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a.
\(\frac{1}{-2}=\frac{x}{-6}=\frac{-5}{y}=\frac{z}{12}\)
<=> x=-6.1/-2=3
<=>y=-5.-2/1=10
<=> z=12.1/-2=-6
b.
\(\frac{x}{-10}=\frac{-7}{y}=\frac{z}{-24}\)
hình như đề thiếu.
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Ta có: \(\left(5x-7\right)^2\ge0\) \(\left(6-5y\right)^2\ge0\) \(\left(5z\right)^2\ge0\) Nên để \(\left(5x-7\right)^2+\left(6-5y\right)^2+\left(5z\right)^2=0\)
Mình bổ sung: \(\Rightarrow\left(5x-7\right)^2=0\) ; \(\left(6+5y\right)^2=0\) \(\left(5z\right)^2=0\) \(\rightarrow\) 5x-7=0 . \(\rightarrow5x=7\Rightarrow x=\dfrac{7}{5}\) \(\left(6-5y\right)^2=0\rightarrow6-5y=0\rightarrow5y=6\rightarrow y=\dfrac{6}{5}\)
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theo mk nghĩ ;
(5x - 7)2 + (6-5y)2 + (5z)2 = 10
chi xay ra khi 4+4+2 (loại) vi k có so nao binh phuong =2
vay 9+1+0 =10 ta co;
x = 2
y=1
z=0
(hop ly va duy nhat)
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Bài 1
a) (x + 3)(x + 2) = 0
x + 3 = 0 hoặc x + 2 = 0
*) x + 3 = 0
x = 0 - 3
x = -3 (nhận)
*) x + 2 = 0
x = 0 - 2
x = -2 (nhận)
Vậy x = -3; x = -2
b) (7 - x)³ = -8
(7 - x)³ = (-2)³
7 - x = -2
x = 7 + 2
x = 9 (nhận)
Vậy x = 9
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a: \(\dfrac{-4}{8}=\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}\)
=>\(\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)
=>\(\left\{{}\begin{matrix}x=\left(-10\right)\cdot\dfrac{\left(-1\right)}{2}=5\\y=\dfrac{-7\cdot2}{-1}=14\\z=\dfrac{-24\cdot\left(-1\right)}{2}=\dfrac{24}{2}=12\end{matrix}\right.\)
b: \(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)
=>\(\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)
=>\(\dfrac{x}{2}=\dfrac{18}{y}=\dfrac{z}{24}=\dfrac{1}{2}\)
=>\(x=2\cdot\dfrac{1}{2}=1;y=18\cdot\dfrac{2}{1}=36;z=\dfrac{24}{2}=12\)