\(x^4-10x^3+35x^2+24>0\)

\(\Leftrightarrow x^4-2.5.x^3+\left(5x\right)^2+10x^2+24>0\)

\(\Leftrightarrow\left(x^2-5x\right)^2+10x^2+24>0\)

\(\Leftrightarrow x^2\left(x-5\right)^2+10x^2+24>0\)(luôn đúng)

Vậy nghiệm của bất phương trình \(x\in R\)

Ta có 27^5=3^3^5=3^15
243^3=3^5^3=3^15
Vậy A=B
2^300=2^(3.100)=2^3^100=8^100
3^200=3^(2.100)=3^2^100=9^100
Vậy A<B

Câu 1: 

a) Ta có: 7x+21=0

\(\Leftrightarrow7x=-21\)

hay x=-3

Vậy: S={-3}

b) Ta có: 3x-2=2x-3

\(\Leftrightarrow3x-2-2x+3=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}

c) Ta có: 5x-2x-24=0

\(\Leftrightarrow3x=24\)

hay x=8

Vậy: S={8}

Câu 2: 

a) Ta có: \(\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{2};1\right\}\)

b) Ta có: \(\left(2x-3\right)\left(-x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=7\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};7\right\}\)

c) Ta có: \(\left(x+3\right)^3-9\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-9\right]=0\)

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

\(\Leftrightarrow x\left(x+3\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-6\end{matrix}\right.\)

Vậy: S={0;-3;-6}

b: ĐKXD: x<>1/5; x<>3

PT\(\Leftrightarrow\dfrac{3}{5x-1}-\dfrac{2}{x-3}=\dfrac{-4}{\left(5x-1\right)\left(x-3\right)}\)

=>3x-9-10x+2=-4

=>-7x-7=-4

=>-7x=3

=>x=-3/7

a: ĐKXĐ: x<>2/3; x<>-2/3

\(PT\Leftrightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x\)

=>9x^2+12x+4-18x+12-9x=0

=>9x^2-15x+16=0

=>\(x\in\varnothing\)

c: ĐKXĐ: x<>1/4; x<>-1/4

PT =>-3(4x+1)=2(4x-1)-6x-8

=>-12x-3=8x-2-6x-8

=>-12x-3=2x-10

=>-14x=-7

=>x=1/2

d: ĐKXĐ: x<>0; x<>2

\(\Leftrightarrow\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)

=>2(5-x)+7(x-2)=4(x-1)+x

=>10-2x+7x-14=4x-4+x

=>5x-4=5x-4

=>0x=0(luôn đung)

Vậy: S=R\{0;2}

e: DKXĐ: x<>0

PT \(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

=>x(x^3+1-x^3+1)=3

=>2x=3

=>x=3/2

\(a,\dfrac{y-1}{y-2}-\dfrac{5}{y+2}=\dfrac{12}{y^2-4}+1\left(ĐKXĐ:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{\left(y-1\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}-\dfrac{5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}-\dfrac{12}{\left(y-2\right)\left(y+2\right)}-\dfrac{\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}=0\)

\(\Leftrightarrow\dfrac{y^2+y-2}{\left(y-2\right)\left(y+2\right)}-\dfrac{5y-10}{\left(y-2\right)\left(y+2\right)}-\dfrac{12}{\left(y-2\right)\left(y+2\right)}-\dfrac{y^2-4}{\left(y-2\right)\left(y+2\right)}=0\)

\(\Leftrightarrow\dfrac{y^2+y-2-5y+10-12-y^2+4}{\left(y-2\right)\left(y+2\right)}=0\)

\(\Rightarrow-4y=0\)

\(\Leftrightarrow y=0\left(tm\right)\)

\(b,\dfrac{1}{4z^2-12z+9}-\dfrac{3}{9-4z^2}=\dfrac{4}{4z^2+12z+9}\left(ĐKXĐ:z\ne\pm\dfrac{3}{2}\right)\)

\(\Leftrightarrow\dfrac{1}{\left(2z-3\right)^2}+\dfrac{3}{\left(2z-3\right)\left(2z+3\right)}-\dfrac{4}{\left(2z+3\right)^2}=0\)

\(⇔\dfrac{\left(2z+3\right)^2}{\left(2z-3\right)^2\left(2z+3\right)^2}+\dfrac{3\left(2z-3\right)\left(2z+3\right)}{\left(2z-3\right)^2\left(2z+3\right)^2}-\dfrac{4\left(2z-3\right)^2}{\left(2z-3\right)^2\left(2z+3\right)^2}=0\)

\(\Leftrightarrow\dfrac{4z^2+12z+9}{\left(2z-3\right)^2\left(2z+3\right)^2}+\dfrac{12z^2-27}{\left(2z-3\right)^2\left(2z+3\right)^2}-\dfrac{16z^2-48z+36}{\left(2z-3\right)^2\left(2z+3\right)^2}=0\)

