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\(\dfrac{5x-10}{77x^2+1}=0\)
Mà: \(77x^2+1\ge1>0\forall x\)
\(\Rightarrow5x-10=0\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=\dfrac{10}{5}\)
\(\Rightarrow x=2\)
\(\left(5x-10\right):\left(77x^2+1\right)=0\)
\(TH1:5x-10=0\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=10:5\)
\(\Rightarrow x=2\)
\(TH2:77x^2+1=0\)
\(\Rightarrow77x^2=-1\) \(\left(vô.lý\right)\)
\(\Rightarrow\left(-1\ne0\right)\)
Vậy nghiệm của đa thức \(\left(5x-10\right):\left(77x^2+1\right)=0\) là: \(x=2\)
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`(5x-10)(3x-21)=0`
`=>5x-10=0` hoặc `3x-21=0`
`=>x=2` hoặc `x=7`
\(\left(5x-10\right)\cdot\left(3x-21\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-10=0\\3x-21=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=10\\3x=21\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)
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1) Ta có: \(\left(-5+x\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-5+x=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{5;7\right\}\)
2) Ta có: \(\left(30-x\right)\left(2x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}30-x=0\\2x-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-30\\2x=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=8\end{matrix}\right.\)
Vậy: \(x\in\left\{30;8\right\}\)
3) Ta có: \(\left(-5-x\right)\left(17+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-5-x=0\\17+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=5\\x=0-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-17\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;-17\right\}\)
4) Ta có: \(\left(-3x+18\right)\left(-5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+18=0\\-5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-18\\-5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{6;-2\right\}\)
Bài nay ta có hai vế bạn hãy đặt giả sử một trong hai vế bằng 0 rồi giải phương trình cho mỗi vế bằng o
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a) (x + 5)(2x - 4) = 0
x + 5 = 0 hoặc 2x - 4 = 0
*) x + 5 = 0
x = -5
*) 2x - 4 = 0
2x = 4
x = 4 : 2
x = 2
Vậy x = -5; x = 2
b) (x - 3)(5x - 10) = 0
x - 3 = 0 hoặc 5x - 10 = 0
*) x - 3 = 0
x = 3
*) 5x - 10 = 0
5x = 10
x = 10 : 5
x = 2
Vậy x = 2; x = 3
A , x+5 hoặc 2x-4 =0
tương tự bài b
CHÚC BẠN HỌC TỐT
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\(\left(x+1\right)\left(x+7\right)< 0\)
thì \(x+1;x+7\)khác dấu
th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)
th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)
vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)
a) (2x - 3) = 5
<=> 2x - 3 = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
=> x = 4
b) (5x - 3) = 1/2
<=> 5x - 3 = 1/2
<=> 5x = 1/2 + 3
<=> 5x = 7/2
<=> x = 7/10
=> x = 7/10
c) (x + 1)(x + 7) < 0
<=> x = -1; -7
<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)
<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)
<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)
Vậy: -7 < x < -1
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\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
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( x - 3) - ( 10 - 5x ) = 0
x - 3 = 10 - 5x
-10 - 3 = -5x - x
-13 = -6x
=> x = 13/6
\(\Leftrightarrow5x-10=0\)
hay x=2