Toán Lớp 7: Mọi người giúp em với ạ !!!!!!!!!!
Tìm x , biết :
a, l 2x + 3 l = x +2
b, l 2x + 3 l – 2.l 4-x l = 5
c, l 2x – 3 l – x = l 2 – x l
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Toán Lớp 7: Mọi người giúp em với ạ !!!!!!!!!!
Tìm x , biết :
a, l 2x + 3 l = x +2
b, l 2x + 3 l – 2.l 4-x l = 5
c, l 2x – 3 l – x = l 2 – x l
Comments ( 1 )
a)Dkxd:x \ge – 2\\
\left| {2x + 3} \right| = x + 2\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 3 = x + 2\\
2x + 3 = – x – 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\left( {tm} \right)\\
3x = – 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = – \dfrac{5}{3}\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = – 1;x = – \dfrac{5}{3}\\
b)\left| {2x + 3} \right| – 2\left| {4 – x} \right| = 5\\
\Leftrightarrow \left| {2x + 3} \right| – 2\left| {x – 4} \right| = 5\\
+ Khi:x \ge 4\\
\Leftrightarrow 2x + 3 – 2\left( {x – 4} \right) = 5\\
\Leftrightarrow 2x + 3 – 2x + 8 = 5\\
\Leftrightarrow 11 = 5\left( {ktm} \right)\\
+ Khi: – \dfrac{3}{2} \le x < 4\\
\Leftrightarrow 2x + 3 – 2\left( {4 – x} \right) = 5\\
\Leftrightarrow 2x + 3 – 8 + 2x = 5\\
\Leftrightarrow 4x = 10\\
\Leftrightarrow x = \dfrac{5}{2}\left( {tm} \right)\\
+ Khi:x < – \dfrac{3}{2}\\
\Leftrightarrow – 2x – 3 – 2\left( {4 – x} \right) = 5\\
\Leftrightarrow – 2x – 3 – 8 + 2x = 5\\
\Leftrightarrow – 11 = 5\left( {ktm} \right)\\
Vậy\,x = \dfrac{5}{2}\\
c)\left| {2x – 3} \right| – x = \left| {2 – x} \right|\\
\Leftrightarrow \left| {2x – 3} \right| – x = \left| {x – 2} \right|\\
+ Khi:x \ge 2\\
\Leftrightarrow 2x – 3 – x = x – 2\\
\Leftrightarrow x – 3 = x – 2\\
\Leftrightarrow – 3 = – 2\left( {ktm} \right)\\
+ Khi:\dfrac{3}{2} \le x < 2\\
\Leftrightarrow 2x – 3 – x = 2 – x\\
\Leftrightarrow 2x = 5\\
\Leftrightarrow x = \dfrac{5}{2}\left( {ktm} \right)\\
+ Khi:x < \dfrac{3}{2}\\
\Leftrightarrow 3 – 2x – x = 2 – x\\
\Leftrightarrow 2x = 1\\
\Leftrightarrow x = \dfrac{1}{2}\left( {tm} \right)\\
Vậy\,x = \dfrac{1}{2}
\end{array}$