\(\Leftrightarrow\dfrac{4z^2+12z+9+12z^2-27-16z^2+48z-36}{\left(2z-3\right)^2\left(2z+3\right)^2}=0\)

\(\Rightarrow60z-54=0\)

\(\Leftrightarrow60z=54\)

\(\Leftrightarrow z=\dfrac{9}{10}\left(tm\right).\)

\(a,\dfrac{y-1}{y-2}-\dfrac{5}{y+2}=\dfrac{12}{y^2-4}+1\left(dkxd:y\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{\left(y-1\right)\left(y+2\right)-5\left(y-2\right)-12-y^2+4}{y^2-4}=0\)
\(\Leftrightarrow y^2+2y-y-2-5y+10-12-y^2+4=0\)

\(\Leftrightarrow-4y=0\)

\(\Leftrightarrow y=0\left(tmdk\right)\)

Vậy \(S=\left\{0\right\}\)

\(b,\dfrac{1}{4z^2-12z+9}-\dfrac{3}{9-4z^2}=\dfrac{4}{4z^2+12z+9}\)

\(\Leftrightarrow\dfrac{1}{\left(2z-3\right)^2}-\dfrac{3}{\left(2z-3\right)\left(2z+3\right)}=\dfrac{4}{\left(2z+3\right)^2}\left(dkxd:z\ne\pm\dfrac{3}{2}\right)\)

\(\Leftrightarrow\left(2z+3\right)^2-3\left(4z^2-9\right)-4\left(2z-3\right)^2=0\)

\(\Leftrightarrow4z^2+12z+9-12z^2+27-4\left(4z^2-12z+9\right)=0\)

\(\Leftrightarrow4z^2+12z+9-12z^2+27-16z^2+48z-36=0\)

\(\Leftrightarrow-24z^2+60z=0\)

\(\Leftrightarrow-12z\left(2z-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-12z=0\\2z-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}z=0\left(tmdk\right)\\z=\dfrac{5}{2}\left(tmdk\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;\dfrac{5}{2}\right\}\)

Em tự biểu diễn nha.

\(a,5x+15>0\\ \Leftrightarrow5x>-15\\ \Leftrightarrow x>-3\)

\(b,-4x+1>17\\ \Leftrightarrow-4x>16\\ \Leftrightarrow x< -4\)

\(c,-5x+10< 0\\ \Leftrightarrow-5x< -10\Leftrightarrow x>2\)

1: \(=\dfrac{-\left[\left(x+5\right)^2-9\right]}{\left(x+2\right)^2}=\dfrac{-\left(x+5-3\right)\left(x+5+3\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+8\right)}{x+2}\)

2: \(=\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\dfrac{2x}{x+4}\)

3: \(=\dfrac{5x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{5x}{x^2-1}\)

4: \(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)

5: \(=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(c+d\right)\left(a-b\right)}=\dfrac{2a}{c+d}\)

6: \(=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\cdot\left(-1\right)=\dfrac{-x}{x+y}\)

7: \(=\dfrac{2\left(1-a\right)}{-\left(1-a^3\right)}=\dfrac{-2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=-\dfrac{2}{1+a+a^2}\)

8: \(=\dfrac{x^4\left(x^3-1\right)}{\left(x^3-1\right)\left(x^3+1\right)}=\dfrac{x^4}{x^3+1}\)

9: \(=\dfrac{\left(x+2-x+2\right)\left(x+2+x-2\right)}{16x}=\dfrac{4\cdot2x}{16x}=\dfrac{1}{2}\)

10: \(=\dfrac{0.5\left(49x^2-y^2\right)}{0.5x\left(7x-y\right)}=\dfrac{1}{x}\cdot\dfrac{\left(7x-y\right)\left(7x+y\right)}{7x-y}\)

\(=\dfrac{7x+y}{x}\)

ĐKXĐ:\(x\ne-2\)

\(\dfrac{1}{x+2}-1=\dfrac{5x+7}{x+2}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{5x+7}{x+2}=1\\ \Leftrightarrow\dfrac{1-5x-7}{x+2}=1\\ \Leftrightarrow-5x-6=x+2\\ \Leftrightarrow x+2+5x+6=0\\ \Leftrightarrow6x+8=0\\ \Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)

 thank

Để \(2x^3-4x^2+6x+a⋮x+2\)

\(\Leftrightarrow2x^3-4x^2+6x+a=\left(x+2\right)\cdot a\left(x\right)\)

Thay \(x=-2\)

\(\Leftrightarrow2\left(-2\right)^3-4\left(-2\right)^2+6\left(-2\right)+a=0\\ \Leftrightarrow-16-16-12+a=0\\ \Leftrightarrow-44+a=0\Leftrightarrow a=44\)