bài gì thế bn

Mih có đăng lên r đó !

1) ĐKXĐ: \(x\ge0\)

2) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge4\)

4) ĐKXĐ: \(x>16\)

5) ĐKXĐ: \(\left[{}\begin{matrix}x\le-2\\x\ge0\end{matrix}\right.\)

6) ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\x\ge4\end{matrix}\right.\)

7) ĐKXĐ: \(\left[{}\begin{matrix}1\le x\\x< 3\end{matrix}\right.\)

8) ĐKXĐ: \(\left[{}\begin{matrix}x\le-2\\x>3\end{matrix}\right.\)

9) ĐKXĐ: \(x\in R\)

10) ĐKXĐ: \(x\in R\)

11) ĐKXĐ: \(x\in R\)

12) ĐKXĐ: \(x\in R\)

13) ĐKXĐ: \(x\in R\)

14) ĐKXĐ: \(x\in R\)

15) ĐKXĐ: \(x\in R\)

16) ĐKXĐ: \(x\ne-\dfrac{1}{2}\)

17) ĐKXĐ: \(x\ge7\)

18) ĐKXĐ: \(x\ge-5\)

\(1,\Leftrightarrow x^2-8x+16-x^2+x+12=7\\ \Leftrightarrow-7x=-21\\ \Leftrightarrow x=3\\ 2,\Leftrightarrow\left(x-4\right)^2-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

IV: Để M nguyên thì \(3x^3-2x^2-6x+5\) ⋮3x-2

=>\(x^2\left(3x-2\right)-6x+4+1\) ⋮3x-2

=>1⋮3x-2

=>3x-2∈{1;-1}

=>3x∈{3;1}

=>x∈{1;1/3}

mà x nguyên

nên x=1

III:

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

=>(x-4)(x-4-x-3)=7

=>-7(x-4)=7

=>x-4=-1

=>x=3

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

=>\(\left(x-4\right)^2-\left(x-4\right)=0\)

=>(x-4)(x-4-1)=0

=>(x-4)(x-5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=5\end{array}\right.\)

a) \(\Leftrightarrow x^2+10x+25-x^2+8x-15=-8\\ \Leftrightarrow18x=-18\\ \Leftrightarrow x=-1\)

b) \(\Leftrightarrow\left(2x+1\right)^2-3\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(2x+1-3\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)

\(1,\\ a,\dfrac{8x}{2xy}=\dfrac{4x}{y}\\ b,\dfrac{2xy}{6y}=\dfrac{x}{3}\\ c,\dfrac{3\left(x+2\right)}{2x}=\dfrac{6\left(x+2\right)}{4x}\\ d,\dfrac{4\left(x-2\right)}{3\left(x+1\right)}=\dfrac{8\left(x-2\right)x}{6\left(x+1\right)x}\\ 2,\\ \dfrac{x^2+3x+2}{x^2+x}=\dfrac{x^2+x+2x+2}{x\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)}=\dfrac{x+2}{x}\\ 3,\\ \dfrac{x^2-3x}{x^2-9}=\dfrac{x}{x+3}\)

Bài 3: 

Ta có: \(x^2-2x+4=\left(x-1\right)^2+3\ge3\forall x\)

\(\Leftrightarrow P=\dfrac{15}{x^2-2x+4}=\dfrac{15}{\left(x-1\right)^2+3}\le5\forall x\)

Dấu '=' xảy ra khi x=1

1B

2B

3C

4A

5A

6A

7A

8B

B: 

Bài 1:

a: =>6x=12

=>x=2

b: =>(5x+6)/(x+2)=1

=>5x+6=x+2

=>4x=-4

=>x=-1

c: =>4x(3-2x)-5(3-2x)=0

=>(3-2x)(4x-5)=0

=>x=5/4 hoặc x=3/2