2x=10
x=?
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12 tháng 7 2021
Ta có: \(\left(2x-5\right)\left(4x^2+10x+25\right)\left(2x+5\right)\left(4x^2-10x+25\right)-64x^6\)
\(=\left(8x^3-125\right)\left(8x^3+125\right)-64x^6\)
\(=64x^6-15625-64x^6\)
=-15625
21 tháng 1 2020
\(\frac{4xy-5}{10x^3y}-\frac{6y^2-5}{10x^3y}=\frac{\left(4xy-5\right)-\left(6y^2-5\right)}{10x^3y}=\frac{4xy-6y^2}{10x^3y}=\frac{2y\left(2x-3y\right)}{2y.5x^3}=\frac{2x-3y}{5x^3}\)
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\)
\(=\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x+1\right)\left(2x-1\right)}:\frac{4x}{10x-5}\)
\(=\frac{\left(2x+1+2x-1\right)\left(2x+1-2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\times\frac{10x-5}{4x}\)
\(=\frac{4x.2}{\left(2x+1\right)\left(2x-1\right)}\times\frac{5\left(2x-1\right)}{4x}\)
\(=\frac{10}{2x+1}\)
21 tháng 1 2020
\(a,\frac{4xy-5}{10x^3y}-\frac{6y^2-5}{10x^3y}=\frac{\left(4xy-5\right)-\left(6y^2-5\right)}{10x^3y}=\frac{4xy-5-6y^2+5}{10x^3y}=\frac{4xy-6y^2}{10x^3y}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\)
\(=\left(\frac{2x+1}{2x-1}+\frac{2x-1}{2x-1}\right):\frac{4x}{10x-5}\)
\(=\frac{2x+1+2x-1}{2x-1}:\frac{4x}{10x-5}\)
\(=\frac{4x}{2x-1}.\frac{10x-5}{4x}\)
\(=\frac{10x-5}{2x-1}\)
\(=\frac{5\left(2x-1\right)}{2x-1}\)
\(=\frac{5}{1}=5\)
26 tháng 10 2021
1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
2) ĐKXĐ: \(x\ge3\)
\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)
4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
16 tháng 10 2023
a) \(\dfrac{4n^2}{17n^4}\cdot\dfrac{-7n^2}{12n}\) \(\left(n\ne0\right)\)
\(=\dfrac{4n^2\cdot-7n^2}{17n^4\cdot12n}\)
\(=\dfrac{-28n^4}{204n^5}\)
\(=\dfrac{-7}{51n}\)
b) \(\dfrac{3x-1}{10x^2+2x}\cdot\dfrac{25x^2+10x+1}{1-9x^2}\) \(\left(x\ne\pm\dfrac{1}{3};x\ne0;x\ne-\dfrac{1}{5}\right)\)
\(=\dfrac{3x-1}{2x\left(5x+1\right)}\cdot\dfrac{\left(5x+1\right)^2}{\left(1-3x\right)\left(3x+1\right)}\)
\(=\dfrac{-\left(1-3x\right)\left(5x+1\right)^2}{2x\left(5x+1\right)\left(1-3x\right)\left(1+3x\right)}\)
\(=\dfrac{-\left(5x+1\right)}{2x\left(1+3x\right)}\)
\(=-\dfrac{5x+1}{6x^2+2x}\)
c) \(\dfrac{27-a^3}{5a+10}:\dfrac{a-3}{3a+6}\) \(\left(a\ne-2;a\ne3\right)\)
\(=\dfrac{\left(3-a\right)\left(9+3a+a^2\right)}{5\left(a+2\right)}\cdot\dfrac{3\left(a+2\right)}{a-3}\)
\(=\dfrac{-\left(a-3\right)\left(a^2+3a+9\right)\cdot3\left(a+2\right)}{5\left(a+2\right)\left(a-3\right)}\)
\(=\dfrac{-3\left(a^2+3x+9\right)}{5}\)
\(=-\dfrac{3x^2+9x+27}{5}\)
d) \(\dfrac{x^2-1}{x^2+2x-15}:\dfrac{x^2+5x+4}{x^2-10x+21}\) \(\left(x\ne3;x\ne-5;x\ne-1;x\ne-4\right)\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x+5\right)}:\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-3\right)\left(x-7\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x+5\right)}\cdot\dfrac{\left(x-3\right)\left(x-7\right)}{\left(x+1\right)\left(x+4\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)
NT
1
XO
29 tháng 6 2023
Ta có : \(x^4+2x^3-10x^2+10x-3=y^2\)
\(\Leftrightarrow\left(x^4+2x^3-3\right)-\left(10x^2-10x\right)=y^2\)
\(\Leftrightarrow\left(x-1\right).\left(x^3+3x^2-7x+3\right)=y^2\)
\(\Leftrightarrow\left(x-1\right)^2.\left(x^2+4x-3\right)=y^2\)
Vì \(x,y\inℤ\) nên y2 là số chính phương khi
x2 + 4x - 3 là số chính phương
Đặt x2 + 4x - 3 = t2
\(\Leftrightarrow\left(x+t+2\right).\left(x-t+2\right)=7\)
Ta có bảng
| x + t + 2 | 1 | 7 | -1 | -7 |
| x - t + 2 | 7 | 1 | -7 | -1 |
| x | 2 | 2 | -6 | -6 |
| t | -3 | 3 | 3 | -3 |
Ta được x = 2 ; x = -6 thỏa
Với x = 2 <=> y = \(\pm3\)
Với x = -6 <=> y = \(\pm21\)
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