\)
Làm tiếp nhé
ĐKXĐ: \(x\ge-1\)
- Với \(x=-1\) ko phải nghiệm
- Với \(x>-1\)
\(\Leftrightarrow x^2-11x+24+\left(x-5\right)\left(x+7-5\sqrt{x+1}\right)=0\)
\(\Leftrightarrow x^2-11x+24+\frac{\left(x-5\right)\left(x^2-11x+24\right)}{x+7+5\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x^2-11x+24\right)\left(1+\frac{x-5}{x+7+5\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-11x+24=0\Rightarrow x=...\\1+\frac{x-5}{x+7+5\sqrt{x+1}}=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow x+7+5\sqrt{x+1}=5-x\)
\(\Leftrightarrow2\left(x+1\right)+5\sqrt{x+1}=0\) (vô nghiệm do \(x>-1\))
Vậy ...
\(\sqrt{x^2-6x+9}+2x=4\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=4-2x\)
\(\Leftrightarrow\left|x-3\right|=4-2x\)
\(\left|x-3\right|=\left\{{}\begin{matrix}4-2xkhix\ge2\\-4+2xkhix< 2\end{matrix}\right.\)
Với \(x\ge2\Rightarrow x-3=4-2x\Rightarrow3x=7\Rightarrow x=\dfrac{7}{3}\left(tm\right)\)
Với \(x< 2\Rightarrow x-3=-4+2x\Rightarrow-x=-1\Rightarrow x=1\left(tm\right)\)
Vậy \(S=\left\{-1;\dfrac{7}{3}\right\}\)
ĐKXĐ: `x\inRR`
`pt<=>sqrt(x^2-6x+9)=4-2x`
`<=>sqrt((x-3)^2)=4-2x`
`<=>|x-3|=4-2x(**)`
Ta thấy rằng `VT(**)>=0AAx\inRR` nên `4-2x>=0<=>x<=2`
Khi đó `|x-3|=3-x`
Suy ra `3-x=4-2x`
`<=>x=1(TM)`
Vậy `S={1}`