\(a)\)

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

\(\Leftrightarrow f\left(x\right)=-x^3+\left(3x^2-x^2\right)+x-3+2^3\)

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

\(\Leftrightarrow f\left(x\right)=-x^3+2x^2+x+5\)

\(g\left(x\right)=-3x^3-x^2+2x^3+5x-3-4x\)

\(\Leftrightarrow g\left(x\right)=\left(-3x^3+2x^3\right)-x^2+\left(5x-4x\right)-3\)

\(\Leftrightarrow g\left(x\right)=-x^3-x^2+x-3\)

\(b)\)

Theo đề ra: \(h\left(x\right)=f\left(x\right)+g\left(x\right)\)

\(f\left(x\right)=-x^3+2x^2+x+5\)

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

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

tính góc gì thế bạn

ACD^ ah

a, \(\left|x-1\right|=3x+2\)ĐK : x >= -2/3 

TH1 : \(x-1=3x+2\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)( ktm )

TH2 : \(x-1=-3x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)( tm )

b, \(\left|5x\right|=x-12\)ĐK : x > = 12 

TH1 : \(5x=x-12\Leftrightarrow4x=-12\Leftrightarrow x=-3\)( ktm )

TH2 : \(5x=12-x\Leftrightarrow6x=12\Leftrightarrow x=2\)( ktm )

c, \(\left|7-x\right|=5x+1\)ĐK : x >= -1/5

TH1 : \(7-x=5x+1\Leftrightarrow6x=-6\Leftrightarrow x=-1\)( ktm )

TH2 : \(7-x=5x+1\Leftrightarrow6x=6\Leftrightarrow x=1\)( tm )

a) Điều kiện : \(x\ge-\frac{2}{3}\)

Xét : 

• \(-\frac{2}{3}\le x< 1\)

\(\Rightarrow x-1< 0\)

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

\(\Rightarrow1-x=3x+2\)

\(\Rightarrow4x=-1\)

\(\Rightarrow x=-\frac{1}{4}\left(TM\right)\)

• \(x\ge1\)

\(\Rightarrow x-1\ge0\)

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

\(\Rightarrow x-1=3x+2\)

\(\Rightarrow2x=-3\)

\(\Rightarrow x=-\frac{3}{2}\left(KTM\right)\)

Vậy \(x=-\frac{1}{4}\)

b) Điều kiện : \(x\ge12\)

\(\Rightarrow5x>0\)

\(\Rightarrow\left|5x\right|=5x\)

\(\Rightarrow5x=x-12\)

\(\Rightarrow4x=-12\)

\(\Rightarrow x=-3\left(KTM\right)\)

Vậy không có x

c)Điều kiện : \(x\ge-\frac{1}{5}\)

Xét :

• \(-\frac{1}{5}\le x\le7\)

\(\Rightarrow7-x\ge0\)

\(\Rightarrow\left|7-x\right|=7-x\)

\(\Rightarrow7-x=5x+1\)

\(\Rightarrow6x=6\)

\(\Rightarrow x=1\left(TM\right)\)

• \(x>7\)

\(\Rightarrow7-x< 0\)

\(\Rightarrow\left|7-x\right|=x-7\)

\(\Rightarrow x-7=5x+1\)

\(\Rightarrow4x=-8\)

\(\Rightarrow x=-2\left(KTM\right)\)

Vậy \(x=1\)

a, \(M=-\left(3x-5\right)^2-18\le-18\)

Dấu ''='' xảy ra khi x = 5/3 

Vậy GTLN của M bằng -18 tại x = 5/3 

b, \(N=-\left(x-5\right)^2-4\left(x-y+2\right)^2+5\le5\)

Dấu ''='' xảy ra khi x = 5 ; y = 7

Vậy GTLN của N bằng 5 tại x = 5 ; y = 7 

c, Ta có : \(\left|2x-4\right|+\left(y-2\right)^2+8\ge8\)

\(\Rightarrow P\le\frac{30}{8}=\frac{15}{4}\)

Dấu ''='' xảy ra khi x = y = 2

Vậy GTLN của P bằng 15/4 tại x = y = 2 

d, \(Q=-\left|3x-2\right|^{2021}-\frac{2}{3}\left(y-7\right)^{2022}+5\le5\)

Dấu ''='' xảy ra khi x = 2/3 ; y = 7

Vậy GTLN của Q bằng 5 tại x = 2/3 ; y = 7

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

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

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

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

\(\Leftrightarrow\frac{5}{2}x=3\)

\(\Leftrightarrow x=3:\frac{5}{2}\)

\(\Leftrightarrow x=3.\frac{2}{5}\)

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

b. \(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)  

= \(\frac{100^4}{100^5}\) 

=\(\frac{1}{100}\)

\(\frac{2^{27}.9^8}{6^{15}.8^{15}}\)

\(=\frac{2^{27}.3^{16}}{2^{60}.3^{15}}\)

\(=\frac{3}{2^{33}}\)