Bài 4 :

\(\left(5x-20\right)+\left(3x^2-12x\right)=0\)

\(\Leftrightarrow5\left(x-4\right)+3x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(5+3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Vậy \(x=4\) hoặc \(x=-\dfrac{5}{3}\)

Bài 5 :

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\1+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x=1\) hoặc \(x=-\dfrac{1}{3}\)

Bài 6 :

\(\left(4x+20\right)-\left(x+5\right)^2=0\)

\(\Leftrightarrow4\left(x+5\right)-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)\left(4-x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(-x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\-x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)

Vậy \(x=-5\) hoặc \(x=-1\)

a)\(\left(x+8\right)-11=20-15\)

\(\left(x+8\right)-11=5\)

\( x+8=5+11\)

\(x+8=16\)

\(x=8\)

b) \(2x-\left(3+x\right)=5-7\)

\(2x-\left(3+x\right)=-2\)

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

\(x=1\)

c) \( \left(3x-2^4\right)\times7^5=2\times7^6\)

\(3x-2^4=2\times\left(7^6:7^5\right) \)

\(\left(3x-2^4\right)=2\times7^2\)

\(3x-2^4=2\times49\)

\(3x-16=98\)

\(3x=114\)

\(x=38\)

a,  3(x+3)-2(x-5)=11

=> 3x+9-2x+10=11

=> 3x-2x=11-10-9

=>  x=-8

Vậy.........

b,   14-4|x|=-6

=>  -4|x|=8

=>   |x|=-2(VL vì trị tuyệt đối luôn lớn hơn hoặc = 0)

Vậy......

\(a,5^{x-2}=125\)

\(\Rightarrow5^{x-2}=5^3\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=5\)

\(b,3^{x+4}=243\)

\(\Rightarrow3^{x+4}=3^5\)

\(\Rightarrow x+4=5\)

\(\Rightarrow x=1\)

\(5^{x-2}=125\)

\(\Rightarrow5^{x-2}=5^3\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=3+2\)

\(\Rightarrow x=5\)

\(5^{x-2}=125\)

\(\Rightarrow5^{x-2}=5^3\)

\(\Rightarrow x-2=3\)

\(x=5\)

\(3^{x+4}=243\)

\(\Rightarrow3^{x+4}=3^5\)

\(\Rightarrow x+4=5\)

\(\Rightarrow x=1\)

a) |3x - 2| - 5 = 7

=> |3x - 2| = 7 + 5

=> |3x - 2| = 12

=> \(\orbr{\begin{cases}3x-2=12\\3x-2=-12\end{cases}}\)

=> \(\orbr{\begin{cases}3x=14\\3x=-10\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{14}{3}\\x=-\frac{10}{3}\end{cases}}\)

b) |3x - 1| - |x + 2| = 0

=> |3x - 1| = |x + 2|

=> \(\orbr{\begin{cases}3x-1=x+2\\3x-1=-x-2\end{cases}}\)

=> \(\orbr{\begin{cases}2x=3\\4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{4}\end{cases}}\)

thanks bn

c) |3x - 3| = x - 3

ĐKXĐ: x - 3 \(\ge\)0 <=>  x \(\ge\)3

=> \(\orbr{\begin{cases}3x-3=x-3\\3x-3=3-x\end{cases}}\)

=> \(\orbr{\begin{cases}2x=0\\4x=6\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\left(ktm\right)\\x=\frac{3}{2}\left(ktm\right)\end{cases}}\)

4) 4 - |4x + 3| = 5x

=> |4x + 3| = 4 - 5x

ĐKXĐ: 4 - 5x \(\ge\)0 <=> x \(\le\)4/5

=> \(\orbr{\begin{cases}4x+3=4-5x\\4x+3=5x-4\end{cases}}\)

=> \(\orbr{\begin{cases}9x=1\\-x=-7\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{9}\left(tm\right)\\x=7\left(ktm\right)\end{cases}}\)

Mk ko biết  kb nha 

a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)

ta có:  f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)

                          \(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)

                      \(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)

\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)

Chúc bn học tốt !!!!!!

Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............