\(\frac{15^{10}.5^{10}}{75^{10}}\)

\(=\frac{15^{10}.5^{10}}{\left(15.5\right)^{10}}\)

\(=\frac{15^{10}.5^{10}}{15^{10}.5^{10}}\)

\(=1\)

          Bài làm :

Ta có :

\(\frac{15^{10}\times5^{10}}{75^{10}}=\frac{\left(15\times5\right)^{10}}{75^{10}}=\frac{75^{10}}{75^{10}}=1\)

nhanh giúp em với ạ

\(\frac{x-2}{3}=\frac{4}{5}\)

\(\Leftrightarrow x-2=\frac{12}{5}\)

\(\Leftrightarrow x=\frac{22}{5}\)

ĐKXĐ : \(x\ne4\)

\(\frac{2x-3}{4-x}>0\)

+) TH1 : \(\hept{\begin{cases}2x-3>0\\4-x>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>\frac{3}{2}\\x< 4\end{cases}\Leftrightarrow}\frac{3}{2}< x< 4}\)

+) TH2 : \(\hept{\begin{cases}2x-3< 0\\4-x< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< \frac{3}{2}\\x>4\end{cases}\left(ktm\right)}}\)

Vậy với \(\frac{3}{2}< x< 4\)thì \(\frac{2x-3}{4-x}>0\)

\(\frac{2x-3}{4-x}>0\)

ĐK : x khác 4

Xét hai trường hợp :

1. \(\hept{\begin{cases}2x-3>0\\4-x>0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x>3\\-x>-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{3}{2}\\x< 4\end{cases}}\Leftrightarrow\frac{3}{2}< x< 4\)

2. \(\hept{\begin{cases}2x-3< 0\\4-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x< 3\\-x< -4\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{3}{2}\\x>4\end{cases}}\)( loại )

=> Với \(\frac{3}{2}< x< 4\)thì thỏa mãn đề bài

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Rightarrow3A=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}\)

\(2A=3A-A\)

\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{2007}}-\frac{1}{3^{2008}}\)

\(=1-\frac{1}{3^{2008}}\)

\(2A=1-\frac{1}{3^{2008}}\Rightarrow A=\frac{1-\frac{1}{3^{2008}}}{2}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\)

\(\Leftrightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(\Leftrightarrow2A=1-\frac{1}{3^{2008}}\)

\(\Leftrightarrow2A=\frac{3^{2008}-1}{3^{2008}}\)

\(\Leftrightarrow A=\frac{3^{2008}-1}{3^{2008}}\div2\)

\(\Leftrightarrow A=\frac{3^{2008}-1}{2.3^{2008}}\)

a) Vì ax là tia phân giác của góc bac nên bax=xac(1)

Vì ax//cd => xac và dca là hai góc so le trong=>xac=acd (2)

Vì bax và adc là hai góc đồng vị =>bax=adc(3)

Từ (1), (2) và (3) => xab=adc=acd (đpcm)

Xin lỗi vì chỉ mới làm đc câu a nhé =))

a) ( x - 1/5 )² = 0

<=> x - 1/5 = 0

<=> x = 1/5

b) ( x - 2 )² = 1

<=> ( x - 2 )² = ( ±1 )²

<=> x - 2 = 1 hoặc x - 2 = -1

<=> x = 3 hoặc x = 1

c) ( 2x - 1 )³ = -8

<=> ( 2x - 1 )³ = (-2)³

<=> 2x - 1 = -2

<=> 2x = -1

<=> x = -1/2

d) ( x⁴ )² = x¹²/x⁵

<=> x⁸ = x⁷

<=> x⁸ - x⁷ = 0

<=> x⁷( x - 1 ) = 0

<=> x⁷ = 0 hoặc x - 1 = 0

<=> x = 0 hoặc x = 1

e) x¹⁰ = 25x⁸

<=> x¹⁰ - 25x⁸ = 0

<=> x⁸( x² - 25 ) = 0

<=> x⁸ = 0 hoặc x² - 25 = 0

<=> x = 0 hoặc x = ±5

f) ( 2x + 3 )² = 9/121

<=> ( 2x + 3 )² = ( ±3/11 )²

<=> 2x + 3 = 3/11 hoặc 2x + 3 = -3/11

<=> x = -15/11 hoặc x = -18/11

a) \(\left(x-\frac{1}{5}\right)^2=0\Leftrightarrow x-\frac{1}{5}=0\Leftrightarrow x=\frac{1}{5}\)

b) \(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

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

c) \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3+8=0\)

\(\Leftrightarrow\left(2x-1+8\right)\left[\left(2x-1\right)^2-8\left(2x-1\right)+64\right]=0\)

\(\Leftrightarrow2x+7=0\)

\(\Leftrightarrow x=\frac{-7}{2}\)

d) ĐKXĐ : \(x\ne0\)

 \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)

\(\Leftrightarrow x^8=x^7\)

\(\Leftrightarrow x^8-x^7=0\)

\(\Leftrightarrow x^7\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=1\left(tm\right)\end{cases}\Leftrightarrow x=1}\)

e) ĐKXĐ : x khác 0 

 \(x^{10}=25x^8\)

\(\Leftrightarrow x^2=25\Leftrightarrow x=5\)

f) \(\left(2x+3\right)^2=\frac{9}{121}\)

\(\Leftrightarrow\left(2x+3+\frac{3}{11}\right)\left(2x+3-\frac{3}{11}\right)=0\)

\(\Leftrightarrow\left(2x+\frac{36}{11}\right)\left(2x+\frac{30}{11}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-18}{11}\\x=-\frac{15}{11}\end{cases}}\)

đéo biết làm

            Bài làm :

Sửa đề bài : 5a+3b / 5a-3b = 5c+3d/5c-3d

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

Ta có :

\(\hept{\begin{cases}\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\\\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\end{cases}}\)

\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)

=> Điều phải chứng minh