a: Đặt A=0

=>-2/3x=5/9

hay x=-5/6

b: Đặt B(x)=0

=>(x-2/5)(x+2/5)=0

=>x=2/5 hoặc x=-2/5

c: Đặt C(X)=0

\(\Leftrightarrow x^3\cdot\dfrac{1}{2}=-\dfrac{4}{27}\)

\(\Leftrightarrow x^3=-\dfrac{8}{27}\)

hay x=-2/3

\(f\left(1-x\right)+f\left(x\right)=\dfrac{9^{1-x}}{9^{1-x}+3}+\dfrac{9^x}{9^x+3}=\dfrac{9}{9+3.9^x}+\dfrac{9^x}{9^x+3}=\dfrac{3}{9^x+3}+\dfrac{9^x}{9^x+3}=1\)

\(\Rightarrow f\left(x\right)=1-f\left(1-x\right)\)

\(\Rightarrow f\left(cos^2x\right)=1-f\left(sin^2x\right)\)

Do đó:

\(f\left(3m+\dfrac{1}{4}sinx\right)+f\left(cos^2x\right)=1\)

\(\Leftrightarrow f\left(3m+\dfrac{1}{4}sinx\right)=f\left(sin^2x\right)\) (1)

Hàm \(f\left(x\right)=\dfrac{9^x}{9^x+3}\) có \(f'\left(x\right)=\dfrac{3.9^x.ln9}{\left(9^x+3\right)^2}>0\Rightarrow f\left(x\right)\) đồng biến trên R

\(\Rightarrow\left(1\right)\Leftrightarrow3m+\dfrac{1}{4}sinx=sin^2x\)

Đến đây chắc dễ rồi, biện luận để pt \(sin^2x-\dfrac{1}{4}sinx=3m\) có 8 nghiệm trên khoảng đã cho

Lời giải:

a) \((5x-1)^6=729=3^6=(-3)^6\)

\(\Rightarrow \left[\begin{matrix} 5x-1=3\\ 5x-1=-3\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{4}{5}\\ x=\frac{-2}{5}\end{matrix}\right.\)

b)

\(\frac{8}{25}=\frac{2^x}{5^{x-1}}=\frac{2^x}{5^x:5}=5.(\frac{2}{5})^x\)

\(\Rightarrow \frac{8}{125}=(\frac{2}{5})^x\)

\(\Rightarrow (\frac{2}{5})^3=(\frac{2}{5})^x\Rightarrow x=3\)

c)

\((\frac{1}{16})^x=(\frac{1}{2})^{10}\)

\(\Rightarrow (\frac{1}{2^4})^x=(\frac{1}{2})^{10}\)

\(\Rightarrow (\frac{1}{2})^{4x}=(\frac{1}{2})^{10}\Rightarrow 4x=10\Rightarrow x=\frac{5}{2}\)

d)

\(9^{x}:3^x=3\Rightarrow (\frac{9}{3})^x=3\)

\(\Rightarrow 3^x=3^1\Rightarrow x=1\)

a) \(\left(5x-1\right)^6=729\)

\(\Leftrightarrow5x-1=3\)

\(\Leftrightarrow5x=4\)

\(\Leftrightarrow x=\dfrac{4}{5}\)

b: \(\Leftrightarrow\dfrac{2^3}{5^2}=\dfrac{2^x}{5^{x-1}}\)

=>x=3 và x-1=2

=>x=3

c: \(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{10}\)

=>4x=10

=>x=5/2

d: =>3^x=3

=>x=1

a) \(\left(5x-1\right)^6=729\)

\(\Rightarrow\left[{}\begin{matrix}\left(5x-1\right)^6=3^6\\\left(5x-1\right)^6=\left(-3\right)^6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=3\\5x-1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=4\\5x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

b) \(\dfrac{8}{25}=\dfrac{2^x}{5^{x-1}}\)

\(\Rightarrow\left[{}\begin{matrix}2^x=2^3\\5^{x-1}=5^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x-1=2\end{matrix}\right.\)

\(\Rightarrow x=3\)

Vậy x = 3

c) \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{2}\right)^{10}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{3x}=\left(\dfrac{1}{2}\right)^{10}\)

\(\Rightarrow3x=10\)

\(\Rightarrow x=\dfrac{10}{3}\)

d) \(9^x:3^x=3\)

\(\Rightarrow\left(9:3\right)^x=3\)

\(\Rightarrow3^x=3^1\)

\(\Rightarrow x=1\)

a,\(\left(5x-1\right)^6=729\)

\(\left(5x-1\right)^6=3^6\)

\(5x-1=3\)

\(5x=4\)

\(x=\dfrac{4}{5}\)

b Ta có \(8=2^3\),\(25=5^2\)

Mà \(\dfrac{8}{25}=\dfrac{2^x}{5^{x-1}}\)

=> \(2^3=2^x,5^2=5^{x-1}\)

=> x=3

c Ta có \(16=2^4\)

hay \(\left(\dfrac{1}{2^4}\right)^x=\left(\dfrac{1}{2}\right)^{10}=>x\cdot4=10=>x=\dfrac{10}{4}=\dfrac{5}{2}\)

B

\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)

\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)