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ta có 5^x+5^x*25=650 suy ra 5^x *26=650 ,5^x=25 suy ra x=2
Mk giúp phần đầu nhé!
Có 5x+5x+2=650
=> 5x.(1+52)=650
5x. 26=650
5x=650:26
5x=25
=> x=2
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
câu 1
=> x+1/2+x+1/3+x+1/4-x-1/5-x-1/6=0
=> (x+x+x-x-x)+(1/2+1/3+1/4-1/5-1/6)=0
=> x+43/60=0
=> x = -43/60
câu dưới làm tương tự bạn nhé!
\(1,\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y+4z}{2-18+12}=\dfrac{24}{-4}=-6\\ \Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-36\\z=-18\end{matrix}\right.\\ 2,\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50-34}{8}=\dfrac{16}{8}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)
\(3,6x=10y=15z\Leftrightarrow\dfrac{6x}{30}=\dfrac{10y}{30}=\dfrac{15z}{30}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{x+y-z}{5+3-2}=\dfrac{90}{6}=15\\ \Leftrightarrow\left\{{}\begin{matrix}x=75\\y=45\\z=30\end{matrix}\right.\)
\(1.a)\dfrac{2^3+3.26-4^3}{2^3.3^2}\)
\(=\dfrac{2^3.3.2.13-\left(2^2\right)^3}{2^3.3^2}\)
\(=\dfrac{2^4.3.13-2^6}{2^3.3^2}\)
\(=\dfrac{2^3\left(2.3.13-2^3\right)}{2^3.3^2}\)
\(=\dfrac{78-8}{9}\)
\(=\dfrac{70}{9}\)
\(b)\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^4.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{13}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}\left(1+2.5\right)}{2^{11}.3^{11}\left(2.3\right)}\)
\(=\dfrac{2.11}{3.6}\)
\(=\dfrac{11}{9}\)
\(2.3^{x-1}-3^{x+1}=90\)
\(\Leftrightarrow3^x:3-3^x.3=90\)
\(\Leftrightarrow3^x\left(\dfrac{1}{3}-3\right)=90\)
\(\Leftrightarrow3^x.\dfrac{-8}{3}=90\)
\(\Leftrightarrow3^x=\dfrac{-135}{4}\)
\(\Leftrightarrow\) \(x\) không có giá trị nào để thỏa mãn đề bài.
Vậy \(x\in\varnothing\)
nữ thám tử nổi tiếng
Đề bài câu 2 sai thì phải, nếu đề bài đc sửa lại là \(3^{x-1}+3^{x+1}=90\) thì \(x=3\) có lẽ là đúng
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Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)