\(21+32,2x3=\dfrac{x+40,4}{x}+96,6\)
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21 + 32,2 * 3 = x + 40,4 + 96,6
21 + 96,6 = ( x+ 40,4 ) + 96,6
21 = x + 40,4
x = 21 - 40,4
x =-19,4.
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21 + 32,2 * 3 = x + 40,4 + 96,6
21 + 96,6 = ( x+ 40,4 ) + 96,6
21 = x + 40,4
x = 21 - 40,4
x =-19,4.
x+40,4:x=21
1:x+40,4:x=21
(1+40,4):x=21
41,4:x=21
x=41,4:21
x=1,9714...
\(20,22\times18,6+20,22\times0,394+40,4\times21\)
\(=20,22\times\left(18,6+0,394\right)+40,4\times21\)
\(\approx384,06+848,4=1232,46\)
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giúp mình nhé
ĐKXĐ: -21\(\le x\le\)21
Đặt \(\left\{{}\begin{matrix}\sqrt{21+x}=a\\\sqrt{21-x}=b\end{matrix}\right.\left(a,b\ge0\right)\) (a\(\ne\)b)
Ta có \(\left\{{}\begin{matrix}21+x=a^2\\21-x=b^2\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}a^2+b^2=42\\a^2-b^2=2x\end{matrix}\right.\)
Pt đã cho trở thành \(\dfrac{a+b}{a-b}=\dfrac{a^2+b^2}{a^2-b^2}\)
<=> \(\left(a+b\right)^2\)(a-b)=(\(a^2+b^2\))(a-b)
<=> (a-b)2ab=0
\(\text{}\text{}\left[{}\begin{matrix}a=b\left(loai\right)\\a=0\left(tm\right)\\b=0\left(tm\right)\end{matrix}\right.\)
Thay vào ta tìm dc S=\(\left\{21,-21\right\}\)
\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)
\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)
\(x-2023=1\)
\(x=2024\)
Vậy..............
\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)
\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)
Lời giải:
$21+32,2\times 3=\frac{x+40,4}{x}+96,6$
$21+96,6=\frac{x+40,4}{x}+96,6$
$21=\frac{x+40,4}{x}$
$21\times x=x+40,4$
$21\times x-x=40,4$
$x\times (21-1)=40,4$
$x\times 20=40,4$
$x=40,4:20=2,02$