\(c,\Rightarrow\left|x-\dfrac{1}{9}\right|=-\dfrac{4}{5}\\ \Rightarrow x\in\varnothing\left(\left|x-\dfrac{1}{9}\right|\ge0>-\dfrac{4}{5}\right)\\ d,\Rightarrow\left\{{}\begin{matrix}3x-2=0\\4y-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{7}{4}\end{matrix}\right.\\ e,\Rightarrow\left\{{}\begin{matrix}2x+1=0\\x-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=y=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x=y=-\dfrac{1}{2}\)

a: \(\left(x-1.2\right)^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1.2=2\\x-1.2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.2\\x=-0.8\end{matrix}\right.\)

b: Ta có: \(\left(x+1\right)^3=-125\)

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

hay x=-6

c) 3^(4-x)=27

3^(4-x) = 3^3

4-x = 3

x = 1

d) \(​​\dfrac{5x+2}{6}\) +\(\dfrac{3-4x}{2}\) = 2-\(\dfrac{x+7}{3}\) 

=>5x+2+3(3-4x)=12-2(x+7)

5x+2+9-12x=12-2x-14

-5x=-13

x=\(\dfrac{13}{5}\) 

e) \(\dfrac{-20}{9}x +4=\dfrac{8}{3}x-40\)

=>-20x+36=24x-360

-44x=-396

x=9

f) 3x(2x-5)-4X+10=0

6X² -15X-4X+10=0

2x(3x-2)-5(3x-2)=0

(3x-2)(2x-5)=0

\(\left[\begin{array}{} Biểu thức (3x-2=0)\\ Biểu thức (2x-5=0) \end{array} \right.\)\(\left[\begin{array}{} (x=\dfrac{2}{3})\\ (x=\dfrac{5}{2}) \end{array} \right.\)

j) \(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)

\(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)

\(\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)

\(\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)

(x-100)(\(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\))=0

=> x-100=0(\(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\) >0)

=> x= 100

d) \(y=4sinx-2cos2x-1\)

\(=4sinx-2\left(1-2sin^2x\right)-1\)

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

Đặt \(t=sinx,t\in\left[-1;1\right]\)

\(y=f\left(t\right)=4t^2+4t-3\) \(\Leftrightarrow f'\left(t\right)=8t+4\)

\(f'\left(t\right)=0\Leftrightarrow t=-\dfrac{1}{2}\)

Vẽ BBT với \(t\in\left[-1;1\right]\) ta được 

\(minf\left(t\right)=miny=-4\Leftrightarrow t=-\dfrac{1}{2}\)\(\Leftrightarrow sinx=-\dfrac{1}{2}\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) ( k thuộc Z)

\(maxf\left(t\right)=miny=5\Leftrightarrow t=1\)\(\Leftrightarrow sinx=1\) \(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\) ( k thuộc Z)

Vậy...

e) \(y=3sin2x+8cos^2x-1\)

\(=3sin2x+4\left(2cos^2x-1\right)+3\)

\(=3sin2x+4cos2x+3\)

\(=5\left(\dfrac{3}{5}sin2x+\dfrac{4}{5}cos2x\right)+3\)

Đặt \(cosu=\dfrac{3}{5}\Leftrightarrow sinu=\dfrac{4}{5}\)

\(y=5\left(sin2x.cosu+cos2x.sinu\right)+3=5.sin\left(2x+u\right)+3\)

Có \(-1\le sin\left(2x+u\right)\le1\) \(\Leftrightarrow-2\le y\le8\)

\(maxy=8\Leftrightarrow sin\left(2x+u\right)=1\) \(\Leftrightarrow2x+u=\dfrac{\pi}{2}+k2\pi\) \(\Leftrightarrow x=-\dfrac{u}{2}+\dfrac{\pi}{4}+k\pi\)\(\Leftrightarrow x=-\dfrac{1}{2}.arccos\dfrac{3}{5}+\dfrac{\pi}{4}+k\pi\) ( k thuộc Z)

\(miny=-2\Leftrightarrow sin\left(2x+u\right)=-1\)\(\Leftrightarrow x=-\dfrac{1}{2}.\dfrac{arccos3}{5}-\dfrac{\pi}{4}+k\pi\) ( k thuộc Z)

