Question

Tìm x biết:

a, \(\sqrt{x^2-4x+4}=3\)

b, \(\sqrt{x^2-10x+25}=x+3\)

c, \(\sqrt{x+1+2\sqrt{x}}-\sqrt{x+16-8\sqrt{x}}=3\)

Answer 1

a, \(\sqrt{x^2-4x+4}=3\Leftrightarrow\sqrt{\left(x-2\right)^2}=3\)

\(\Leftrightarrow x-2=3\Leftrightarrow x=5\)

b, \(\sqrt{x^2-10x+25}=x+3\Leftrightarrow\sqrt{\left(x-5\right)^2}=x+3\)

\(\Leftrightarrow x-5=x+3\Leftrightarrow0\ne8\)( vô nghiệm ) 

Answer 2

câu c nữa bạn!!!!!!!!!!

Answer 3

a) Đk: \(\forall x\in R\)

a) \(\sqrt{x^2-4x+4}=3\) <=> \(\sqrt{\left(x-2\right)^2}=3\) <=> \(\left|x-2\right|=3\)

<=> \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\) <=> \(\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

Vậy S = {5; -1}

b) Đk: \(\forall x\in R\)

Ta có: \(\sqrt{x^2-10x+25}=x+3\)

<=> \(\sqrt{\left(x-5\right)^2}=x+3\)

<=> \(\left|x-5\right|=x+3\)

<=> \(\orbr{\begin{cases}x-5=x+3\\5-x=x+3\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=8\left(vl\right)\\2=2x\end{cases}}\) <=> x = 1

Vậy S = {1}

c)Đk: x \(\ge\)0

 \(\sqrt{x+1+2\sqrt{x}}-\sqrt{x+16-8\sqrt{x}}=3\)

<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}-\sqrt{\left(\sqrt{x}-4\right)^2}=3\)

<=> \(\left|\sqrt{x}+1\right|-\left|\sqrt{x}-4\right|=3\)

Do \(x\ge0\) => \(\sqrt{x}+1>0\)

<=> \(\sqrt{x}+1-\left|\sqrt{x}-4\right|=3\)

<=> \(\sqrt{x}-2=\left|\sqrt{x}-4\right|\)

<=> \(\orbr{\begin{cases}\sqrt{x}-2=\sqrt{x}-4\left(đk:x\ge16\right)\\\sqrt{x}-2=4-\sqrt{x}\left(đk:0\le x\le16\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=-2\left(vl\right)\\2\sqrt{x}=6\end{cases}}\) <=> \(x=9\)

Vậy S = {9}

Answer 4

a) \(\sqrt{x^2-4x+4}=3\)

<=> \(\sqrt{\left(x-2\right)^2}=3\)

<=> \(\left|x-2\right|=3\)

<=> \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

b) \(\sqrt{x^2-10x+25}=x+3\)

<=> \(\sqrt{\left(x-5\right)^2}=x+3\)

<=> \(\left|x-5\right|=x+3\)

<=> \(\orbr{\begin{cases}x-5=x+3\left(x\ge5\right)\\5-x=x+3\left(x< 5\right)\end{cases}}\Leftrightarrow x=1\)

c) \(\sqrt{x+1+2\sqrt{x}}-\sqrt{x+16-8\sqrt{x}}=3\)( ĐK : x ≥ 0 )

<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}-\sqrt{\left(\sqrt{x}-4\right)^2}=3\)

<=> \(\left|\sqrt{x}+1\right|-\left|\sqrt{x}-4\right|=3\)(1)

Vì \(\sqrt{x}+1\ge1>0\forall x\ge0\)

=> (1) = \(\sqrt{x}+1-\left|\sqrt{x}-4\right|=3\)

<=> \(\sqrt{x}-2=\left|\sqrt{x}-4\right|\)

<=> \(\orbr{\begin{cases}\sqrt{x}-2=\sqrt{x}-4\left(x\ge16\right)\\\sqrt{x}-2=4-\sqrt{x}\left(0\le x< 16\right)\end{cases}}\Leftrightarrow x=9\left(tm0\le x< 16\right)\)