Tìm x
a,2x+2=12+14
b,18+3x=12.3
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a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow80x=480\Rightarrow x=6\)
b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow4x=0\Rightarrow x=0\)
c) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
a: \(x\left(1-2x\right)+2x^2=14\)
=>\(x-2x^2+2x^2=14\)
=>x=14
b: \(x\left(x-5\right)+3x-15=0\)
=>\(\left(x-5\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
a: ta có: \(2\left(4-3x\right)+2x=5\left(2x-3\right)\)
\(\Leftrightarrow8-6x+2x-10x+15=0\)
\(\Leftrightarrow-14x=-23\)
hay \(x=\dfrac{23}{14}\)
b: Ta có: \(\dfrac{1}{2}-\left(2x-\dfrac{1}{3}\right)^2=\dfrac{7}{18}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{3}\\2x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{2}{3}\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=0\end{matrix}\right.\)
\(a,=3x-9-4x+12=-x+3=0\)
\(\Leftrightarrow x=3\)
Vậy ..
\(b,=\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy ..
\(c,=x^3-3x^2+3x-1=\left(x-1\right)^3=0\)
\(\Leftrightarrow x=1\)
Vậy ..
\(d,\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy ..
\(e,=\left(2x-3-5\right)\left(2x-3+5\right)=\left(2x-8\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)
Vậy ...
a) Ta có: 3(x-3)-4x+12=0
\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
hay x=3
Vậy: S={3}
b) Ta có: \(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4=0\)
\(\Leftrightarrow4x=-8\)
hay x=-2
Vậy: S={-2}
c) Ta có: \(x^3+3x=3x^2+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
Vậy: S={1}
d) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: S={0;2;-2}
Lời giải:
a.
$|2x-5|=12-3x$
Nếu $x\geq \frac{5}{2}$ thì $2x-5=12-3x$
$\Leftrightarrow x=3,4$ (thỏa mãn)
Nếu $x< \frac{5}{2}$ thì: $5-2x=12-3x$
$\Leftrightarrow x=7$ (loại)
Vậy......
b.
$4x=|x+1|+|x+2|+|x+3|\geq 0$
$\Rightarrow x\geq 0$
Do đó: $|x+1|+|x+2|+|x+3|=(x+1)+(x+2)+(x+3)=3x+6$
Vậy: $3x+6=4x$
$\Leftrightarrow x=6$ (thỏa mãn)
c.
$|x^2+|x+2||=x^2+3$
$\Leftrightarrow x^2+|x+2|=x^2+3$
$\Leftrightarrow |x+2|=3$
$\Leftrightarrow x+2=3$ hoặc $x+2=-3$
$\Leftrightarrow x=1$ hoặc $x=-5$
d.
$|x^2-3|=6$
$\Leftrightarrow x^2-3=6$ hoặc $x^2-3=-6$
$\Leftrightarrow x^2=9$ (chọn) hoặc $x^2=-3< 0$ (loại)
$\Leftrightarrow x=\pm 3$
a) \(18-\left(2x+5\right)=9\)
\(2x+5=18-9\)
\(2x+5=9\)
\(2x=9-5\)
\(2x=4\)
\(x=2\)
a) \(18-\left(2x+5\right)=9\)
\(\Rightarrow2x+5=18-9=9\)
\(\Rightarrow2x=9-5=4\Rightarrow x=4:2=2\)
b) \(23x-4=32\Rightarrow23x=32+4=36\Rightarrow x=\dfrac{36}{23}\)
c) \(\left(3x+2\right)^2=64\)
\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)
d) \(x\left(2x-12\right)=0\Rightarrow6x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)
\(\Rightarrow6x^2+2x+12x-6x^2=7\)
\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)
b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(-80x=-480\)
x = 6
c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)
\(\Rightarrow18x+16=7\)
18x = -9
x = -2
18x =
a,2x+2=12+14
2x+2=26
2x=26-2
2x=24
x=24:2
x=12
Vậy x=12
b,18+3x=12.3
18+3x=36
3x=36-18
3x=18
x=18:3
x=6
Vậy x=6
a, 2x+2=12+14
<=> 2x=12+14-2=24
<=> x=24/2=12
b, 18+3x=12.3
<=> 3x=12.3-18=18
<=> x=18/3=6