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a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
b. \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)
c, \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\frac{7}{5}\)
e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }
| x - 2 | 1 | -1 | 7 | -7 |
| x | 3 | 1 | 9 | -5 |
Vậy....
a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy : ....
b) \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)
c) \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
Vậy :...
a) -4/5 + 5/2x = -3/10
5/2x = -3/10 + 4/5
5/2x = 1/5
5/2x = 1/2
x = 1/2 : 5/2
x = 1/5
b) 4/3 + 5/8 : x = 1/12
5/8x = 1/12 - 4/3
5/8x = -5/4
5 = -5/4.8x
5 = -10x
5/-10 = x
-1/2 = x
x = -1/2
c) (x - 1/3)(x - 2/5) = 0
x - 1/3 = 0 hoặc x - 2/5 = 0
x = 0 + 1/3 x = 0 + 2/5
x = 1/3 x = 2/5
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0
a, \(\left(2-x\right)\left(x+3\right)>0\Leftrightarrow\left(x-2\right)\left(x+3\right)< 0\)
Vì \(x+3>x-2\)
nên \(\hept{\begin{cases}x+3>0\\x-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-3\\x< 2\end{cases}\Leftrightarrow-3< x< 2}\)
c, \(\left(5-2x\right)\left(x+4\right)>0\)
TH1 : \(\hept{\begin{cases}5-2x>0\\x+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{5}{2}\\x>-4\end{cases}}\Leftrightarrow-4< x< \frac{5}{2}\)
TH2 : \(\hept{\begin{cases}5-2x< 0\\x+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{5}{2}\\x< -4\end{cases}}\)( vô lí )
bạn làm tương tự nhé
a) TH1:
\(\left\{{}\begin{matrix}x< 0\\8-x>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 0\\x< 8\end{matrix}\right.\) \(\Rightarrow x< 0\)
TH2:
\(\left\{{}\begin{matrix}x>0\\8-x< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>0\\x>8\end{matrix}\right.\)\(\Rightarrow x>8\)
Vậy \(\left[{}\begin{matrix}x< 0\\x>8\end{matrix}\right.\)
b) TH1:
\(\left\{{}\begin{matrix}2-x>0\\x+3>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 2\\x>-3\end{matrix}\right.\)\(\Rightarrow-3< x< 2\)
TH2:
\(\left\{{}\begin{matrix}2-x< 0\\x+3< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\)(vô nghiệm)
Vậy \(-3< x< 2\)
c) TH1:
\(\left\{{}\begin{matrix}2x-4>0\\5-x< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2x>4\\x>5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>2\\x>5\end{matrix}\right.\Rightarrow x>5\)
TH2:
\(\left\{{}\begin{matrix}2x-4< 0\\5-x>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x< 4\\x< 5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 2\\x< 5\end{matrix}\right.\Rightarrow x< 2\)
Vậy \(\left[{}\begin{matrix}x>5\\x< 2\end{matrix}\right.\)
a, \(x\left(8-x\right)< 0\\ 8x-x^2< 0\)
Có x2 ≥ 0 ∀ x (1)
\(\Rightarrow\) - x2 ≤ 0 ∀ x
Mà 8x - x2 < 0
\(\Rightarrow\) 8x < x2 (2)
Thay (1) vào (2) \(\Rightarrow\) 8x < 0
\(\Rightarrow\) x < 0
Vậy x < 0
b, \(\left(2-x\right)\left(x+3\right)>0\\ 2x+6-x^2-3x>0\\ \Rightarrow\left(6-x\right)-x^2>0\)
Có x2 ≥ 0 ∀ x (1)
⇒ - x2 ≤ 0 ∀ x
Mà (6 - x) - x2 > 0
\(\Rightarrow6-x>x^2\left(2\right)\)
Thay (1) vào (2) \(\Rightarrow6-x>0\\ \Rightarrow x< 0\)
Vậy x < 0
c, \(\left(2x-4\right)\left(5-x\right)< 0\\ 10x-2x^2-20+4x< 0\\ \Rightarrow\left(-20+14x\right)-2x^2< 0\)
Có x2 ≥ 0 ∀ x
⇒ 2x2 ≥ 0 ∀ x (1)
⇒ - 2x2 ≤ 0 ∀ x
Mà (-20 + 14x) - 2x2 < 0
\(\left(-20+14x\right)< 2x^2\left(2\right)\)
Thay (1) vào (2) \(\Rightarrow-20+14x< 0\\ \Rightarrow14x< 20\\ \Rightarrow x< \frac{10}{7}\)
Vậy \(x< \frac{10}{7}\)