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Ta có: \(\left\{{}\begin{matrix}\left|2y-27\right|^{2023}\ge0\forall y\\\left(3x+10\right)^{2024}\ge0\forall x\end{matrix}\right.\)
\(\Rightarrow\left|2y-27\right|^{2023}+\left(3x+10\right)^{2024}\ge0\forall x,y\)
Mà: \(\left|2y-27\right|^{2023}+\left(3x+10\right)^{2024}=0\)
nên: \(\left\{{}\begin{matrix}2y-27=0\\3x+10=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\dfrac{27}{2}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: ...
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1:
a: AB=căn 25^2-20^2=15cm
b: HC^2-HB^2
=AC^2-AH^2-(AB^2-AH^2)
=AC^2-AB^2
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A)Xét tam giác DME và tam giác DMF
Có:DE=DF(gt)
ME=MF(gt)
DM cạnh chung
Do đó:tam giác DME=tam giácDMF
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d. \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
\(\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Rightarrow x^2+5x-14=x^2+3x-4\)
\(\Rightarrow x^2+5x-x^2-3x=-4+14\)
\(\Rightarrow2x=10\) \(\Rightarrow x=\dfrac{10}{3}\) \(\Rightarrow x=5\)
\(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
⇔ \(\dfrac{\left(x-2\right)\left(x+7\right)}{\left(x-1\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x-1\right)}{\left(x+7\right)\left(x-1\right)}\)
⇔ (x - 2)(x + 7) = (x + 4)(x - 1)
⇔ x2 + 7x - 2x - 14 = x2 - x + 4x - 4
⇔ x2 - x2 + 7x - 2x + x - 4x = 14 - 4
⇔ 2x = 10
⇔ x = 10/2 = 5
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