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a) \(\sqrt{1-4x+4x^2}=5\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)
\(\Leftrightarrow\left|1-2x\right|=5\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b) \(\sqrt{x^2+6x+9}=3x-1\)
\(\Leftrightarrow\sqrt{\left(x+3\right)^2=3x-1}\)
\(\Leftrightarrow\left|x+3\right|=3x-1\)
\(\Leftrightarrow x+3=3x-1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
\(a,\sqrt{1-4x+4x^2}=5\\ \Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\\ \Leftrightarrow\left|1-2x\right|=5\)
\(TH_1:x\le\dfrac{1}{2}\)
\(1-2x=5\\ \Leftrightarrow x=-2\left(tm\right)\)
\(TH_2:x\ge\dfrac{1}{2}\)
\(-1+2x=5\\ \Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{-2;3\right\}\)
\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left|x+3\right|=3x-1\)
\(TH_1:x\ge-3\\ x+3=3x-1\\ \Leftrightarrow-2x=-4\Leftrightarrow x=2\left(tm\right)\)
\(TH_2:x< 3\\ -x-3=3x-1\\ \Leftrightarrow-4x=2\\ \Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)
Vậy \(S=\left\{2;-\dfrac{1}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)
e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)
c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)
\(\Leftrightarrow x-4=0\)
hay x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1)\) 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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{4x^2}=3\left(ĐK:4x^2\ge0\forall x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x\right)^2}=3\\ \Leftrightarrow\left|2x\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}2x=-3\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\x=\dfrac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{2};\dfrac{3}{2}\right\}\)
\(\sqrt{x^2-6x+9}=2\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=2\left(ĐK:\left(x-3\right)^2\ge0\forall x\in R\right)\\ \Leftrightarrow\left|x-3\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+3\\x=-2-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left(\pm5\right)\)
\(\sqrt{\left(2x-3\right)^2}=6\left(ĐK:\left(2x-3\right)^2\ge0\forall x\in R\right)\\ \Leftrightarrow\left|2x-3\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=3+6\\2x=-6+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4,5\left(tm\right)\\x=-1,5\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{4,5;-1,5\right\}\)
\(\sqrt{25x^2}=100\\ \sqrt{\left(5x\right)^2}=100\left(ĐK:\left(5x\right)^2\ge0\forall x\in R\right)\\\Leftrightarrow \left|5x\right|=100\\ \Leftrightarrow\left[{}\begin{matrix}5x=100\\5x=-100\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=20\left(tm\right)\\x=-20\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\pm20\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) điều kiện xác định : \(x\ge1\)
ta có : \(\sqrt{\dfrac{x-1}{4}}-3=\sqrt{\dfrac{4x-4}{9}}\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-3=\dfrac{2}{3}\sqrt{x-1}\)
\(\Leftrightarrow\dfrac{1}{6}\sqrt{x-1}=-3\left(vôlí\right)\) vậy phương trình vô nghiệm
b) điều kiện xác định \(x\ge3\)
ta có : \(\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}=x-3\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+3\right)^2}=x-3\) \(\Leftrightarrow\left|x-2\right|+\left|x+3\right|=x-3\)
\(\Leftrightarrow x-2+x+3=x-3\Leftrightarrow x=-4\left(L\right)\) vậy phương trình vô nghiệm
c) điều kiện xác định : \(\left[{}\begin{matrix}x\ge\dfrac{3}{2}\\x< 1\end{matrix}\right.\)
ta có : \(\sqrt{\dfrac{2x-3}{x-1}}=2\) \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tmđk\right)\) vậy \(x=\dfrac{1}{2}\)
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a. ĐKXĐ: $x\geq 2$ hoặc $x=1$
PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$
$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)
b.
PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$
$\Leftrightarrow |x-2|=|2x-3|$
\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)
c. ĐKXĐ: $x=2$ hoặc $x\geq 3$
PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)
d.
PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$
$\Leftrightarrow |2x-1|=|x-3|$
\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
pt tên tương đương với
\(\left|x-2\right|+\left|x-3\right|=1\) (*)
- Xét \(x< 2\)
(*) \(\Rightarrow2-x+3-x=1\Leftrightarrow x=2\) (loại)
- Xét \(2\le x< 3\)
(*) \(\Rightarrow x-2+3-x=1\Leftrightarrow1=1\) (luôn đúng)
Vậy \(\forall x\) sao cho \(2\le x< 3\) đều thõa mãn (*)
- Xét \(x\ge3\)
(*) \(\Rightarrow x-2+x-3=1\Leftrightarrow x=3\) (Nhận)
Vậy \(2\le x\le3\) là tập nghiệm của pt đã cho.
Cái này áp dụng BĐT cũng được
\(\left|x-2\right|+\left|x-3\right|=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1\)
Dấu ''='' xảy ra khi
\(\left\{{}\begin{matrix}x-2\ge0\\3-x\ge0\end{matrix}\right.\)\(\Rightarrow2\le x\le3\)
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a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
\(ĐK:\forall x\in R\)
\(PT\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\)
\(\Leftrightarrow\left|2x-1\right|=x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x-3\\1-2x=x-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{4}{3}\end{matrix}\right.\)(t/m)