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a/ 2x(x-5)-x(3+2x)=26
=>2x2-10x-2x2-3x=26
=>-13x=26
=>x=-2
b/ 49x2-81=0
=>49x2=81
=>x2=\(\frac{49}{81}\)
\(\Rightarrow x^2=\left(-\frac{9}{7}\right)^2\)hoặc \(\left(\frac{9}{7}\right)^2\)
\(\Rightarrow x=\pm\frac{9}{7}\)
BÀI \(1\):
\(\left(-1005\right).\left(x+2\right)=0\)
\(x+2=0:\left(-1005\right)\)
\(x+2=0\)
\(x=0-2\)
\(x=-2\)
BÀI \(2\):
\(x+x+x+91=-2\)
\(3x=\left(-2\right)-91\)
\(3x=-93\)
\(x=\left(-93\right):3\)
\(x=-31\)
BÀI \(3\):
\(\left|5x+1\right|=11\)
Có hai trường hợp:
\(TH^{ }1:_{ }5x+1=11\)
\(5x=11-1\)
\(5x=10\)
\(x=10:5\)
\(x=2\)
\(TH2:^{ }5x+1=-11\)
\(5x=\left(-11\right)-1\)
\(5x=-12\)
\(x=\left(-12\right):5\)
\(x=-2,4\)
\(k\)\(minh\)\(nhe\)\(.\)
\(\left(7-x\right)\left(2019.2019.1024....\right)=0\)
\(\Leftrightarrow7-x=0\)
\(\Leftrightarrow x=7\)
a) Ta có: \(x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
Vậy: x∈{0;-7}
b) Ta có: \(\left(x+12\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
Vậy: x∈{-12;3}
c) Ta có: \(\left(-x+5\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-5\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy: x∈{3;5}
d) Ta có: \(x\left(2+x\right)\left(7-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy: x∈{-2;0;7}
e) Ta có: \(\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=3\end{matrix}\right.\)
Vậy: x∈{-2;1;3}
g) Ta có: \(\left(x-5\right)\left(x^2-81\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-81=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\\x=-9\end{matrix}\right.\)
Vậy: x∈{-9;5;9}
h) Ta có: \(x^3+27=0\)
\(\Leftrightarrow x^3=-27\)
hay x=-3
Vậy: x=-3
Bài làm
a) x( 2 + x )( 7 - x ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy x = 0 hoặc x = -2 hoặc x = 7
b) ( x - 1 )( x + 2 )( x - 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=3\end{matrix}\right.\)
Vậy x = 1 hoặc x = -2 hoặc x = 3
c) ( x - 5 )( x2 - 81 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-81=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=81\Leftrightarrow x=\pm9\end{matrix}\right.\)
Vậy x = 5 hoặc x = \(\pm\) 9 .
# Học tốt #
a) \(x\left(2+x\right)\left(7-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{0;-2;7\right\}\)
b) \(\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;-2;3\right\}\)
c) \(\left(x-5\right)\left(x^2-81\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x^2-81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x^2=81\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=9\\x=-9\end{matrix}\right.\)
Vậy \(x\in\left\{5;9;-9\right\}\)
(x - 2)(2x - 6) = 0
=> x - 2 = 0 hoac 2x - 6 = 0
=> x = 2 hoac x = 3
vay_
(3x + 9)(1 - 3x) = 0
=> 3x + 9 = 0 hoac 1 - 3x = 0
=> x = -3 hoac x = 1/3
vay_
|2 - x| + 2 = x
=> |2 - x| = x - 2
=> 2 - x = x - 2 hoac 2 - x = 2 - x
=> -2x = -4 hoac x thuoc tap hop rong
=> x = 2
\(a,\left(x-2\right).\left(2x-6\right)=0\Leftrightarrow2\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(b,\left(3x+9\right).\left(1-3x\right)=0\Leftrightarrow3\left(x+3\right).\left(1-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\1-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
\(c,\left(x^2+1\right).\left(81-x^2\right)=0\Leftrightarrow\left(x^2+1\right).\left(9-x\right).\left(9+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9-x=0\\9+x=0\end{cases}}\) ( vì \(x^2+1\ne0\forall x\)) \(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
(x + 1) + (x + 3) + (x + 5) + ... + (x + 81) = 0
=> (x + x + x + ... + x) + (1 + 3 + 5 + ... + 81) = 0
Đặt A = 1 + 3 + 5 + ... + 81
Số số hạng của A là: (81 - 1) : 2 + 1 = 41 (số)
=> 41x + (81 + 1).41:2 = 0
=> 41x + 1681 = 0
=> 41x = -1681
=> x = -41