Vẽ k đc đẹp lm xl

\(\left(2x-3\right)^5=\left(2x-3\right)^{2020}\)

\(\Leftrightarrow\left(2x-3\right)^{2020}-\left(2x-3\right)^5=0\)

\(\Leftrightarrow\left(2x-3\right)^5\left[\left(2x-3\right)^{2015}-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(2x-3\right)^5=0\\\left(2x-3\right)^{2015}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-3=0\\\left(2x-3\right)^{2015}=1\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=3\\2x-3=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=2\end{cases}}}\)

           Vậy\(x=\frac{3}{2};2\)

Vì ( 2x - 3 )⁵ = ( 2x - 3 )²⁰²⁰

\(\Rightarrow\orbr{\begin{cases}2x-3=1\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=1+3\\2x=0+3\end{cases}\Rightarrow}\orbr{\begin{cases}2x=4\\2x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=4:2\\x=3:2\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}}\)

Vậy \(x\in\left\{2;\frac{3}{2}\right\}\)thì ( 2x - 3 )⁵ = ( 2x - 3 )²⁰²⁰

help me,Please

Câu hỏi của Thiên Hà Milky Way - Toán lớp 6 - Học toán với OnlineMath

Tham khảo

dể v bạn, cần mik giải giúp 0

a) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\)

\(\Rightarrow3+\frac{b}{a}=3+\frac{d}{c}\Rightarrow\frac{3a+b}{a}=\frac{3c+d}{c}\)

\(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)

b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)

\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{\left(ck\right)^2-\left(dk\right)^2}{c^2-d^2}=k^2\)

và \(\frac{ab}{cd}=\frac{ck.dk}{cd}=k^2\)

Vậy \(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\left(đpcm\right)\)

Ta có:

\(\frac{x}{2x+y}=\frac{2}{5}\Leftrightarrow5x=2\left(2x+y\right)=4x+2y\Leftrightarrow x=2y\Leftrightarrow\frac{x}{2}=y\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{2}=y=\frac{3x}{6}=\frac{4y}{4}=\frac{3x+4y}{6+4}=\frac{50}{10}=5\)

Vậy\(\hept{\begin{cases}x=5\cdot2=10\\y=5\end{cases}}\)

a) 

\(=\frac{3}{5}.\frac{3}{7}+\frac{3}{5}.\frac{4}{7}-\left(1+\frac{3}{5}\right)\)

\(=\frac{3}{5}\left(\frac{3}{7}+\frac{4}{7}\right)-1-\frac{3}{5}\)

\(=\frac{3}{5}-1-\frac{3}{5}\)

\(=-1\)

b) \(=\frac{2^2.5.7.5^2.7^3}{2^2.5^2.7^{2.2}}\)

\(=\frac{2^2.5^{1+2}.7^{3+1}}{2^2.5^2.7^4}=\frac{2^2.5^3.7^4}{2^2.5^2.7^4}=2^{2-2}.5^{3-2}.7^{4-4}=2^0.5^1.7^0=1.5.1=5\)