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Lời giải:
\(\frac{6^{x+3}-6^{x+1}+6^x}{211}=\frac{7^{2x}+7^{2x+1}+7^{2x-3}}{8\frac{1}{49}}\)
\(\Leftrightarrow \frac{6^x(6^3-6+1)}{211}=\frac{7^{2x}(1+7+\frac{1}{7^3})}{\frac{393}{49}}\)
\(\Leftrightarrow 6^x=7^{2x}.\frac{915}{917}\)
\(\Leftrightarrow (\frac{6}{49})^x=\frac{915}{917}\)
\(\Rightarrow x=\log_{\frac{6}{49}}\frac{915}{917}\)
Trần Linh: cách giải này gây khó hiểu cho bạn ở dòng cuối đúng không? Nếu không dùng log thì không thể tìm ra kết quả cuối cùng theo cách lớp 7 do nghiệm quá xấu. Do đó, bạn hãy xem lại đề xem có nhầm dấu hay viết sai ở chỗ nào không.
#)Giải :
\(2x-3=x+\frac{1}{2}\)
\(\Leftrightarrow2x-3-x+\frac{1}{2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\x=-\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}}\)
a) \(2x-3=x+\frac{1}{2}\)
\(\Leftrightarrow2x-x=\frac{1}{2}+3\)
\(\Leftrightarrow x=\frac{7}{2}\)
Vậy...
b) \(4x-\left(2x+1\right)=3-\frac{1}{3}+x\)
\(\Leftrightarrow4x-2x-1=3-\frac{1}{3}+x\)
\(\Leftrightarrow4x-2x-x=3-\frac{1}{3}+1\)
\(\Leftrightarrow x=\frac{11}{3}\)
Vậy ...
c) \(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=7-\frac{1}{50}+x\)
\(\Leftrightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(1-\frac{1}{50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\frac{49}{50}=\frac{349}{50}+x\)
\(\Leftrightarrow2x-x=\frac{349}{50}+\frac{49}{50}\)
\(\Leftrightarrow x=\frac{199}{25}\)
Vậy ...
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)
\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)
Vậy A = \(\frac{1}{2}\)
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)
Vậy B = \(\frac{1}{2}\)
1) \(\frac{x-1}{x-5}=\frac{6}{7};\left(x-1\right).7=\left(x-5\right).6\)
7x - 7 = 6x - 30
=> 7x - 6x = -30 - (-7)
x = -23
2) \(\frac{x-1}{3}=\frac{x+3}{5};\left(x-1\right).5=\left(x+3\right).3\)
5x - 5 = 3x + 9
=> 5x - 3x = 9 - (-5)
2x = 14
x = 7
3) \(\frac{3}{7}=\frac{2x+1}{3x+5};\left(3x+5\right).3=\left(2x+1\right).7\)
9x + 15 = 14x + 7
9x - 14x = 7-15
5x = -8
x = -8/5
1) =>\(\hept{\begin{cases}x-1=6\\x-5=7\end{cases}=>\hept{\begin{cases}x=6+1=7\\x=7+5=13\end{cases}}}\)
Vậy x\(\varepsilon\){7;13}
2)
Ta có : \(\frac{x+1}{5}=\frac{x+2}{6}\)
\(\Rightarrow\left(x+1\right)6=5\left(x+2\right)\)
\(\Leftrightarrow6x+6=5x+10\)
\(\Leftrightarrow6x-5x=10-6\)
\(\Rightarrow x=4\)
\(\frac{x+1}{2}\)= \(\frac{8}{x+1}\)
x + 1 . x + 1 = 2 . 8
x . 2 = 16
x = 16 : 2
x = 8
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)