số nghiệm phương trình \(\left(2\sqrt[3]{x}+3\right)\left(2\sqrt[3]{x}-5\right)=21\)
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1 tháng 8 2016
Đặt \(2\sqrt[3]{x}+3=a\). Khi đó biểu thức trên trở thành: \(a\left(a+2\right)=21\)
Mà \(\hept{\begin{cases}\left(a+2\right)-a=2\\\left(a+2\right)+a=k\end{cases}\Rightarrow\hept{\begin{cases}a+2=\frac{k+2}{2}\\a=\frac{k-2}{2}\end{cases}}}\) ( với k là hằng số )
\(\Rightarrow a\left(a+2\right)=\frac{k-2}{2}\cdot\frac{k+2}{2}\)
\(\Rightarrow\frac{\left(k-2\right)\left(k+2\right)}{4}=21\)
\(\Rightarrow k^2-4=84\)
\(\Rightarrow k^2=88\)
\(\Rightarrow\hept{\begin{cases}k=\sqrt{88}=2\sqrt{22}\\k=-\sqrt{88}=-2\sqrt{22}\end{cases}}\)
TH1: Nếu k > 0 thì
\(\Rightarrow a=\frac{2\sqrt{22}-2}{2}=\frac{2\left(\sqrt{22}-1\right)}{2}=\sqrt{22}-1\)
Thế lại vào ta có:
\(2\sqrt[3]{x}+3=\sqrt{22}-1\)
\(\Rightarrow2\sqrt[3]{x}=\sqrt{22}-4\)
\(\Rightarrow\sqrt[3]{x}=\sqrt{\frac{11}{2}}-2\)
\(\Rightarrow x=\left(\sqrt{\frac{11}{2}}-2\right)^3\)
\(\Rightarrow x=\left(\sqrt{\frac{11}{2}}\right)^3-3\cdot\left(\sqrt{\frac{11}{2}}\right)^2\cdot2+3\cdot\sqrt{\frac{11}{2}}\cdot2^2-2^3\)
\(\Rightarrow x=\sqrt{\left(\frac{11}{2}\right)^2\cdot\frac{11}{2}}-3\cdot\frac{11}{2}\cdot2+3\cdot\sqrt{\frac{11}{2}}\cdot4-8\)
\(\Rightarrow x=\frac{11}{2}\sqrt{\frac{11}{2}}-33+12\sqrt{\frac{11}{2}}-8\)
\(\Rightarrow x=\left(\frac{11}{2}\sqrt{\frac{11}{2}}+12\sqrt{\frac{11}{2}}\right)-\left(33+8\right)\)
\(\Rightarrow x=\frac{35}{2}\sqrt{\frac{11}{2}}-41\)
1 tháng 8 2016
TH2: Nếu k < 0 thì:
\(\Rightarrow a=\frac{-2\sqrt{22}-2}{2}=\frac{-2\left(\sqrt{22}+1\right)}{2}=-\left(\sqrt{22}+1\right)\)
Thế lại vào ta có:
\(2\sqrt[3]{x}+3=-\left(\sqrt{22}+1\right)\)
\(\Rightarrow2\sqrt[3]{x}=-\left(\sqrt{22}+4\right)\)
\(\Rightarrow\sqrt[3]{x}=-\left(\sqrt{\frac{11}{2}}+2\right)\)
\(\Rightarrow x=-\left(\sqrt{\frac{11}{2}}+2\right)^3\)
\(\Rightarrow x=-\left[\left(\sqrt{\frac{11}{2}}\right)^3+3\cdot\left(\sqrt{\frac{11}{2}}\right)^2\cdot2+3\cdot\sqrt{\frac{11}{2}}\cdot2^2+2^3\right]\)
\(\Rightarrow x=-\left[\sqrt{\left(\frac{11}{2}\right)^2\cdot\frac{11}{2}}+3\cdot\frac{11}{2}\cdot2+3\cdot\sqrt{\frac{11}{2}}\cdot4+8\right]\)
\(\Rightarrow x=-\left[\left(\frac{11}{2}\sqrt{\frac{11}{2}}+12\sqrt{\frac{11}{2}}\right)+\left(33+8\right)\right]\)
\(\Rightarrow x=-\left[\frac{35}{2}\sqrt{\frac{11}{2}}+41\right]\)
\(\Rightarrow x=-\frac{35}{2}\sqrt{\frac{11}{2}}-41\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 12 2020
\(\sqrt{2-f\left(x\right)}=f\left(x\right)\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)\ge0\\f^2\left(x\right)+f\left(x\right)-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)=1\\f\left(x\right)=-2< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow f\left(1\right)=f\left(2\right)=f\left(3\right)=1\)
\(\sqrt{2g\left(x\right)-1}+\sqrt[3]{3g\left(x\right)-2}=2.g\left(x\right)\)
\(VT=1.\sqrt{2g\left(x\right)-1}+1.1\sqrt[3]{3g\left(x\right)-2}\)
\(VT\le\dfrac{1}{2}\left(1+2g\left(x\right)-1\right)+\dfrac{1}{3}\left(1+1+3g\left(x\right)-2\right)\)
\(\Leftrightarrow VT\le2g\left(x\right)\)
Dấu "=" xảy ra khi và chỉ khi \(g\left(x\right)=1\)
\(\Rightarrow g\left(0\right)=g\left(3\right)=g\left(4\right)=g\left(5\right)=1\)
