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\(\dfrac{x-2}{2}=\dfrac{y-4}{3}=\dfrac{z-8}{5}\)
\(\Rightarrow\dfrac{x-2}{2}+2=\dfrac{y-4}{3}+2=\dfrac{z-8}{5}+2\)
\(\Rightarrow\dfrac{x+2}{2}=\dfrac{y+2}{3}=\dfrac{z+2}{5}\)
\(\Rightarrow\left(\dfrac{x+2}{2}\right)^2=\left(\dfrac{y+2}{3}\right)^2=\left(\dfrac{z+2}{5}\right)^2\)
\(\Rightarrow\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}=\dfrac{3.\left(y+2\right)^2}{27}\dfrac{\left(x+2\right)^2+3\left(y+2\right)^2-\left(z+2\right)^2}{4+27-25}=\dfrac{24}{6}=4\)\(\Rightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=16\\\left(y+2\right)^2=36\\\left(z+2\right)^2=100\end{matrix}\right.\)
Bạn chia trường hợp rồi tìm x,y,z nhé
Ta có: 2x = 8y+1 => 2x = (23)y+1 => 2x = 23y+3 => x=3y+3
9y = 3x-9 => (32)y = 3x-9 => 32y = 3x-9 => 2y = x-9
Do x=3y+3 => 2y = 3y+3-9 => 2y=3y-6 => y=6
=> x = 3.6+3 = 18+3=21
=>x+y=21+6=27
Ta có :
\(2^x=8^{y+1}\Rightarrow2^x=\left(2^3\right)^{y+1}\Rightarrow2^x=2^{3y+3}\Rightarrow x=3y+3\)
\(9^y=3^{x-9}\Rightarrow\left(3^2\right)^y=3^{x-9}\Rightarrow3^{2y}=3^{x-9}\Rightarrow2y=x-9\)
Do : \(3y+3\Rightarrow2y=3y+3-9\Rightarrow2y=3y-6\Rightarrow y=6\)
\(\Rightarrow3.6+3=18+3=21\)
\(\Rightarrow x+y=21+6=27\)
a, Với x = 3 và y = -2 ta có:
\(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-\left|3\right|\right)+\left(-2\right)\)
\(A=\dfrac{3}{2}+\dfrac{4}{9}.\left(6-3\right)-2\)
\(A=\dfrac{3}{2}+\dfrac{4}{9}.3-2\)
\(A=\dfrac{3}{2}+\dfrac{4}{3}-2\)
\(A=\dfrac{5}{6}\)
Với x = 3 và y = -3 ta có:
\(B=\left|2.3-1\right|+\left|3.\left(-3\right)+2\right|\)
\(B=\left|5\right|+\left|-7\right|\)
\(B=5+7=12\)
Hoctot ! ko hiểu chỗ nào cứ hỏi cj nhé
Ta có:
\(2^x=8^{y+1}\Rightarrow2^x=2^{3\left(y+1\right)}\Rightarrow x=3\left(y+1\right)\) (1)
\(9^y=3^{x-9}\Rightarrow3^{2y}=3^{x-9}\Rightarrow2y=x-9\) (2)
Thay (1) vào (2) ta có:
\(2y=3y+3-9\\ 2y=3y-6\\ 2y-3y=-6\\ -y=6\\ \Rightarrow y=6\)
Thay \(y=6\) vào \(2y=x-9\), ta có:
\(26=x-9\\ \Rightarrow x=26+9\\ \Rightarrow x=35\)
\(\Rightarrow x+y=6+35=41\)
Vậy: \(x+y=41\)
Mình nhầm, xin lỗi
Chỗ mà thay y=6 vào 2y = x-9 á, đổi 26 = x - 9 thành: 2.6 = x - 9 nha! Phần còn lại mình nghĩ bạn tự tính cũng được :)
C1: \(\left(x-1\right)^2=5^4=625\)
\(\Rightarrow\left[{}\begin{matrix}x-1=25\\x-1=-25\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=26\\x=-24\end{matrix}\right.\) => Chọn C
C2: \(\left(4x^2-9\right)\left(2^{x-1}-1\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(2x+3\right)\left(2^{x-1}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\\x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\) => Chọn A
C3: \(3^x=9^3.27^5\)
\(\Rightarrow3^x=3^6.3^{15}=3^{21}\Rightarrow x=21\) => Chọn B
\(2^x=2^{3\left(y+1\right)}\Rightarrow x=3y+3\)
\(3^{2y}\Rightarrow3^{x-9}\Rightarrow2y=x-9\Rightarrow x=2y+9\)
\(\Rightarrow3y+3=2y+9\Rightarrow y=6\Rightarrow x=21\Rightarrow x+y=27\)
Ta có:\(2^x=8^{y+1}\Rightarrow2^x=2^{3\left(y+1\right)}\Rightarrow2^x=2^{3y+3}\Rightarrow x=3y+3\)
\(\Rightarrow9^y=3^{x-9}\Rightarrow3^{2y}=3^{3y+3-9}\Rightarrow3^{2y}=3^{3y-6}\Rightarrow2y=3y-6\)
\(\Rightarrow2y-3y=-6\Rightarrow-y=-6\Rightarrow y=6\)
\(\Rightarrow x=6\cdot3+3=21\)
\(\Rightarrow x+y=21+6=27\)