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\(\text{a)}\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{3}{7}\)
\(\dfrac{3}{4}x\) \(=\dfrac{3}{7}+\dfrac{1}{2}=\dfrac{13}{14}\)
\(x\) \(=\dfrac{13}{14}.\dfrac{4}{3}=\dfrac{26}{21}\)
\(\text{b)}\dfrac{x}{27}=\dfrac{-2}{3,6}\)
\(x=\dfrac{-2.27}{3,6}=-15\)
\(\text{c)}\dfrac{x}{2}=\dfrac{18}{x}\)
\(\Leftrightarrow x.x=18.2=36\)
\(\Leftrightarrow x^2=36\)
\(\Rightarrow x\in\left\{6;\left(-6\right)\right\}\)
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Cách 1:
(x3 – 2x2 + x – 1) (3x – 2)
= x3 . (3x – 2) + (-2x2) .(3x – 2) + x .(3x – 2) + (-1) . (3x – 2)
= x3 . 3x + x3 . (-2) + (-2x2). 3x + (-2x2) . (-2) + x . 3x + x. (-2) + (-1). 3x + (-1). (-2)
= 3x4 – 2x3 – 6x3 + 4x2 + 3x2 – 2x – 3x + 2
= 3x4 + (-2x3 -6x3) + (4x2 + 3x2 ) + (-2x – 3x) + 2
= x4 + (-8x3) + 7x2 + (-5x) + 2
= x4 – 8x3 +7x2 – 5x + 2
Cách 2:
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- \(x^2\) + 5\(x\) - 4 = 0
-\(x^2\) + \(x\) + 4\(x\) - 4 = 0
(- \(x^2\) + \(x\)) + (4\(x\) - 4) = 0
-\(x\)(\(x-1\)) + 4\(\times\)( \(x\) -1) = 0
(\(x-1\))( -\(x\) +4) = 0
\(\left[{}\begin{matrix}x-1=0\\-x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
\(x\) \(\in\) { 1; 4}
`-x^2+5x-4 =0`
`\Rightarrow x^2-5x+4=0`
`\Rightarrow x^2-4x-x+4=0`
`\Rightarrow (x^2-4x)-(x-4)=0`
`\Rightarrow x(x-4)-(x-4)=0`
`\Rightarrow (x-4)(x-1)=0`
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=0+4\\x=0+1\end{matrix}\right.\)
``\Rightarrow `\(\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Vậy, nghiệm của đa thức là `x={4; 1}.`
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ta xét thấy 3^n+1 = 27
nếu n bằng 0 thì tức là 3^0+1 =3^1 =3
nếu n bằng 1 thì tức là 3^1+1=3^2=9
neu n bang 2 thi tuc la 3^2+1=3^3=27
..........................
vậy n =2
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( x - 3 ) . ( 2x - 5 ) > 3
=> x > 3 . Vì nếu x bé hơn hoặc = 3 thì x - 3 = 0 kéo theo ( x - 3 ) ( 2x - 5 ) = 0
\(\widehat{A1}+\widehat{A4}=180\)
\(\Rightarrow\)\(\widehat{A1}=180-\widehat{A4}\)
\(\Rightarrow\)\(\widehat{A1}=50\)
mà \(\widehat{A1}=\widehat{B1}(Đồng vị)\)
\(\Rightarrow\)a // b (ĐPCM)