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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)
b. Tự đặt đk
\(x^{^2}+5\sqrt{x-3}=21\\\Leftrightarrow x^{^2}-9+5\sqrt{x-3}=12 \)
Đặt \(a=\sqrt{x-3}\) \(\left(a\ge0\right)\) Phương trình trở thành:
\(a^{^2}\left(a^{^2}+6\right)+5a=12\\ \Leftrightarrow a^{^4}+6a^{^2}+5a-12=0\\ \Leftrightarrow a^{^4}-a^{^3}+a^{^3}-a^{^2}+7a^{^2}-7a+12a-12=0\\ \Leftrightarrow\left(a-1\right)\left(a^{^3}+a^{^2}+7a+12\right)=0\\ \Leftrightarrow a=1\left(tmdk\right)\)
Ta có: vì \(a\ge0\) nên \(a^{^3}+a^{^2}+7a+12\ne0\)
Với a = 1 ta có x=4 (tmdk)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x^4+4x^3+x^2+3x\ge0\)
\(4x^4+4x^2+1-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)
\(\Leftrightarrow\left(2x^2+1\right)^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)
\(2x^2+1=u;\sqrt{4x^4+4x^3+x^2+3x}=v\left(u>0;v>0\right)\)
\(\hept{\begin{cases}u^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)v\\v^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)u\end{cases}\Rightarrow u^2-v^2=\left(x^2-x+1\right)\left(v-u\right)\Leftrightarrow\orbr{\begin{cases}u=v\\u+v+x^2-x+1=0\end{cases}}}\)
- \(u+v+x^2-x+1=0\Leftrightarrow u+v+\left(x-\frac{1}{2}\right)^2=-\frac{3}{4}\)
- \(u=v\Leftrightarrow4x^4+4x^2+1=4x^4+4x^3+x^2+3x\Leftrightarrow\left(x-1\right)^3=-3x^3\Leftrightarrow x-1=-x\sqrt[3]{3}\Leftrightarrow x=\frac{1}{1+\sqrt[3]{3}}\)Đối chiếu điều kiện ta thu được nghiệm duy nhất \(x=\frac{1}{1+\sqrt[3]{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
ĐKXĐ : \(x\ne2\)
\(PT\Leftrightarrow\left(x^2+\left(\frac{2x}{x-2}\right)^2+\frac{4x^2}{x-2}\right)-\frac{4x^2}{x-2}-5=0\)
\(\Leftrightarrow\left(x+\frac{2x}{x-2}\right)^2-\frac{4x^2}{x-2}-5=0\)
\(\Leftrightarrow\frac{x^4}{\left(x-2\right)^2}-\frac{4x^2}{x-2}-5=0\)
\(\Leftrightarrow\frac{x^4}{\left(x-2\right)^2}-\frac{5x^2}{x-2}+\frac{x^2}{x-2}-5=0\)
\(\Leftrightarrow\left(\frac{x^2}{x-2}-5\right)\left(\frac{x^2}{x-2}+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-5x+10=0\left(\Delta=25-40< 0;l\right)\\x^2+x-2=0\end{cases}}\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)(TMĐKXĐ)
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\(c,\frac{x^2+\sqrt{3}}{x+\sqrt{x^2+\sqrt{3}}}+\frac{x^2-\sqrt{3}}{x+\sqrt{x^2+\sqrt{3}}}=x\)
\(\Rightarrow\frac{x^2}{x+\sqrt{x^2+\sqrt{3}}}=x\)
\(\Rightarrow2x^2=x^2+x\sqrt{x^2+\sqrt{3}}\)
\(\Rightarrow x^2=x\sqrt{x^2+\sqrt{3}}\)
\(\Rightarrow x^4=x^3+x\sqrt{3}\)
\(\Rightarrow x\left(x^2-x+\sqrt{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-x+\sqrt{3}=0\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:4x-1\ge0\Leftrightarrow x\ge\frac{1}{4}\)
\(pt\Leftrightarrow\frac{x}{\sqrt{4x-1}}-2+\frac{\sqrt{4x-1}}{x}=0\)
\(\Leftrightarrow\frac{x^2-2\sqrt{4x-1}.x+4x-1}{x\sqrt{4x-1}}=0\Leftrightarrow\frac{\left(x-\sqrt{4x-1}\right)^2}{x\sqrt{4x-1}}=0\)
\(\Rightarrow x=\sqrt{4x-1}\Rightarrow x^2=4x-1\Leftrightarrow x^2-4x+1=0\)
\(\Leftrightarrow\left(x-2\right)^2=3\Rightarrow\orbr{\begin{cases}x-2=\sqrt{3}\\x-2=-\sqrt{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\)
Nguyễn Hưng Phát ĐKXĐ : \(x>\frac{1}{4}\) mới đúng nha nhok :v
\(\Leftrightarrow\left(x^2-4x+6\right)\cdot\left(x^2-4x+10\right)=21\)
\(\Leftrightarrow\left(x^2-4x+6\right)\cdot\left(x^2-4x+10\right)-21=0\)
\(\Leftrightarrow x^4-4x^3+10x^2-4x^3+16x^2-40x+6x^2-24x+60-21=0\)
\(\Leftrightarrow x^4-8x^3+32x^2-64x+39=0\)
\(\Leftrightarrow x^4-x^3-7x^3+7x^2+25x^2-25x-39x+39=0\)
\(\Leftrightarrow x^3\left(x-1\right)-7x^2\cdot\left(x-1\right)+25x\left(x-1\right)-39x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x^3-7x^2+25x-39\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-4x^2+12x+13x-39\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)-4x\cdot\left(x-3\right)+13\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-4x+13\right)=0\)
\(\hept{\begin{cases}x-1=0\\x-3=0\\x^2-4x+13=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\x=3\\x\notin R\end{cases}}\)
Vậy phương trình của tập nghiệm là S={1;3}