-2/3 = x/-6 = 10/-y = z/9
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\(-\dfrac{2}{3}=\dfrac{x}{-6}\Rightarrow x=\left(-\dfrac{2}{3}\right)\left(-6\right)=4\)
\(-\dfrac{2}{3}=\dfrac{10}{-y}\Rightarrow y=\left(-10\right):\left(-\dfrac{2}{3}\right)=15\)
\(-\dfrac{2}{3}=\dfrac{z}{9}\Rightarrow z=\left(-\dfrac{2}{3}\right).9=-6\)
\(\dfrac{-2}{3}=\dfrac{x}{-6}=\dfrac{10}{-y}=\dfrac{z}{9}\)
\(x=\left(-6.-2\right):3=4;y=\left(-6.10\right):-4=15;z=\left(10.9\right):-15=-6\)
\(\Leftrightarrow9x\left(x+2\right)+9y\left(y-\dfrac{2}{3}\right)=10\\ \Leftrightarrow9x^2+18x+9y^2-6y-10=0\\ \Leftrightarrow\left(9x^2+18x+9\right)+\left(9y^2-6y+1\right)=0\\ \Leftrightarrow9\left(x+1\right)^2+\left(3y-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Ta có: x=9
nên x+1=10
Ta có: \(x^{14}-10x^{13}+10x^{12}-...+10x^2-10x+10\)
\(=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-...+x^3+x^2-x^2-x+x+1\)
=1
\(\dfrac{1}{8}< \dfrac{x}{18}< \dfrac{2}{9}\\ \Rightarrow\dfrac{9}{4}< x< 4\\ \Rightarrow2,25< x< 4\\ \Rightarrow x=3\)
=>9/72<4x/72<16/72
=>9<4x<16
mà x là số nguyên
nên 4x=12
hay x=3
Lời giải:
a.
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=60\\ y=45\\ z=40\end{matrix}\right.\)
b)
Từ đkđb suy ra \(\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{10x-5y+z}{1-\frac{1}{3}+\frac{1}{6}}=\frac{25}{\frac{5}{6}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=3\\ y=2\\ z=5\end{matrix}\right.\)
Do x, y nguyên => \(\left\{{}\begin{matrix}\left(x+2\right)^2nguyên\ge0\\y-1nguyên\end{matrix}\right.\)
(x+2)2 . (y-1) = -9
Ta có bảng:
(x+2)2 | 1 | 3 | 9 |
y-1 | -9 | -3 | -1 |
x | \(\left[{}\begin{matrix}x=-1\left(TM\right)\\x=-3\left(TM\right)\end{matrix}\right.\) | \(\left[{}\begin{matrix}x=\sqrt{3}-2\left(L\right)\\x=-\sqrt{3}-2\left(L\right)\end{matrix}\right.\) | \(\left[{}\begin{matrix}x=1\left(TM\right)\\x=-5\left(TM\right)\end{matrix}\right.\) |
y | -8 (TM) | 0 |
\(\frac{-2}{3}=\frac{x}{-6}=\frac{10}{-y}=\frac{z}{9}\)
\(\frac{-2}{3}=\frac{x}{-6}\Rightarrow\left(-2\right)\left(-6\right)=3x\Rightarrow12=3x\Rightarrow x=4\)
\(\frac{4}{-6}=\frac{10}{-y}\Rightarrow-4y=-60\Rightarrow y=15\Rightarrow-y=-15\)
\(\frac{10}{-15}=\frac{z}{9}\Rightarrow-15z=90\Rightarrow z=-6\)
\(\frac{-2}{3}=\frac{x}{-6}=\frac{10}{-y}=\frac{z}{9}\)
\(\frac{-2}{3}=\frac{x}{-6}\Leftrightarrow\frac{4}{-6}=\frac{x}{-6}\Leftrightarrow x=4\)
\(\frac{4}{-6}=\frac{10}{-y}\Leftrightarrow4y=-60\Leftrightarrow y=-15\)
\(\frac{10}{15}=\frac{z}{9}\Leftrightarrow\frac{90}{135}=\frac{15z}{135}\Leftrightarrow15z=90\Leftrightarrow z=6\)