Vậy...

f)\(y=4+sin^4x+cos^4x\)

\(=4+\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)

\(=4+1-\dfrac{1}{2}\left(2sinx.cosx\right)^2\)

\(=5-\dfrac{1}{2}.\left(sin2x\right)^2\)

\(\left(sin2x\right)^2\in\left[0;1\right]\Leftrightarrow y\in\left[\dfrac{9}{2};\dfrac{11}{2}\right]\)

\(maxy=\dfrac{11}{2}\Leftrightarrow sin2x=0\Leftrightarrow2x=k\pi\Leftrightarrow x=\dfrac{k\pi}{2}\) ( k thuộc Z )

\(miny=\dfrac{9}{2}\Leftrightarrow\left(sin2x\right)^2=1\)\(\Leftrightarrow cos2x=0\)\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\) ( k thuộc Z )

Vậy...

a) \(N\left(x\right)=9-x^3+4x^3-7x+3x^2+x^2\)

\(N\left(x\right)=-\left(x^3-4x^3\right)+\left(3x^2+x^2\right)-7x+9\)

\(N\left(x\right)=3x^3+4x^2-7x+9\)

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

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

\(M\left(x\right)=3x^3+4x^2+3x+4\)

b) \(P\left(x\right)=N\left(x\right)-M\left(x\right)\)

\(P\left(x\right)=\left(3x^3+4x^2-7x+9\right)-\left(3x^2+4x^2+3x+4\right)\)

\(P\left(x\right)=3x^3+4x^2-7x+9-3x^3-4x^2-3x-4\)

\(P\left(x\right)=-10x+5\)

\(Q\left(x\right)=N\left(x\right)+M\left(x\right)\)

\(Q\left(x\right)=\left(3x^3+4x^2-7x+9\right)+\left(3x^3+4x^2+3x+4\right)\)

\(Q\left(x\right)=3x^3+4x^2-7x+9+3x^3+4x^2+3x+4\)

\(Q\left(x\right)=6x^3+8x^2-4x+13\)

c) Nghiệm của đa thức \(P\left(x\right)\)

\(P\left(x\right)=-10x+5=0\)

\(\Rightarrow-10x=-5\)

\(\Rightarrow10x=5\)

\(\Rightarrow x=\dfrac{5}{10}=\dfrac{1}{2}\)

Nghiệm của đa thức \(Q\left(x\right)\)

Vì: \(Q\left(x\right)=6x^3+8x^2+4x+13\ge0\)

\(\Rightarrow Q\left(x\right)\ge0\)

Vậy đa thức vô nghiệm

d) \(Q\left(x\right)\left(1-2x\right)\)

\(=\left(6x^3+8x^2-4x+13\right)\left(1-2x\right)\)

\(=6x^3+8x^2-4x+13-12x^4-16x^3+8x^2-26x\)

\(=-12x^4-10x^3+16x^2-30x+13\)

no

c) 108(12 + 13 ) + 25 . 92

= 2700 + 2300

= 5000

d) 2.169.12 - 3.68.8 - 24 

= ( 2 .12 ) . 169 - (3.8) . 68 - 24

= 24 . 169 - 24 . 68 - 24

= 24(169 - 68 ) - 24

= 2424 - 24 = 2400

e) 2.56.24 - 3.36.16 + 4.12.95 + 6.3.8.5

= ( 2 . 24 ) . 56 - ( 3.16 ) . 36 + (4.12) .95 + (6.8) . 3 . 5

= 48 . 56 - 48 . 36 + 48 . 95 + 48 . 15

= 48(56 - 36 + 95 + 15 )

= 6240

c)   108.12+25.92+13.108

=(108+12)+(25.13)+(108+92)

=120+325+200

=645

a: \(A=\dfrac{x\left(x+2\right)}{\left(x-2\right)^2}:\dfrac{x^2-4+x+6-x^2}{x\left(x-2\right)}\)

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

c: A<0

=>x-2<0

=>x<2

d: B nguyên

=>x^2-4+4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;6}