ĐKXĐ: \(x\ne-2\).

\(PT\Leftrightarrow\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{x+3}{x+2}\)

\(\Rightarrow12=\left(x+3\right)\left(x^2-2x+4\right)\)

\(\Leftrightarrow x^3+x^2-2x+12=12\)

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

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

Vậy x = 0; x = 1.

Ở B(x) thiếu hạng tử 2x^2 kìa

a) \(A\left(x\right)=3x^5-x^4-2x^3-2x^2+3x\)

\(B\left(x\right)=-x^5+5x^4+2x^3+2x^2-9\)

b) \(A\left(x\right)+B\left(x\right)=\left(3x^5-x^4-2x^3-2x^2+3x\right)+\left(-x^5+5x^4+2x^3+2x^2-9\right)=2x^5+4x^4+3x-9\)

\(A\left(x\right)-B\left(x\right)=\left(3x^5-x^4-2x^3-2x^2+3x\right)-\left(-x^5+5x^4+2x^3+2x^2-9\right)=4x^5-6x^4-4x^3-4x^2+3x+9\)

ĐKXĐ: \(x\ne-2;x\ne2\).

Ta có \(PT\Leftrightarrow\dfrac{12}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow2x=4\Leftrightarrow x=2\) (loại)

Vậy PT đã cho vô nghiệm

a) A = \(\left(x^2y-2x^2y\right)+3xy-2x^3+5y^4=3xy-x^2y-2x^3+5y^4\).

b) Với x = 1; y = 2 ta có \(A=3.1.2-1^2.2-2.1^3+5.2^4=82\)

\(a^2+9b^2=10ab\Leftrightarrow\left(a-b\right)\left(a-9b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=9b\end{matrix}\right.\).

+) a = b: 

\(2log_2\left(a+3b\right)-4=log_2a+log_2b\Leftrightarrow2log_24a-4=2log_2a\Leftrightarrow2\left(log_2a+2\right)-4=2log_2a\Leftrightarrow4-4=0\) (luôn đúng). Chọn D.

+) a = 9b: \(2log_2\left(a+3b\right)-4=log_2a+log_2b\Leftrightarrow2log_212b-4=log_29b+log_2b\Leftrightarrow2\left(log_2b+log_23+2\right)-4=2log_2b+log_29\Leftrightarrow2log_23=log_29\) (luôn đúng). Chọn D.

Vậy ...

ĐKXĐ: \(x\ne\pm1\).

\(PT\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow\left(x+1\right)^2+\left(x-1\right)^2=16\)

\(\Leftrightarrow2x^2=14\Leftrightarrow x=\pm\sqrt{7}\). (TMĐK)

Vậy...

\(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)

\(\Leftrightarrow2x+12=\left|x+21\right|\) (*)

Do đó 2x + 12 \(\ge0\Leftrightarrow x\ge-6\).

Khi đó (*) \(\Leftrightarrow\left[{}\begin{matrix}x+21=2x+12\\x+21=-\left(2x+12\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\left(TM\right)\\x=-11\left(\text{\left\{loại\right\}}\right)\end{matrix}\right.\)

\(\left|3x-2\right|\le2\Leftrightarrow-2\le3x-2\le2\Leftrightarrow0\le x\le\dfrac{4}{3}\).

Chọn D

Áp dụng bất đẳng thức AM - GM ta có:

\(f\left(x\right)=x+\left(2x+\dfrac{1}{2x}\right)\ge1+2\sqrt{2x.\dfrac{1}{2x}}=3\).

Dấu "=" xảy ra khi và chỉ khi x = 1.

Vậy...