a, 2x-5=-x-17 b, 4-|2x-1|=-11 c, (2x-3)^2-17=2^3 ( “^”=mũ ) 06/07/2021 Bởi Kennedy a, 2x-5=-x-17 b, 4-|2x-1|=-11 c, (2x-3)^2-17=2^3 ( “^”=mũ )
Đáp án: a) x=-4 b) x=-7 hoặc x=8 c) \(\left[ \begin{array}{l}x=4\\x=-1\end{array} \right.\) Giải thích các bước giải: a) 2x-5=-x-17 ⇔ 3x=-12 ⇔ x=-4. Vậy x=-4. b) Ta có: |2x-1|=2x-1 khi x$\geq$ $\frac{1}{2}$ |2x-1|=1-2x khi x<$\frac{1}{2}$ +) Với x$\geq$ $\frac{1}{2}$ , ta có: 4-(2x-1)+11=0 ⇔ 2x=16 ⇔ x=8 +) Với x<$\frac{1}{2}$ , ta có: 4-(1-2x)+11=0 ⇔ 2x=-14 ⇔ x=-7 Vậy x=8 hoặc x=-7. c) (2x-3)²-17=8 ⇔ 4x²-12x+9-25=0 ⇔ 4x²-12x-16=0 ⇔ (4x²-16x)+(4x-16)=0 ⇔ 4x(x-4)+4(x-4)=0 ⇔ (x-4)(4x+4)=0 ⇔ \(\left[ \begin{array}{l}x-4=0\\4x+4=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=4\\x=-1\end{array} \right.\) . Vậy \(\left[ \begin{array}{l}x=4\\x=-1\end{array} \right.\) . Bình luận
a, $2x – 5 = -x – 17$ $\Leftrightarrow 2x + x = -17 + 5$ $\Leftrightarrow 3x = -12$ $\Leftrightarrow x = (-12) : 3$ $\Leftrightarrow x = -4$ b, $4 – \left | 2x – 1 \right | = -11$ $\Leftrightarrow \left | 2x – 1 \right | = 4 + 11$ $\Leftrightarrow \left | 2x – 1 \right | = 15$ $\Leftrightarrow \left[ \begin{array}{l}2x+1=15\\2x+1=-15\end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l}x=7\\x=-8\end{array} \right.$ c) $\left ( 2x – 3 \right )^{2} – 17 = 2^{3}$ $\Leftrightarrow \left ( 2x – 3 \right )^{2} – 17 = 8$ $\Leftrightarrow \left ( 2x – 3 \right )^{2} = 8 + 17$ $\Leftrightarrow \left ( 2x – 3 \right )^{2} = 25$ $\Leftrightarrow \left | 2x – 3 \right | = 5$ $\Leftrightarrow \left[ \begin{array}{l}2x-3=5\\2x-3=-5\end{array} \right.$ $\Leftrightarrow \left[ \begin{array}{l}x=4\\x=-1\end{array} \right.$ Bình luận
Đáp án:
a) x=-4
b) x=-7 hoặc x=8
c) \(\left[ \begin{array}{l}x=4\\x=-1\end{array} \right.\)
Giải thích các bước giải:
a) 2x-5=-x-17
⇔ 3x=-12
⇔ x=-4.
Vậy x=-4.
b) Ta có: |2x-1|=2x-1 khi x$\geq$ $\frac{1}{2}$
|2x-1|=1-2x khi x<$\frac{1}{2}$
+) Với x$\geq$ $\frac{1}{2}$ , ta có: 4-(2x-1)+11=0
⇔ 2x=16
⇔ x=8
+) Với x<$\frac{1}{2}$ , ta có: 4-(1-2x)+11=0
⇔ 2x=-14
⇔ x=-7
Vậy x=8 hoặc x=-7.
c) (2x-3)²-17=8
⇔ 4x²-12x+9-25=0
⇔ 4x²-12x-16=0
⇔ (4x²-16x)+(4x-16)=0
⇔ 4x(x-4)+4(x-4)=0
⇔ (x-4)(4x+4)=0
⇔ \(\left[ \begin{array}{l}x-4=0\\4x+4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4\\x=-1\end{array} \right.\) .
Vậy \(\left[ \begin{array}{l}x=4\\x=-1\end{array} \right.\) .
a, $2x – 5 = -x – 17$
$\Leftrightarrow 2x + x = -17 + 5$
$\Leftrightarrow 3x = -12$
$\Leftrightarrow x = (-12) : 3$
$\Leftrightarrow x = -4$
b, $4 – \left | 2x – 1 \right | = -11$
$\Leftrightarrow \left | 2x – 1 \right | = 4 + 11$
$\Leftrightarrow \left | 2x – 1 \right | = 15$
$\Leftrightarrow \left[ \begin{array}{l}2x+1=15\\2x+1=-15\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x=7\\x=-8\end{array} \right.$
c) $\left ( 2x – 3 \right )^{2} – 17 = 2^{3}$
$\Leftrightarrow \left ( 2x – 3 \right )^{2} – 17 = 8$
$\Leftrightarrow \left ( 2x – 3 \right )^{2} = 8 + 17$
$\Leftrightarrow \left ( 2x – 3 \right )^{2} = 25$
$\Leftrightarrow \left | 2x – 3 \right | = 5$
$\Leftrightarrow \left[ \begin{array}{l}2x-3=5\\2x-3=-5\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x=4\\x=-1\end{array} \right.$