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B = \(\dfrac{x^2-2x+1}{x+1}\)

Với \(x\in\)Z, để B là số nguyên thì \(x^2-2x+1\)⋮ \(x+1\)

Theo Bezout ta có:   F(\(x\)) =  \(x^2\) - 2\(x\) + 1  ⋮ \(x+1\) ⇔ F(\(-1\))  ⋮ \(x+1\)

⇒ (-1)² - 2.(-1) + 1 ⋮ \(x\) + 1 ⇔ 4 ⋮ \(x\) + 1

⇔ \(x\) + 1 \(\in\) Ư(4) = { -4; -2; -1; 1; 2; 4}

\(\Leftrightarrow\) \(x\) \(\in\) { -5; -3; -2; 0; 1; 3}

Lời giải:
$7^{3x-2}-3.7^3=7^3.4$

$7^{3x-2}=3.7^3+7^3.4=7^3(3+4)=7^3.7=7^4$
$\Rightarrow 3x-2=4$

$\Rightarrow 3x=6$

$\Rightarrow x=2$

\(\left(3x-\dfrac{1}{3}\right)^2=16=4^2=\left(-4\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{3}=4\\3x-\dfrac{1}{3}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{13}{3}\\3x=-\dfrac{11}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{9}\\x=-\dfrac{11}{9}\end{matrix}\right.\)

`(2x-1/3)^2=4`

\(< =>\left[{}\begin{matrix}2x-\dfrac{1}{3}=2\\2x-\dfrac{1}{3}=-2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=2+\dfrac{1}{3}\\2x=-2+\dfrac{1}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=-\dfrac{5}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}:2\\x=-\dfrac{5}{3}:2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}\cdot\dfrac{1}{2}\\x=-\dfrac{5}{3}\cdot\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\)

\(\left(2x-\dfrac{1}{3}\right)^2=4\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=2\\2x-\dfrac{1}{3}=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=2+\dfrac{1}{3}\\2x=-2+\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=-\dfrac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

`7(x-1/2)^2=9`

`(x-1/2)^2=9/7`

\(=>\left[{}\begin{matrix}x-\dfrac{1}{2}=\sqrt{\dfrac{9}{7}}\\x-\dfrac{1}{2}=-\sqrt{\dfrac{9}{7}}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\\x=-\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{6+\sqrt{7}}{2\sqrt{7}}\\x=\dfrac{-6+\sqrt{7}}{2\sqrt{7}}\end{matrix}\right.\)

7.(x-\(\dfrac{1}{2}\))²=9

7.x+\(\dfrac{1}{4}\) =9

7.x=\(\dfrac{37}{4}\)

x=\(\dfrac{37}{28}\)

a) 120 - 20 : x = 110

20:x= 120 - 110

20:x= 10

x= 20:10=2

b) 213 - ( x - 13 ) = 1442 : 14

213 - (x-13)=103

x-13= 213 - 103

x-13=110

x=110+13

x=123

c) 2550 : [ 120 - ( x - 5 ) ] = 25

120 - (x-5) = 2550:25

120 - (x-5)=102

x-5 = 120 - 102

x-5= 18

x=18+5

x=23

a,120-20:X=110

20:X=120-110

20:X=10

     X=20:10

     X=2