a, Ta có :  \(f\left(x\right)-g\left(x\right)=h\left(x\right)\)hay 

\(4x^2+3x+1-3x^2+2x-1=h\left(x\right)\)

\(\Rightarrow h\left(x\right)=x^2+5x\)

b, Đặt \(h\left(x\right)=x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Vậy nghiệm của đa thức h(x) là x = -5 ; x = 0 

Đặt \(k\left(x\right)=7x^2-35x+42=0\)

\(\Leftrightarrow7\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow7\left(x^2+2x+3x+6\right)=0\Leftrightarrow7\left(x+2\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}\)

Vậy nghiệm của đa thức k(x) là x = -3 ; x = -2

xin lỗi mọi người 1 tý nha cái phần c) ý ạ đề thì vậy như thế nhưng có cái ở phần biểu thức ở dưới ý là 

\(\left(\frac{3^2}{6}-81\right)^3\) chuyển thành \(\left(\frac{3^3}{6}81\right)^3\)

bị sai mỗi thế thôi ạ mọi người giúp em với ạ

a,

\(\left(4x-\dfrac{1}{3}\right)^6=1\\ \Rightarrow\left[{}\begin{matrix}4x-\dfrac{1}{3}=1\\4x-\dfrac{1}{3}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{-2}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{-1}{6}\end{matrix}\right.\)

b,

\(\left(5x-\dfrac{2}{3}\right)^2=0\\ \Rightarrow5x-\dfrac{2}{3}=0\\ 5x=\dfrac{2}{3}\\ x=\dfrac{2}{15}\)

c,

\(\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=-8\\ \Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}=-2\\ \dfrac{1}{3}x=\dfrac{-3}{2}\\ x=\dfrac{-9}{2}\)

d,

\(\dfrac{81}{3^n}=3\\ \Leftrightarrow3^4:3^n=3^1\\\Leftrightarrow3^{4-n}=3^1 \\ \Rightarrow n=3\)

e,

\(\dfrac{\left(-2\right)^x}{64}=-2\\ \Leftrightarrow\left(-2\right)^x:\left(-2\right)^6=\left(-2\right)^1\\ \Leftrightarrow\left(-2\right)^{x-6}=\left(-2\right)^1\\ \Rightarrow x=7\)

f,

\(\left(-20\right)^n:10^n=16\\ \left[\left(-20\right):10\right]^n=16\\ \left(-2\right)^n=\left(-2\right)^4\\ \Rightarrow n=4\)

Bài 1:

a) \(\left(4x-\dfrac{1}{3}\right)^6=1\)

\(\Rightarrow4x-\dfrac{1}{3}=1\)

\(4x=1+\dfrac{1}{3}\)

\(4x=\dfrac{4}{3}\)

\(x=\dfrac{4}{3}:4\)

\(x=\dfrac{1}{3}\)

b) \(\left(5x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow5x-\dfrac{2}{3}=0\)

\(5x=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}:5\)

\(x=\dfrac{2}{15}\)

c) \(\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=-8\)

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

\(\dfrac{1}{3}x-\dfrac{1}{2}=-2\)

\(\dfrac{1}{3}x=-2+\dfrac{1}{2}\)

\(\dfrac{1}{3}x=\dfrac{-3}{2}\)

\(x=\dfrac{-3}{2}:\dfrac{1}{3}\)

\(x=\dfrac{-9}{2}\)

d) \(\dfrac{81}{3^n}=3\)

\(\Rightarrow\dfrac{3^4}{3^n}=3\)

\(\Rightarrow3^n.3=3^4\)

\(3^{n+1}=3^4\)

n + 1 = 4

n = 4 - 1

n = 3

e) \(\dfrac{\left(-2\right)^x}{64}=-2\)

\(\Rightarrow\dfrac{\left(-2\right)^x}{\left(-2\right)^6}=-2\)

