giải pt
a) 3x lx + 1l – 2x lx + 2l = 12
b) $\frac{x^2 – 4 – | x + 1|}{2}$ = x( x + 1 )
c) $\frac{7}{8x}$ – $\frac{x – 1}{2x ( x – 2)}$ – $\frac{1}{x – 16}$ = $\frac{x – 5}{4x ^2 – 8x}$
d) $\frac{x^2 – x}{x + 3}$ – $\frac{x^2}{x- 3}$ = $\frac{7x^2 – 3x}{9 – x^2}$
Đáp án:
$\begin{array}{l}
a)3x\left| {x + 1} \right| – 2x\left| {x + 2} \right| = 12\\
+ Khi:x \ge – 1\\
\Leftrightarrow 3x\left( {x + 1} \right) – 2x\left( {x + 2} \right) = 12\\
\Leftrightarrow 3{x^2} + 3x – 2{x^2} – 4x – 12 = 0\\
\Leftrightarrow {x^2} – x – 12 = 0\\
\Leftrightarrow \left( {x – 4} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow x = 4\left( {do:x \ge – 1} \right)\\
+ Khi: – 2 \le x < – 1\\
3x\left( { – x – 1} \right) – 2x\left( {x + 2} \right) = 12\\
\Leftrightarrow – 3{x^2} – 3x – 2{x^2} – 4x = 12\\
\Leftrightarrow 5{x^2} + 7x + 12 = 0\left( {vn} \right)\\
+ Khi:x < – 2\\
\Leftrightarrow 3x\left( { – x – 1} \right) + 2x\left( {x + 2} \right) = 12\\
\Leftrightarrow – 3{x^2} – 3x + 2{x^2} + 4x = 12\\
\Leftrightarrow {x^2} – x + 12 = 0\left( {vn} \right)\\
\text{Vậy}\,x = 4\\
b)\dfrac{{{x^2} – 4 – \left| {x + 1} \right|}}{2} = x\left( {x + 1} \right)\\
\Leftrightarrow {x^2} – 4 – \left| {x + 1} \right| = 2x\left( {x + 1} \right)\\
+ Khi:x \ge – 1\\
\Leftrightarrow {x^2} – 4 – x – 1 = 2{x^2} + 2x\\
\Leftrightarrow {x^2} + 3x + 5 = 0\left( {vn} \right)\\
+ Khi:x < – 1\\
\Leftrightarrow {x^2} – 4 + x + 1 = 2{x^2} + 2x\\
\Leftrightarrow {x^2} + x + 3 = 0\left( {vn} \right)\\
\text{Vậy phương trình vô nghiệm}\\
c)Dkxd:x\# 0;x\# 2;x\# 16\\
\dfrac{7}{{8x}} – \dfrac{{x – 1}}{{2x\left( {x – 2} \right)}} – \dfrac{1}{{x – 16}} = \dfrac{{x – 5}}{{4{x^2} – 8x}}\\
\Leftrightarrow \dfrac{{7\left( {x – 2} \right) – 4\left( {x – 1} \right)}}{{8x\left( {x – 2} \right)}} – \dfrac{{x – 5}}{{4x\left( {x – 2} \right)}} = \dfrac{1}{{x – 16}}\\
\Leftrightarrow \dfrac{{3x – 10 – 2x + 10}}{{8x\left( {x – 2} \right)}} = \dfrac{1}{{x – 16}}\\
\Leftrightarrow \dfrac{x}{{8x\left( {x – 2} \right)}} = \dfrac{1}{{x – 16}}\\
\Leftrightarrow \dfrac{1}{{8\left( {x – 2} \right)}} = \dfrac{1}{{x – 16}}\\
\Leftrightarrow 8x – 16 = x – 16\\
\Leftrightarrow x = 0\left( {ktm} \right)\\
\text{Vậy phương trình vô nghiệm}\\
d)Dkxd:x\# 3;x\# – 3\\
\dfrac{{{x^2} – x}}{{x + 3}} – \dfrac{{{x^2}}}{{x – 3}} = \dfrac{{7{x^2} – 3x}}{{9 – {x^2}}}\\
\Leftrightarrow \dfrac{{\left( {{x^2} – x} \right)\left( {x – 3} \right) – {x^2}\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x – 3} \right)}}\\
= \dfrac{{ – 7{x^2} + 3x}}{{\left( {x + 3} \right)\left( {x – 3} \right)}}\\
\Leftrightarrow {x^3} – 4{x^2} + 3x – {x^3} – 3{x^2} = – 7{x^2} + 3x\\
\Leftrightarrow 0 = 0\left( {tm} \right)
\end{array}$
Vậy pt nghiệm đúng với mọi x#3; x#-3