Bài 4:

a: xy=-2

=>\(x\cdot y=1\cdot\left(-2\right)=\left(-2\right)\cdot1=\left(-1\right)\cdot2=2\cdot\left(-1\right)\)

=>\(\left(x,y\right)\in\left\{\left(1;-2\right);\left(-2;1\right);\left(-1;2\right);\left(2;-1\right)\right\}\)

b: \(\left(x-1\right)\left(y+2\right)=-3\)

=>\(\left(x-1\right)\cdot\left(y+2\right)=1\cdot\left(-3\right)=\left(-3\right)\cdot1=-1\cdot3=3\cdot\left(-1\right)\)

=>\(\left(x-1;y+2\right)\in\left\{\left(1;-3\right);\left(-3;1\right);\left(-1;3\right);\left(3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;-5\right);\left(-2;-1\right);\left(0;1\right);\left(4;-3\right)\right\}\)

Bài 3:

a: \(x\left(x+9\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x+9=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)

b: \(\left(x-5\right)^2=9\)

=>\(\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3+5=8\\x=-3+5=2\end{matrix}\right.\)

c: \(\left(7-x\right)^2=-64\)

mà \(\left(7-x\right)^2>=0\forall x\)

nên \(x\in\varnothing\)

Bài 2:

a: \(\left(-31\right)\cdot x=-93\)

=>\(31\cdot x=93\)

=>\(x=\dfrac{93}{31}=3\)

b: \(\left(-4\right)\cdot x=-20\)

=>\(4\cdot x=20\)

=>\(x=\dfrac{20}{4}=5\)

c: \(5x+1=-4\)

=>\(5x=-4-1=-5\)

=>\(x=-\dfrac{5}{5}=-1\)

d: \(-12x+1=-4\)

=>\(-12x=-4-1=-5\)

=>\(12x=5\)

=>\(x=\dfrac{5}{12}\)

Câu 1 : a . \(lim\dfrac{9n^2-3n-1}{7n^3+3n^2}=lim\dfrac{\dfrac{9}{n}-\dfrac{3}{n^2}-\dfrac{1}{n^3}}{7+\dfrac{3}{n}}=0\)

b. \(lim_{x\rightarrow2}\dfrac{\sqrt{4x+1}-3}{4-x^2}=lim_{x\rightarrow2}\dfrac{4x+1-9}{\left(\sqrt{4x+1}+3\right)\left(4-x^2\right)}\) 

\(=lim_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(\sqrt{4x+1}+3\right)\left(2-x\right)\left(2+x\right)}\)

\(=lim_{x\rightarrow2}\dfrac{-4}{\left(\sqrt{4x+1}+3\right)\left(2+x\right)}=\dfrac{-4}{\left(3+3\right)\left(2+2\right)}=-\dfrac{1}{6}\)

Câu 2 : Ta có : f(x) = \(\left\{{}\begin{matrix}2x^2+x\left(x< 2\right)\\mx-1\left(x\ge2\right)\end{matrix}\right.\)

TXĐ : D = R   .  Với x < 2 ; hàm số liên tục

Với x > 2 ; hàm số liên tục 

Với x = 2  , ta có :  \(lim_{x\rightarrow2^-}f\left(x\right)=lim_{x\rightarrow2^-}2x^2+x=2.2^2+2=10\)

\(lim_{x\rightarrow2^+}f\left(x\right)=lim_{x\rightarrow2^+}mx-1=2m-1\) 

Hàm số liên tục trên R <=> Hàm số liên tục tại x = 2 

\(\Leftrightarrow lim_{x\rightarrow2^-}f\left(x\right)=lim_{x\rightarrow2^+}f\left(x\right)\)

\(\Leftrightarrow10=2m-1\) \(\Leftrightarrow m=\dfrac{11}{2}\)

Vậy ...

Bài 4:

$A+2=1+2+2^2+2^3+...+2^{11}$

$=(1+2)+(2^2+2^3)+....+(2^{10}+2^{11})$

$=(1+2)+2^2(1+2)+....+2^{10}(1+2)$

$=(1+2)(1+2^2+....+2^{10})$

$=3(1+2^2+...+2^{10})\vdots 3$

Vậy $A+2\vdots 3$ nên $A$ không chia hết cho $3$

Bài 5:

$n^2+n+1=n(n+1)+1$
Vì $n,n+1$ là hai số tự nhiên liên tiếp nên sẽ tồn tại một số chẵn và 1 số lẻ

$\Rightarrow n(n+1)$ chẵn 

$\Rightarrow n^2+n+1=n(n+1)+1$ lẻ (điều phải chứng minh) 

Ta có: \(a\left(b+1\right)+b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\)

\(\Leftrightarrow a+ab+b+ab=a+ab+b+1\)

\(\Leftrightarrow\left(a+ab+b+ab\right)-\left(a+ab+b\right)=1\)

\(\Leftrightarrow ab=1\left(ĐPCM\right)\)

Chúc bạn hok tot

Ta có: \(a\left(b+1\right)+b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\Leftrightarrow ab+a+ba+b=ab+a+b+1\)

\(\Leftrightarrow2ab+a+b=ab+a+b+1\Leftrightarrow ab=1\left(đpcm\right).\)

\(1,\\ a,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x+1}=\dfrac{2}{3}\Leftrightarrow2x+1=\dfrac{4}{9}\Leftrightarrow x=-\dfrac{5}{18}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=2\Leftrightarrow\left[{}\begin{matrix}x-3=2\\3-x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\\ 2,\\ a,=\left|5-x\right|=x-5\\ b,=\sqrt{4a\cdot44a}=\sqrt{176a^2}=4\left|a\right|\sqrt{11}=4a\sqrt{11}\\ c,=\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)