\(\left(2x-1\right)\cdot\left(15-x\right)=0\)
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a1.
$\cot (2x+\frac{\pi}{3})=-\sqrt{3}=\cot \frac{-\pi}{6}$
$\Rightarrow 2x+\frac{\pi}{3}=\frac{-\pi}{6}+k\pi$ với $k$ nguyên
$\Leftrightarrow x=\frac{-\pi}{4}+\frac{k}{2}\pi$ với $k$ nguyên
a2. ĐKXĐ:...............
$\cot (3x-10^0)=\frac{1}{\cot 2x}=\tan 2x$
$\Leftrightarrow \cot (3x-\frac{\pi}{18})=\cot (\frac{\pi}{2}-2x)$
$\Rightarrow 3x-\frac{\pi}{18}=\frac{\pi}{2}-2x+k\pi$ với $k$ nguyên
$\Leftrightarrow x=\frac{\pi}{9}+\frac{k}{5}\pi$ với $k$ nguyên.
a3. ĐKXĐ:........
$\cot (\frac{\pi}{4}-2x)-\tan x=0$
$\Leftrightarrow \cot (\frac{\pi}{4}-2x)=\tan x=\cot (\frac{\pi}{2}-x)$
$\Rightarrow \frac{\pi}{4}-2x=\frac{\pi}{2}-x+k\pi$ với $k$ nguyên
$\Leftrightarrow x=-\frac{\pi}{4}+k\pi$ với $k$ nguyên.
a4. ĐKXĐ:.....
$\cot (\frac{\pi}{6}+3x)+\tan (x-\frac{\pi}{18})=0$
$\Leftrightarrow \cot (\frac{\pi}{6}+3x)=-\tan (x-\frac{\pi}{18})=\tan (\frac{\pi}{18}-x)$
$=\cot (x+\frac{4\pi}{9})$
$\Rightarrow \frac{\pi}{6}+3x=x+\frac{4\pi}{9}+k\pi$ với $k$ nguyên
$\Rightarrow x=\frac{5}{36}\pi + \frac{k}{2}\pi$ với $k$ nguyên.
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\(15-\left\{2.\left[\left(2x-4\right).5\right].3.\left(x+1\right)\right\}=12-x\)
\(15-\left\{\left[10x-20\right].6.\left(x+1\right)\right\}=12-x\)
\(15-\left\{10x-20.6x+1\right\}=12-x\)
\(15-\left\{10x-120x+1\right\}=12-x\)
\(15-\left(-110x\right)-1=12-x\)
\(15+110x-1=12-x\)
\(110x+x=12-15+1\)
\(111x=-2\)
\(x=\dfrac{-2}{111}\)
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1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
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a)\((x^2- 4).(x^2 - 10) = 72 Đặt x^2 - 7 = a(1), ta có (a+3)(a-3)=72 a^2-9=72 a^2=81 a=+-9 xét 2 trường hợp a = 9 và -9 khi thay vào (1) ta có..... tự lm nốt nha \)
b) nhóm x+1 vs x+4 và x+2 vs x+3 ta sẽ có (x2+5x+4)(x2+5x+6)(x+5)=40
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a) (2 - x)(2x + 1) > 0
TH1: \(\hept{\begin{cases}2-x>0\\2x+1>0\end{cases}\Rightarrow\hept{\begin{cases}x< 2\\x>-\frac{1}{2}\end{cases}\Rightarrow}-\frac{1}{2}< x< 2}\)
TH2: \(\hept{\begin{cases}2-x< 0\\2x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x>2\\x< -\frac{1}{2}\end{cases}\left(vl\right)}}\)(vô lí)
Vậy: -1/2 < x < 2
b) (2x+3)(x + 1) < 0
TH1: \(\hept{\begin{cases}2x+3>0\\x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-\frac{3}{2}\\x< -1\end{cases}\Rightarrow-\frac{3}{2}< x< -1}}\)
TH2: \(\hept{\begin{cases}2x+3< 0\\x+1>0\end{cases}\Rightarrow\hept{\begin{cases}\left(x< -\frac{3}{2}\right)\\x>-1\end{cases}}\left(vl\right)}\)(vô lí)
Vậy -3/2 < x < -1
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\(a,4x^2-\left(3x+1\right)\left(2x-1\right)=2\left(x-3\right)^2\)
\(\Leftrightarrow4x^2-\left(6x^2-3x+2x-1\right)=2\left(x^2-6x+9\right)\)
\(\Leftrightarrow4x^2-6x^2+x+1-2x^2+12x-18=0\)
\(\Leftrightarrow-4x^2+13x-17=0\)
\(\Leftrightarrow-4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)-\dfrac{103}{16}=0\)
\(\Leftrightarrow-4\left(x-\dfrac{13}{8}\right)^2=\dfrac{103}{16}\)
\(\Leftrightarrow\left(x-\dfrac{13}{8}\right)^2=\dfrac{-103}{64}\Rightarrow\) pt vô nghiệm
\(b,\left(5x-1\right)\left(x+1\right)-\left(2x-1\right)\left(2x+1\right)=x.\left(x+1\right)\)\(\Leftrightarrow5x^2+5x-x-1-\left(4x^2-1\right)=x^2+x\)
\(\Leftrightarrow5x^2+5x-x-1-4x^2+1-x^2-x=0\) \(\Leftrightarrow3x=0\Rightarrow x=0\)
\(c,7x^2-\left(2x-3\right)^2=1+3\left(x+2\right)^2\)
\(\Leftrightarrow7x^2-\left(4x^2-12x+9\right)=1+3\left(x^2+4x+4\right)\)
\(\Leftrightarrow7x^2-4x^2+12x-9=1+3x^2+12x+12\)\(\Leftrightarrow7x^2-4x^2+12x-9-1-3x^2-12x-12=0\)\(\Leftrightarrow-22=0\) ( vô lí)
Vậy phương trình vô nghiệm
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a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)
e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)
g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)
h)Tương tự các câu trên
i) x = 0
k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)
l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)
=> x + 1 = 3 => x = 2
x.(x+1)=0
suy ra x=0 hoac x+1=0
x=0-1
x=-1
vay x=0 hoac x=-1
mấy câu sau cũng làm tương tự
`( 2x - 1 ) . ( 15 - x ) = 0`
`@TH1: 2x - 1 = 0`
`=> 2x = 1`
`=> x = 1 / 2`
`@TH2: 15 - x = 0`
`=> x = 15`
Vậy `x = 1 / 2` hoặc `x = 15`
\(\left(2x-1\right)\cdot\left(15-x\right)=0\)
\(TH1:2x-1=0\)
\(2x=0+1\)
\(2x=1\)
\(x=1:2\)
\(x=\dfrac{1}{2}\)
\(TH2:15-x=0\)
\(x=15-0\)
\(x=15\)
Vậy \(x=\left[{}\begin{matrix}\dfrac{1}{2}\\15\end{matrix}\right.\)