Giúp mik với! Please!!!

Đáp án:

 B1:

d) x=4

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
a)\dfrac{{4\left( {x - 2} \right) + 2\left( {2 - x} \right)}}{{{x^2} + 1}} = 0\\
 \to 4\left( {x - 2} \right) - 2\left( {x - 2} \right) = 0\left( {do:{x^2} + 1 > 0\forall x} \right)\\
 \to 2\left( {x - 2} \right) = 0\\
 \to x = 2\\
c)DK:x \ne  - 5\\
\dfrac{{2x - 5}}{{x + 5}} = 3\\
 \to 2x - 5 = 3x + 15\\
 \to x =  - 20\\
b)DK:x \ne  - 1\\
\dfrac{{{{\left( {x + 1} \right)}^2}}}{{x + 1}} = 0\\
 \to x + 1 = 0\\
 \to x =  - 1\left( l \right)\\
 \to x \in \emptyset \\
d)DK:x \ne 2\\
\dfrac{{4 - 2x + 4}}{{x - 2}} = 0\\
 \to 8 - 2x = 0\\
 \to x = 4\\
B2:\\
a)DK:x \ne 2\\
\dfrac{{{x^2} + 6x - 16}}{{x - 2}} = x + 8\\
 \to {x^2} + 6x - 16 = \left( {x + 8} \right)\left( {x - 2} \right)\\
 \to {x^2} + 6x - 16 = {x^2} + 6x - 16\\
 \to 0x = 0\left( {ld} \right)\\
KL:\forall x \ne 2\\
c)DK:x \ne  - 17\\
\dfrac{{{x^2} - 15x + 1}}{{x + 17}} = x - 2\\
 \to {x^2} - 15x + 1 = \left( {x + 17} \right)\left( {x - 2} \right)\\
 \to {x^2} - 15x + 1 = {x^2} + 15x - 34\\
 \to 1 =  - 34\left( l \right)\\
 \to x \in \emptyset \\
b)DK:x \ne 2\\
\dfrac{{3x\left( {x - 2} \right) - 1}}{{x - 2}} =  - \dfrac{{x - 1}}{{x - 2}}\\
 \to 3{x^2} - 6x - 1 + x - 1 = 0\\
 \to 3{x^2} - 5x - 2 = 0\\
 \to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x =  - \dfrac{1}{3}
\end{array} \right.\\
d)DK:x \ne 2\\
\dfrac{{x - 1 - 3\left( {x - 2} \right) + x\left( {x - 2} \right) - 1}}{{x - 2}} = 0\\
 \to x - 2 - 3x + 6 + {x^2} - 2x = 0\\
 \to {x^2} - 4x + 4 = 0\\
 \to {\left( {x - 2} \right)^2} = 0\\
 \to x = 2\left( l \right)\\
 \to x \in \emptyset 
\end{array}\)