Đề bài tìm dkxđ
\(\sqrt{1-8b+16b^2}\)
\(\sqrt{\frac{3a-4}{-5}}\)
\(\sqrt{2x^2}\)
\(\sqrt{2x^2+1}\)
\(\sqrt{\frac{-5}{x^2+1}}\)
K
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DH
1
Những câu hỏi liên quan
TT
2 tháng 9 2020
Bạn xem lại đề câu b và c nhé !
a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)
\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ
\(\Rightarrow x\ge2\) thỏa mãn đề.
d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)
Pt tương đương :
\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )
e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)
\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)
Phương trình (1) tương đương :
\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )
30 tháng 7 2019
\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)
\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)
\(\Rightarrow3x+2=2\left(x+2\right)\)
\(\Rightarrow3x+2=2x+4\)
\(\Rightarrow3x-2x=4-2\)
\(\Rightarrow x=2\)
\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)
\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)
\(\Rightarrow2\sqrt{x-2}=4\)
\(\Rightarrow\sqrt{x-2}=2\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)
\(\Rightarrow x+4=1-2x+2\sqrt{\left(1-2x\right)\left(1-x\right)}+1-x\)
\(\Rightarrow x+4=2-3x+2\sqrt{1-3x+2x^2}\)
\(\Rightarrow x+4-2+3x=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)
\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)
\(\Rightarrow2x^2+7x=0\)
\(\Rightarrow x\left(2x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)
\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)
\(\frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{2x\left(\sqrt{3}-1\right)}{3-1}=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+\sqrt{3}x-\sqrt{3x}+x=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)
\(\Rightarrow x=1\)
S
25 tháng 7 2019
1, \(x^3=\left(7+\sqrt{\frac{49}{8}}\right)+\left(7-\sqrt{\frac{49}{8}}\right)+3x\sqrt[3]{\left(7+\sqrt{\frac{49}{8}}\right)\left(7-\sqrt{\frac{49}{8}}\right)}\)
\(=14+3x\cdot\frac{7}{2}=14+\frac{21x}{2}\)
\(\Leftrightarrow x^3-\frac{21}{2}x-14=0\)
Ta có: \(f\left(x\right)=\left(2x^3-21-29\right)^{2019}=\left[2\left(x^3-\frac{21}{2}x-14\right)-1\right]^{2019}=\left(-1\right)^{2019}=-1\)
2, ta có: \(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2\) (bạn tự cm)
Áp dụng công thức trên ta được n=2016
3, \(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\frac{\sqrt[3]{\left(\sqrt{5}\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}\)
\(=\frac{\sqrt[3]{\left(\sqrt{5}-2\right)^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{\sqrt{5}+3-\sqrt{5}}=\frac{5-4}{3}=\frac{1}{3}\)
Thay x=1/3 vào A ta được;
\(A=3x^3+8x^2+2=3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2=3\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 10 2019
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 10 2019
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
\(\sqrt{1-8b+16b^2}\)
\(\sqrt{\left(1-4b\right)^2}\)
vậy với mọi giá trị thực của x thì biều thức đxđ
\(\sqrt{\frac{3a-4}{-5}}\sqrt{2x^2}\sqrt{2x^2+1}\)
\(\hept{\begin{cases}2x^2+1\ge0\\2x^2\ge0\\3x-4\le0\end{cases}}\)
\(\hept{\begin{cases}x\\x\ge0\\x\le\frac{4}{3}\end{cases}}\)đoạn x là tất cả các số thực
\(< =>0\le x\le\frac{4}{3}\)
\(\sqrt{\frac{-5}{x^2+1}}\)
\(x^2+1\ne0;x^2+1\le0< =>x^2+1< 0\left(1\right)\)
\(x^2+1\ge1\left(2\right)\)
\(\left(1\right);\left(2\right)< =>\)ko có giá trị x thỏa mãn để pt đc xác định