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1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a: =>(x-7)(x+3)=0
hay \(x\in\left\{7;-3\right\}\)
b: =>2x+7=0
hay x=-7/2
c: \(\Delta=50-4\cdot6\cdot2=50-48=2\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{5\sqrt{2}-\sqrt{2}}{12}=\dfrac{\sqrt{2}}{3}\\x_2=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\begin{array}{l} 2{x^2} - 11x + 21 - 3\sqrt[3]{{4x - 4}} = 0 \\ <=> 2{x^2} - 8x + 6 - 3x + 9 + 6 - 3\sqrt[3]{{4x - 4}} \\ <=> \left( {x - 3} \right)\left( {x - 1} \right) - 3\left( {x - 3} \right) - \frac{{108\left( {x - 3} \right)}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}} = 0 \\ <=> \left( {x - 3} \right)\left[ {x - 4 - \frac{{108}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}}} \right] = 0 \\ <=> x = 3 \\ \end{array} \)
_Học tốt_
\(\begin{array}{l} 2{x^2} - 11x + 21 - 3\sqrt[3]{{4x - 4}} = 0 \\ <=> 2{x^2} - 8x + 6 - 3x + 9 + 6 - 3\sqrt[3]{{4x - 4}} \\ <=> \left( {x - 3} \right)\left( {x - 1} \right) - 3\left( {x - 3} \right) - \frac{{108\left( {x - 3} \right)}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}} = 0 \\ <=> \left( {x - 3} \right)\left[ {x - 4 - \frac{{108}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}}} \right] = 0 \\ <=> x = 3 \\ \end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow x^2-4x+21-6\sqrt{2x+3}=0$
$\Leftrightarrow (x^2-6x+9)+[(2x+3)-6\sqrt{2x+3}+9]=0$
$\Leftrightarrow (x-3)^2+(\sqrt{2x+3}-3)^2=0$
Ta thấy: $(x-3)^2\geq 0; (\sqrt{2x+3}-3)^2\geq 0$ với mọi $x\geq \frac{-3}{2}$
Do đó để tổng của chúng bằng $0$ thì:
$(x-3)^2=(\sqrt{2x+3}-3)^2=0$
$\Leftrightarrow x=3$ (tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(x^2-4x+6\right)\cdot\left(x^2-4x+10\right)=21\)
\(\Leftrightarrow\left(x^2-4x+6\right)\cdot\left(x^2-4x+10\right)-21=0\)
\(\Leftrightarrow x^4-4x^3+10x^2-4x^3+16x^2-40x+6x^2-24x+60-21=0\)
\(\Leftrightarrow x^4-8x^3+32x^2-64x+39=0\)
\(\Leftrightarrow x^4-x^3-7x^3+7x^2+25x^2-25x-39x+39=0\)
\(\Leftrightarrow x^3\left(x-1\right)-7x^2\cdot\left(x-1\right)+25x\left(x-1\right)-39x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x^3-7x^2+25x-39\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-4x^2+12x+13x-39\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)-4x\cdot\left(x-3\right)+13\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-4x+13\right)=0\)
\(\hept{\begin{cases}x-1=0\\x-3=0\\x^2-4x+13=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\x=3\\x\notin R\end{cases}}\)
Vậy phương trình của tập nghiệm là S={1;3}
\(x^2-4x-21>0\)
\(\Leftrightarrow\) \(x^2-4x+4>25\)
\(\Leftrightarrow\) \(\left(x-2\right)^2>25\)
\(\Leftrightarrow\) \(\left|x-2\right|>5\)
\(\Leftrightarrow\orbr{\begin{cases}x-2>5\\x-2>-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x>7\\x>-3\end{cases}}}\)
\(x^2-4x-21>0\)
\(x^2-4x+4-25>0\)
\(\left(x-2\right)^2>25\)
Ta có: \(25=5^2=\left(-5\right)^2\)
TH1: \(\left(x-2\right)^2>5^2\)
\(x-2>5\)
\(x>7\)
TH2: \(\left(x-2\right)^2>\left(-5\right)^2\)
\(x-2>-5\)
\(x>-3\)
Kết hợp cả 2 TH ta đc x>-3
=.= hok tốt!!