\(\Rightarrow\left(-2\right)^x=\left(-2\right)^6.\left(-2\right)\)

\(\left(-2\right)^x=\left(-2\right)^7\)

x = 7

f) (-20)ⁿ : 10ⁿ = 16

(-20 : 10)ⁿ = 16

(-2)ⁿ = 16

(-2)ⁿ = (-2)⁴

n = 4.

a: =>x=(-2/3)^5:(-2/3)^2=(-2/3)^3=-8/27

b: =>x*(-1/3)^3=(-1/3)^4

=>x=-1/3

d: =>3x-2=-3

=>3x=-1

=>x=-1/3

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây

hay đó

a/ \(\dfrac{\left(-3\right)^x}{81}=-27\)

\(\Leftrightarrow\left(-3\right)^x=\left(-27\right).81\)

\(\Leftrightarrow\left(-3\right)^x=-2187\)

\(\Leftrightarrow\left(-3\right)^x=\left(-3\right)^7\)

\(\Leftrightarrow x=7\)

Vậy ...

b/ \(2^{x-1}=16\)

\(\Leftrightarrow2^{x-1}=2^4\)

\(\Leftrightarrow x-1=4\)

\(\Leftrightarrow x=5\)

Vậy ....

c/ \(\left(x-1\right)^2=25\)

\(\Leftrightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Vậy ....

d/ \(0,2-\left|4,2-2x\right|=0\)

\(\Leftrightarrow\left|4,2-2x\right|=0,2\)

\(\Leftrightarrow\left[{}\begin{matrix}4,2-2x=0,2\\4,2-2x=-0,2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=4,4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=2,2\end{matrix}\right.\)

Vậy ............

e, \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)

\(\Leftrightarrow\dfrac{5}{3}.\dfrac{4}{x}=20\)

\(\Leftrightarrow3x=1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

Vậy ..

a) \(\dfrac{\left(-3\right)^x}{81}=-27\)

⇔ \(\dfrac{\left(-3\right)^x}{81}=\dfrac{-2187}{81}\)

⇔ (-3)^x = -2187

⇔ (-3)^x = (-3)⁷

⇔ x = 7

b) 2^x-1 = 16

⇔ 2^x-1 = 2⁴

⇔ x - 1 = 4

⇔ x = 4 + 1

⇔ x = 5

c) (x - 1)² = 25

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

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

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

⇔ \(\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

d) 0,2 - |4,2 - 2x| = 0

⇔ |4,2 - 2x| = 0,2

⇔ \(\left[{}\begin{matrix}4,2-2x=0,2\\4,2-2x=-0,2\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}2x=4,2-0,2\\2x=4,2-\left(-0,2\right)\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}2x=4\\2x=4,4\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=2\\x=2,2\end{matrix}\right.\)

e) \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)

⇔ \(\dfrac{5}{3}.\dfrac{4}{x}=20\)

⇔ \(\dfrac{20}{3x}=20\)

⇔ 3x = 1

⇔ x = \(\dfrac{1}{3}\)

a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)

\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)

\(x=\left(\frac{3}{7}\right)^2\)

\(x=\frac{9}{49}\)

Vậy...

b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)

\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)

\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)

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

Vậy...

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

=>\(x-\frac{1}{2}=\frac{1}{3}\)

\(x=\frac{1}{3}+\frac{1}{2}\)

\(x=\frac{5}{6}\)

Vậy...

d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)

=>\(x+\frac{1}{4}=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{1}{4}\)

\(x=\frac{5}{12}\)

Vậy...

Phù, mãi mới xong, tk cho mk nha bn

a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)

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

\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)

\(\Rightarrow x+3=-3\)

\(\Rightarrow x=-6\)

b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)

\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)

\(\Rightarrow2x+2=-2\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)

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

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

\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)

\(\Rightarrow x+5=-4\)

\(\Rightarrow x=-9\)

d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)

\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)

\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)