1: Sửa đề: \(f\left(x\right)=3x\left(1-3x+2x^3\right)-2x^2\left(-4+3x^2-x\right)\)

\(=3x-9x^2+6x^4+8x^2-6x^4+2x^3\)

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

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

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

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

Bậc là 3

Hệ số cao nhất là 2

Hệ số tự do là 0

2: f(x)=g(x)+h(x)

=>h(x)=f(x)-g(x)

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

3: Đặt h(x)=0

=>3x(x+1)=0

=>x(x+1)=0

=>\(\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

1. `G(x)=-4(x^4+x^2+1)+x^3(4x+2)+4`

`=-4x^4-4x^2-4+4x^4+2x^3+4`

`=(4x^4-4x^4)+2x^3-4x^2+(4-4)`

`=2x^3-4x^2`

Bậc 3

Hệ số cao nhất: 2

Hệ số tự đó: 0

2. `F(x) = G(x) + H(x)`

`=>H(x)=F(x) - G(x)`

`=>H(x)=[3x(1-3x+2x^3)-2x^2(-4+3x^2-x)]-(2x^3-4x^2)

`=>H(x)=3x-9x^2+6x^4+8x^2-6x^4+2x^3-2x^3+4x^2`

`=>H(x)=3x^2+3x`

3. `H(x)=3x^2+3x=0`

`=>3x(x+1)=0`

TH1: `x=0`

TH2: `x+1=0=>x=-1`

BÀI 1: 

a) \(\dfrac{9}{70}>\dfrac{5}{42}\)              b) \(-\dfrac{4}{27}>-\dfrac{10}{63}\)

c) \(\dfrac{999}{556}>\dfrac{1000}{557}\)        d) \(-\dfrac{2}{15}< \dfrac{10}{11}\)

BÀI 2:

\(-1\dfrac{2}{3}< -\dfrac{3}{5}< -\dfrac{5}{9}< 0,5< \dfrac{10}{9}\)

BÀI 3:

a) -3 ϵ Q     b) 10 ϵ N

c) -3/7 ϵ Q     d) -2 ϵ Q

`(-3^x*3^6)/(-27*9^x) = -3`

`(-3^(x+6))/(-3^3*(3^2)^x)=-3`

`(-3^(x+6))/(-3^3*3^(2x))=-3`

`(-3^(x+6))/(-3^(2x+3))=-3`

`3^(x+6-2x-3)=-3`

`3^(3-x)=-3`

`(3^(3-x))/3=-1`

`3^(3-x-1)=-1`

`3^(2-x)=-1`

Vì `3^(2-x) > 0` mà `-1<0`

Nên không có x thoả mãn

Bạn vt lại gt lại \(a\left(a+b\right)=c^2\) bạn áp dụng bổ đề trên ta được a+a+b+c+c là hợp số

\(\dfrac{\left(-\dfrac{1}{3}\right)^6}{\left(\dfrac{1}{6}\right)^2}=\dfrac{\left(\dfrac{1}{3}\right)^6}{\left(\dfrac{1}{6}\right)^2}\\ =\dfrac{\dfrac{1^6}{3^6}}{\dfrac{1^2}{6^2}}=\dfrac{\dfrac{1}{3^6}}{\dfrac{1}{6^2}}\\ =\dfrac{1}{3^6}:\dfrac{1}{6^2}=\dfrac{1}{3^6}.\dfrac{6^2}{1}\\ =\dfrac{6^2}{3^6}=\dfrac{2^2.3^2}{3^2.3^4}\\ =\dfrac{2^2}{3^4}=\dfrac{4}{81}\)

(-1/6)^3/(1/6)^2

= -1/18 / 1/36

= -1/18 ÷ 1/36

= -1/18 × 36/1

= -1/1 × 2/1

= -2

a: \(A=\left|x-3,5\right|+\left|4,1-x\right|=\left|x-3,5\right|+\left|x-4,1\right|\)

3,5<=x<=4,1

=>x-3,5>=0 và x-4,1<=0

=>A=x-3,5+4,1-x=0,6

b: \(A=\left|x-7\right|+\left|1-x\right|=\left|x-7\right|+\left|x-1\right|\)

\(1< =x< =7\)

=>\(x-1>=0;x-7< =0\)

=>A=x-1+7-x=6

c: \(A=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)

\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)

\(-\dfrac{3}{5}< x< \dfrac{1}{7}\)

=>\(x+\dfrac{3}{5}>0;x-\dfrac{1}{7}< 0\)

=>\(A=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{1}{7}+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)

d: \(A=\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+8\dfrac{1}{5}\)

\(=\left|x-2\dfrac{1}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

\(\dfrac{1}{5}< =x< =2\dfrac{1}{5}\)

=>\(x-\dfrac{1}{5}>=0;x-2\dfrac{1}{5}< =0\)

=>\(D=2\dfrac{1}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=2+\dfrac{41}{5}=\dfrac{51}{5}\)

Biết $13a+b+2c$ như thế nào hả bạn?

c: \(A=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)

\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)

\(-\dfrac{3}{5}< x< \dfrac{1}{7}\)

=>\(x+\dfrac{3}{5}>0;x-\dfrac{1}{7}< 0\)

=>\(A=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{1}{7}+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)

d: \(A=\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+8\dfrac{1}{5}\)

\(=\left|x-2\dfrac{1}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

\(\dfrac{1}{5}< =x< =2\dfrac{1}{5}\)

=>\(x-\dfrac{1}{5}>=0;x-2\dfrac{1}{5}< =0\)

=>\(D=2\dfrac{1}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=2+\dfrac{41}{5}=\dfrac{51}{5}\)

1: \(\left|x-3,5\right|>=0\forall x\)

\(\left|4,5-y\right|>=0\forall y\)

