Tìm x:2x-(3x+5)=37-(-26)
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a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
\(\Rightarrow3x-6x^2+6x+14=29\)
\(\Rightarrow-6x^2+9x-15=0\)
\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)
\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)
Vậy \(S=\varnothing\)
*) \(-7+2x=-37-\left(-26\right)\)
⇔ \(-7+2x=-11\)
⇔ \(2x=\left(-11\right)-\left(-7\right)\)
⇔ \(2x=-4\)
⇔ \(x=\left(-4\right):2\)
⇔ \(x=-2\)
\(\text{(3x +9).(11-x ) =0}\)
⇔\(\left[{}\begin{matrix}3x+9=0\\11=x=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}3x=-9\\x=11\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=-3\\x=11\end{matrix}\right.\)
Vậy x= -3 và x= 11
Rin sẽ làm bài 2
a. 4 - 2x = -56
2x = 4 + 56
2x = 60
x = 30
b. -9x + 42 = 66
-9x = 24
x = 24 : -9
x = \(\frac{-24}{9}\)
c. 5 x 1x - 111 = 125
1x - 111 = 25
1x = 136
x = 136
d. (2x +1)2 = 81
(2x +1)2 = 92
=> 2x + 1 = 9
2x = 10
x = 5
b) Ta có: 2x(x – 5) – x(3 + 2x) = 26
⇔ 2x2x2 – 10x – 3x – 2x2x2 =26
⇔ - 13x = 26
⇔ x = - 2
\(a,3x-31=-40\Rightarrow3x=-9\Rightarrow x=-3\)
\(b,-3x+37=\left(-4\right)^2\Rightarrow-3x=-21\Rightarrow x=7\)
\(c,\left|2x+7\right|=5\)
\(\Rightarrow\left\{{}\begin{matrix}2x+7=5\Rightarrow x=-1\\2x+7=-5\Rightarrow x=-6\end{matrix}\right.\)
\(d,-x+21=15+2x\Rightarrow3x=6\Rightarrow x=2\)
a) Ta có: 3x-31=-40
\(\Leftrightarrow3x=-9\)
hay x=-3
Vậy: x=-3
b) Ta có: \(-3x+37=\left(-4\right)^2\)
\(\Leftrightarrow-3x+37=16\)
\(\Leftrightarrow-3x=16-37=-21\)
hay x=7
Vậy: x=7
Bài 2:
c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)