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a, Ta có:
\(\left|2x+4\right|+\left|4x+8\right|\ge0\)
Để \(\left|2x+4\right|+\left|4x+8\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|2x+4\right|=0\\\left|4x+8\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=-2\end{matrix}\right.\Rightarrow x=-2\)
Vậy...........
b, Ta có:
\(\left|x-5\right|+\left|x-7\right|\ge0\)
Để \(\left|x-5\right|+\left|x-7\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x-5\right|=0\\\left|x-7\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=7\end{matrix}\right.\Rightarrow x\in\varnothing\)
Vậy...........
c,\(\left|x+8\right|-\left|2x+2\right|=0\)
\(\Rightarrow\left|x+8\right|=\left|2x+2\right|\)
\(\Rightarrow\left\{{}\begin{matrix}x+8=2x+2\\x+8=-2x-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}-x=-6\\3x=-10\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy...........
Chúc bạn học tốt!!!
\(\begin{cases}3\left(x-7\right)=4\left(y-5\right)\\4x-3y+8=0\end{cases}\)
\(\Leftrightarrow\begin{cases}3x-4y=1\\4x-3y=-8\end{cases}\Leftrightarrow\begin{cases}9x-12y=3\\-16x+12y=32\end{cases}\Leftrightarrow}\begin{cases}-7x=35\\3x-4y=1\end{cases}\Leftrightarrow\begin{cases}x=-5\\y=-4\end{cases}}}\)
\(\begin{cases}3\left(x-7\right)=4\left(y-5\right)\\4x-3y+8=0\end{cases}\)
<=> \(\begin{cases}3x-4y=1\\4x-3y=-8\end{cases}\)<=> \(\begin{cases}9x-12y=3\\-16x+12y=32\end{cases}\)<=> \(\begin{cases}-7x=35\\3x-4y-1\end{cases}\)<=> \(\begin{cases}x=-5\\y=-4\end{cases}\)
a) \(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\)
\(\Leftrightarrow-x=45\Leftrightarrow x=-45\)
b) \(5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\)
\(\Leftrightarrow3x=48\Leftrightarrow x=16\)
c) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(\Leftrightarrow3x-12-8+x=12\)
\(\Leftrightarrow4x-20=12\)
\(\Leftrightarrow4x=32\Leftrightarrow x=8\)
d) \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
\(\Leftrightarrow-21x+35+14x-28=28\)
\(\Leftrightarrow-7x+35=0\Leftrightarrow x=5\)
a)
(- 2)5 + 7(x - 8) = - 167 + 8x
32 + 7x - 56 + 167 = 8x
8x - 7x = 143
x = 143
b)
(7 - 3x)(4x + 4) = 0
4(x + 1)(7 - 3x) = 0
\(\left[\begin{matrix}x+1=0\\7-3x=0\end{matrix}\right.\)
\(\left[\begin{matrix}x=-1\\3x=7\end{matrix}\right.\)
\(\left[\begin{matrix}x=-1\\x=\frac{7}{3}\end{matrix}\right.\)
a) \(\left(\frac{1}{81}\right)^x\cdot27^{2x}=\left(-9\right)^4\)
\(\Leftrightarrow\frac{1}{3^{4x}}\cdot3^{6x}=9^4\)
\(\Leftrightarrow\frac{3^{6x}}{3^{4x}}=3^8\)
\(\Leftrightarrow3^{2x}=3^8\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
b) \(5^x\cdot\left(5^3\right)^2=625\)
\(\Leftrightarrow5^{x+6}=5^4\)
\(\Leftrightarrow x+6=4\)
\(\Leftrightarrow x=-2\)
c) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\Leftrightarrow\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\\left(4x-1\right)^{10}=1=\left(\pm1\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{1}{2}\\x=0\end{matrix}\right.\)
Vậy....
5x.(-x)2+1=6
5x.x+1=6
5x2+1=6
5x2=6-1
5x2=5
x2=5:5
x2=1
x2=12
=>x=1
(15-x)+(x-12)=7-(-8+x)
15-x+x-12=7+8-x
3-x+x=15-x
3=15-x
x=15-3
x=12
4x3=4x
Để 4x3=4x=>x3=x
=>x=1
Câu cuối cùng mình ko bik.
hoc tốt
a)\(10\left(x-7\right)-8\left(x+5\right)=6\cdot\left(-5\right)+24\)
\(10x-10\cdot7-8x-8\cdot5=\left(-30\right)+24\)
\(10x-70-8x-40=-6\)
\(10x-8x=\left(-6\right)+70+40\)
\(2x=104\)
\(x=104\div2\)
\(x=52\)
b)\(2\left(4x-8\right)-7\left(3+x\right)=6\)
\(2\cdot4x-2\cdot8-7\cdot3-7x=6\)
\(8x-16-21-7x=6\)
\(8x-7x=6+16+21\)
\(x=43\)
(4x-8)5-x(4x-8)=0
=>(4x-8).5=x.(4x-8) mà 4x-8=4x-8
=>x=5