Do đó: \(\left|x-3,5\right|+\left|4.5-y\right|>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-3,5=0\\4,5-y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3,5\\y=4,5\end{matrix}\right.\)

2: \(\left\{{}\begin{matrix}\left|x+\dfrac{2}{3}\right|>=0\forall x\\\left|y-\dfrac{3}{4}\right|>=0\forall y\\\left|z-5\right|>=0\forall z\end{matrix}\right.\)

Do đó: \(\left|x+\dfrac{2}{3}\right|+\left|y-\dfrac{3}{4}\right|+\left|z-5\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+\dfrac{2}{3}=0\\y-\dfrac{3}{4}=0\\z-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=\dfrac{3}{4}\\z=5\end{matrix}\right.\)

3: \(\left|x-2\right|+\left|3-x\right|=0\)

=>|x-2|+|x-3|=0(1)

TH1: x<2

Phương trình (1) sẽ trở thành 2-x+3-x=0

=>5-2x=0

=>2x=5

=>x=2,5(loại)

TH2: 2<=x<3

Phương trình (1) sẽ trở thành x-2+3-x=0

=>1=0(loại)

TH3: x>=3

Phương trình (1) sẽ trở thành x-2+x-3=0

=>2x=5

=>x=2,5(loại)

Vậy: Phương trình vô nghiệm

4: \(\left\{{}\begin{matrix}\left|x-\dfrac{2}{3}\right|>=0\forall x\\\left|x+y+\dfrac{3}{4}\right|>=0\forall x,y\\\left|y-z-\dfrac{5}{6}\right|>=0\forall y,z\end{matrix}\right.\)

Do đó: \(\left|x-\dfrac{2}{3}\right|+\left|x+y+\dfrac{3}{4}\right|+\left|y-z-\dfrac{5}{6}\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\x+y+\dfrac{3}{4}=0\\y-z-\dfrac{5}{6}=0\end{matrix}\right.\)

=>\(\begin{matrix}x=\dfrac{2}{3}\\y=-x-\dfrac{3}{4}=-\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{-17}{12}\\z=y-\dfrac{5}{6}=-\dfrac{17}{12}-\dfrac{5}{6}=-\dfrac{27}{12}=-\dfrac{9}{4}\end{matrix}\)

5: \(\left\{{}\begin{matrix}\left|x-\dfrac{2}{3}\right|>=0\forall x\\\left|xy-\dfrac{5}{8}\right|>=0\forall x,y\\\left|yz+\dfrac{3}{4}\right|>=0\forall y,z\end{matrix}\right.\)

Do đó: \(\left|x-\dfrac{2}{3}\right|+\left|xy-\dfrac{5}{8}\right|+\left|yz+\dfrac{3}{4}\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\xy-\dfrac{5}{8}=0\\yz+\dfrac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\xy=\dfrac{5}{8}\\yz=-\dfrac{3}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{5}{8}:x=\dfrac{5}{8}:\dfrac{2}{3}=\dfrac{5}{8}\cdot\dfrac{3}{2}=\dfrac{15}{16}\\z=-\dfrac{3}{4}:\dfrac{15}{16}=-\dfrac{3}{4}\cdot\dfrac{16}{15}=\dfrac{-48}{60}=-\dfrac{4}{5}\end{matrix}\right.\)

6: \(\left\{{}\begin{matrix}\left|xy+\dfrac{2}{3}\right|>=0\forall x,y\\\left|yz-\dfrac{8}{9}\right|>=0\forall y,z\\\left|xz+\dfrac{3}{4}\right|>=0\forall x,z\end{matrix}\right.\)

Do đó: \(\left|xy+\dfrac{2}{3}\right|+\left|yz-\dfrac{8}{9}\right|+\left|xz+\dfrac{3}{4}\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}xy+\dfrac{2}{3}=0\\yz-\dfrac{8}{9}=0\\xz+\dfrac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=-\dfrac{2}{3}\\yz=\dfrac{8}{9}\\xz=-\dfrac{3}{4}\end{matrix}\right.\)

=>\(\left(xyz\right)^2=-\dfrac{2}{3}\cdot\dfrac{8}{9}\cdot\dfrac{-3}{4}=\dfrac{1}{2}\cdot\dfrac{8}{9}=\dfrac{4}{9}\)

=>\(\left[{}\begin{matrix}xyz=\dfrac{2}{3}\\xyz=-\dfrac{2}{3}\end{matrix}\right.\)

TH1: xyz=2/3

=>\(\left\{{}\begin{matrix}z=\dfrac{xyz}{xy}=\dfrac{2}{3}:\dfrac{-2}{3}=-1\\x=\dfrac{xyz}{yz}=\dfrac{2}{3}:\dfrac{8}{9}=\dfrac{2}{3}\cdot\dfrac{9}{8}=\dfrac{18}{24}=\dfrac{3}{4}\\y=\dfrac{xyz}{xz}=\dfrac{2}{3}:\dfrac{-3}{4}=\dfrac{2}{3}\cdot\dfrac{4}{-3}=-\dfrac{8}{9}\end{matrix}\right.\)

TH2: xyz=-2/3

=>\(\left\{{}\begin{matrix}z=\dfrac{xyz}{xy}=-\dfrac{2}{3}:\dfrac{-2}{3}=1\\x=\dfrac{xyz}{yz}=-\dfrac{2}{3}:\dfrac{8}{9}=\dfrac{-2}{3}\cdot\dfrac{9}{8}=\dfrac{-18}{24}=\dfrac{-3}{4}\\y=\dfrac{xyz}{xz}=\dfrac{-2}{3}:\dfrac{-3}{4}=\dfrac{-2}{3}\cdot\dfrac{4}{-3}=\dfrac{8}{9}\end{matrix}\right.\)