Để các căn thức xác định \(\Rightarrow\left\{{}\begin{matrix}f\left(x\right)-1\ge0\\g\left(x\right)-1\ge0\end{matrix}\right.\)
Ta có:
\(\sqrt{f\left(x\right)-1}+\sqrt{g\left(x\right)-1}+f\left(x\right).g\left(x\right)-f\left(x\right)-g\left(x\right)+1=0\)
\(\Leftrightarrow\sqrt{f\left(x\right)-1}+\sqrt{g\left(x\right)-1}+\left[f\left(x\right)-1\right]\left[g\left(x\right)-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(x\right)=1\\g\left(x\right)=1\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Vậy tập nghiệm của pt đã cho có đúng 1 phần tử
HQ
Hà Quang Minh
Giáo viên
30 tháng 9 2023
ĐK: \(x - 1 \ge 0\,\, \Leftrightarrow \,\,x \ge 1\)
\( \Rightarrow \) TXĐ của phương trình là: \(D = \left[ {1; + \infty } \right)\)
Giải phương trình: \(\sqrt {2{x^2} - 3} = x - 1\)
\(\begin{array}{l} \Leftrightarrow \,\,{\left( {\sqrt {2{x^2} - 3} } \right)^2} = {\left( {x - 1} \right)^2}\\ \Leftrightarrow \,\,2{x^2} - 3 = {x^2} - 2x + 1\\ \Leftrightarrow \,\,{x^2} + 2x - 4 = 0\\ \Leftrightarrow \,\,\left[ {\begin{array}{*{20}{c}}{x = - 1 + \sqrt 5 }\\{x = - 1 - \sqrt 5 }\end{array}} \right.\end{array}\)
Ta thấy \(x = - 1 + \sqrt 5 \) thỏa mãn.
Vậy tập nghiệm của phương trình là: \(S = \left\{ { - 1 + \sqrt 5 } \right\}\)
Chọn C.
1 tháng 9 2023
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) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
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) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
10 tháng 12 2023
a: \(\left\{{}\begin{matrix}\sqrt{5}x-y=\sqrt{5}\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\sqrt{15}x-2\sqrt{3}\cdot y=2\sqrt{15}\left(\sqrt{3}-1\right)\\2\sqrt{15}x+15y=21\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2\sqrt{3}y-15y=2\sqrt{45}-2\sqrt{15}-21\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-2\sqrt{3}-15\right)=-15\sqrt{5}-2\sqrt{15}\\2\sqrt{3}\cdot x+3\sqrt{5}\cdot y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{15\sqrt{5}+2\sqrt{15}}{2\sqrt{3}+15}=\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}\cdot y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\sqrt{5}\\2\sqrt{3}x=21-3\sqrt{5}\cdot\sqrt{5}=21-15=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\sqrt{5}\\x=\dfrac{6}{2\sqrt{3}}=\sqrt{3}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}1,7x-2y=3,8\\2,1x+5y=0,4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8,5x-10y=19\\4,2x+10y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8,5x-10y+4,2x+10y=19,8\\2,1x+5y=0,4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}12,7x=19,8\\2,1x+5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{198}{127}\\5y=0,4-2,1x=-\dfrac{365}{127}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{198}{127}\\y=-\dfrac{73}{127}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 12 2020
\(\Leftrightarrow x^2-4x+5+3\sqrt{x^2-4x+5}-2=0\)
Đặt \(\sqrt{x^2-4x+5}=t>0\)
\(\Rightarrow t^2+3t-2=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{-3+\sqrt{17}}{2}\\t=\dfrac{-3-\sqrt{17}}{2}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x^2-4x+5=\dfrac{13-3\sqrt{17}}{2}\)
\(\Leftrightarrow x^2-4x+\dfrac{-3+3\sqrt{17}}{2}=0\)
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=4^2-2\left(\dfrac{-3+3\sqrt{17}}{2}\right)=19-3\sqrt{17}\)
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