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a, x= 21 hoặc x= -21
b, Không có số nào thỏa mãn
Sơ đồ con đường |
Lời giải chi tiết |
Áp dụng định nghĩa giá trị tuyệt đối |
a) − x = 21 ⇔ − x = 21 hoặc − x = − 21 nên x= 21 hoặc x= -21 b) Giá trị tuyệt đối của một số luôn là số dương nên không có x nào thỏa mãn. |
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\(=-\left(215+x\right)+215=-9\\ -\left(215+x\right)=-224\\ x=-9\)
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\(\frac{2}{x}=\frac{y}{-3}\)
\(\Leftrightarrow xy=-6\)
Tự tìm
\(\frac{2}{x}=\frac{y}{-3}\)
=> \(xy=2.\left(-3\right)\)
=>\(xy=-6\)
=>x=+2 hoặc +3 và y=+3 hoặc +2
Nếu x=2 thì y=-3
x=-2 thì y=3
x=3 thì y=-2
x=-3 thì y=2
Chúc bạn học tốt!
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c) (-8).x = (-7).(-6) - 2
-8.x = 42 - 2
-8x = 40
x = 40 : (-8)
x = -5
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k: \(\left(4x-16\right)\left(-72+9x\right)=0\)
=>\(4\cdot\left(x-4\right)\cdot9\left(x-8\right)=0\)
=>\(36\left(x-4\right)\left(x-8\right)=0\)
=>\(\left(x-4\right)\left(x-8\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=8\end{matrix}\right.\)
m: \(\left(20+5x\right)\left(4x-8\right)=0\)
=>\(5\cdot\left(x+4\right)\cdot4\left(x-2\right)=0\)
=>\(\left(x+4\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
n: \(\left(-4x+48\right)\left(2x-24\right)=0\)
=>\(-4\left(x-12\right)\cdot2\left(x-12\right)=0\)
=>\(\left(x-12\right)^2=0\)
=>x-12=0
=>x=12
o: \(\left(4x+16\right)\left(-2x+20\right)\left(-40+x\right)=0\)
=>\(4\cdot\left(x+4\right)\cdot\left(-2\right)\left(x-10\right)\left(x-40\right)=0\)
=>\(\left(x+4\right)\left(x-10\right)\left(x-40\right)=0\)
=>\(\left[{}\begin{matrix}x+4=0\\x-10=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=10\\x=40\end{matrix}\right.\)
p: \(\left(-5x+40\right)\left(-x+2023\right)\left(2x-2\right)=0\)
=>\(-5\left(x-8\right)\cdot\left(-1\right)\cdot\left(x-2023\right)\cdot2\left(x-1\right)=0\)
=>\(\left(x-8\right)\left(x-2023\right)\left(x-1\right)=0\)
=>\(\left[{}\begin{matrix}x-8=0\\x-1=0\\x-2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\x=2023\end{matrix}\right.\)
q: \(2024x\left(4x-8\right)\left(5+5x\right)=0\)
=>\(x\cdot4\left(x-2\right)\cdot5\left(x+1\right)=0\)
=>\(x\left(x-2\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-1\end{matrix}\right.\)
r: \(-4x\left(3x+9\right)\left(2x-16\right)=0\)
=>\(-4x\cdot3\left(x+3\right)\cdot2\left(x-8\right)=0\)
=>\(x\left(x+3\right)\left(x-8\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=8\end{matrix}\right.\)
s: \(\left(-100+5x\right)\left(2x-10\right)\left(6x+6\right)=0\)
=>\(5\cdot\left(x-20\right)\cdot2\left(x-5\right)\cdot6\left(x+1\right)=0\)
=>\(\left(x-20\right)\left(x-5\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-20=0\\x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=5\\x=-1\end{matrix}\right.\)
t: \(\left(-2x+4\right)\left(2x+16\right)\cdot\left(7-x\right)=0\)
=>\(-2\left(x-2\right)\cdot2\left(x+8\right)\cdot\left(-1\right)\cdot\left(x-7\right)=0\)
=>\(\left(x-2\right)\left(x+8\right)\left(x-7\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x-7=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\\x=7\end{matrix}\right.\)
\(\left(\dfrac{16}{5}+\dfrac{8}{5}x\right):\dfrac{1}{3}=\dfrac{3}{4}\)
\(\dfrac{16}{5}+\dfrac{8}{5}x=\dfrac{3}{4}.\dfrac{1}{3}=\dfrac{1}{4}\)
\(x=\left(\dfrac{1}{4}-\dfrac{16}{5}\right):\dfrac{8}{5}=-\dfrac{59}{32}\)