\(7^{2x}+7^{2x+2}=2450\)

\(7^{2x}+7^{2x}\times7^2=2450\)

\(7^{2x}\left(1+49\right)=2450\)

\(7^{2x}=2450\div50\)

\(7^{2x}=49\)

\(7^{2x}=7^2\)

\(\Rightarrow2x=2\Leftrightarrow x=1\)

\(7^{2x}+7^{2x+2}=2450\)

\(\Rightarrow7^{2x}+7^{2x}\cdot7^2=2450\)

\(\Rightarrow7^{2x}\cdot\left(1+49\right)=2450\)

\(\Rightarrow7^{2x}=2450:50=49\)

\(\Rightarrow7^{2x}=7^2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(\frac{1}{4}:\left(\frac{5}{3}\right)^{x-1}=0,09\)

\(\frac{1}{4}:\left(\frac{5}{3}\right)^{x-1}=\frac{9}{100}\)

\(\left(\frac{5}{3}\right)^{x-1}=\frac{1}{4}:\frac{9}{100}\)

\(\left(\frac{5}{3}\right)^{x-1}=\frac{1}{4}\cdot\frac{100}{9}\)

\(\left(\frac{5}{3}\right)^{x-1}=\frac{100}{36}\)

\(\left(\frac{5}{3}\right)^{x-1}=\frac{25}{9}\)

\(\left(\frac{5}{3}\right)^{x-1}=\left(\frac{5}{3}\right)^2\)

\(\Rightarrow x-1=2\Rightarrow x=3\)

Vậy x cần tìm bằng 3

=> (5/3)^x-1=25/9

=> (5/3)^x-1=(5/30)²

=> x-1=2

=> x=3

\(x=\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)

\(\text{Áp dụng t/c dãy tỉ số bằng nhau}\)

\(\Rightarrow\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2b+2c+2a}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

\(\Rightarrow x=\frac{1}{2}\)

\(\left(2x-1\right)^3=-8\)

\(\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\frac{1}{2}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{25}{16}\)

\(\left(x+\frac{1}{2}\right)^2=\left(\frac{5}{4}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\frac{5}{4}\)

\(\Leftrightarrow x=\frac{5}{4}-\frac{1}{2}\)

\(\Leftrightarrow x=\frac{3}{4}\)

khó vãi

ai jup tui với

Trl :

1.1 + 1.2 + 1.3 + ... + 1.9

= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9

= 45

Hok tốt nhé :)

\(\left(3x-1\right)\left(2x-\frac{6}{11}=0\right)\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\2x-\frac{6}{11}=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=1\\2x=\frac{6}{11}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{6}{22}\end{cases}}}\)

\(a,\left[3x-1\right]\left[2x-\frac{6}{11}\right]=0\)

=> \(\orbr{\begin{cases}3x-1=0\\2x-\frac{6}{11}=0\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{1}{3}\\2x=\frac{6}{11}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{6}{11}:2=\frac{6}{22}=\frac{3}{11}\end{cases}}\)

\(b,y^2-4y=0\)

=> \(y\left[y-4\right]=0\)

=> \(\orbr{\begin{cases}y=0\\y-4=0\end{cases}}\)=> \(\orbr{\begin{cases}y=0\\y=4\end{cases}}\)

           Bài làm :

Ta có hình vẽ :

Theo đề bài :

\(\hept{\begin{cases}AB\perp\text{Ox}\\AC\perp Oy\end{cases}}\)

Mà góc xOy vuông =>AB vuông góc với AC

Tham khảo : Câu hỏi của Nguyễn Thị Bích Ngọc - Vật lý lớp 7 | Học trực tuyến