1C

2A

1C        2A

a: ĐKXĐ: x∉{0;2;-2}

\(B=\left(\frac{x^3}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{2+x}\right):\left(x+2+\frac{10-x^2}{x-2}\right)\)

\(=\left(\frac{x^2}{\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\frac{\left(x+2\right)\left(x-2\right)+10-x^2}{x-2}\)

\(=\left(\frac{x^2}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right)\cdot\frac{x-2}{x^2-4+10-x^2}\)

\(=\frac{x^2-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{6}=\frac{x^2-2x-4+x-2}{\left(x+2\right)\cdot6}=\frac{x^2-x-6}{\left(x+2\right)\cdot6}=\frac{\left(x-3\right)\left(x+2\right)}{6\left(x+2\right)}=\frac{x-3}{6}\)

b: \(x^2-5x+6=0\)

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

=>x=2(loại) hoặc x=3(nhận)

Thay x=3 vào B, ta được:

\(B=\frac{3-3}{6}=0\)

c: Để B là số nguyên thì x-3⋮6

=>x-3=6k(k∈Z)

=>x=6k+3(k∈Z)

d: |B|>1

=>B>1 hoặc B<-1

TH1: B>1

=>B-1>0

=>\(\frac{x-3}{6}-1>0\)

=>\(\frac{x-9}{6}>0\)

=>x-9>0

=>x>9

TH2: B<-1

=>\(\frac{x-3}{6}<-1\)

=>x-3<-6

=>x<-3

bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

\(1,\sqrt{3}x-3=\sqrt{27}\)

\(\Leftrightarrow\sqrt{3}x-3=3\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}\left(x-\sqrt{3}\right)=3\sqrt{3}\)

\(\Leftrightarrow x-\sqrt{3}=3\)

\(\Leftrightarrow x=3+\sqrt{3}\)

\(2,\sqrt{2}x-\sqrt{28}=\sqrt{32}\)

\(\Leftrightarrow\sqrt{2}x-2\sqrt{7}=4\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}x=4\sqrt{2}+2\sqrt{7}\)

\(\Leftrightarrow x=\dfrac{\sqrt{2^2}\left(2\sqrt{2}+\sqrt{7}\right)}{\sqrt{2}}\)

\(\Leftrightarrow x=\sqrt{2}\left(2\sqrt{2}+\sqrt{7}\right)\)

\(\Leftrightarrow x=4+\sqrt{14}\)

\(3,\sqrt{6}x-2\sqrt{6}=\sqrt{54}\)

\(\Leftrightarrow\sqrt{6}\left(x-2\right)=3\sqrt{6}\)

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=5\)

\(4,\sqrt{3}x-\sqrt{2}x=\sqrt{3}+\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{3}+\sqrt{2}\)

\(\Leftrightarrow x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)

\(\Leftrightarrow x^2-36-x^2+12x-9=9\)

\(\Leftrightarrow12x=54\)

hay x=9/2

Ta có

( x   –   6 ) ( x   +   6 )   –   ( x   +   3 ) 2   =   9     ⇔   x 2   –   36   –   ( x 2   +   6 x   +   9 )   =   9     ⇔   x 2   –   36   –   x 2   –   6 x   –   9   –   9   =   0

ó - 6x – 54 = 0 ó 6x = -54 ó x = -9

Vậy x = -9

Đáp án cần chọn là: A

3:

a: 3^x*3=243

=>3^x=81

=>x=4

b; 2^x*16^2=1024

=>2^x=4

=>x=2

c: 64*4^x=16^8

=>4^x=4^16/4^3=4^13

=>x=13

d: 2^x=16

=>2^x=2^4

=>x=4

mình cần gấp

a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)

\(\Leftrightarrow2x^2-2x-2x^2=-6\)

\(\Leftrightarrow x=3\)

b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)