Khai triển rồi thu gọn

đối với các câu này bạn hãy khai triển phần nào dài bằng hàng dẳng thức rồi thu gọn lại nếu đúng thì vế trái bằng vế phải

1)

a/ \(12\left(x-5\right)+7\left(3-x\right)=15\)

\(\Rightarrow12x-12.5+7.3-7x=15\)

\(\Rightarrow12x-60+21-7x=15\)

\(\Rightarrow12x-7x=15+60-21\)

\(\Rightarrow5x=54\)

\(\Rightarrow x=\frac{54}{5}=10,8\)

b/ \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)

\(\Rightarrow30x+30.2-6x-6.5-24x=100\)

\(\Rightarrow30x+60-6x+30-24x=100\)

\(\Rightarrow30x-6x-24x=100-60-30\)

\(\Rightarrow0x=10\)

\(\Rightarrow\) k có giá trị \(x\) nào thỏa mãn đề bài

\(12\left(x-5\right)+7\left(3-x\right)=15\)

\(12x-60+21-7x=15\)

\(12x-\left(60-21+7x\right)=15\)

\(12x-\left(39+7x=15\right)\)

\(12x-39-7x=15\)

\(5x=15+39\)

\(5x=54\)

\(x=10,8\)

vì \( (x-3)(2y+1)=7\) nên x-3 và 2y+1 là 2 số nguyên cùng dấu nhau

\(\Rightarrow\left\{\begin{matrix}x-3>0\\2y+1>0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x>3\\2y>1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x\in\left\{4,5,6,7,...\right\}\\y\in\left\{1,2,3,4,...\right\}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x-3< 0\\2y+1< 0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x< 3\\2y< 1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x\in\left\{2,1,0,...\right\}\\y\in\left\{0,-1,...\right\}\end{matrix}\right.\)

b, ( 5/2 - x ) ^2

=25/4-4/5x+x^2

c,( xy/2 - x/3 ) ( xy/2 + x/3)

=(xy/2)^2-(x/3)^2

c: \(\left(\dfrac{xy}{2}-\dfrac{x}{3}\right)\left(\dfrac{xy}{2}+\dfrac{x}{3}\right)=\dfrac{x^2y^2}{4}-\dfrac{x^2}{9}\)

e: \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)

b) \(\left(\dfrac{5}{2}-x\right)^2=\dfrac{25}{4}-5x+x^2\)

c) \(\left(\dfrac{xy}{2}-\dfrac{x}{3}\right)\left(\dfrac{xy}{2}+\dfrac{x}{3}\right)=\dfrac{x^2y^2}{4}-\dfrac{x^2}{9}\)

d) \(\left(3x+\dfrac{2}{3}yz\right)^2=9x^2+4xyz+\dfrac{4}{9}y^2z^2\)

e) \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)

f) \(\left(2x-y+2\right)^2=\left(2x-y\right)^2+4\left(2x-y\right)+2^2=4x^2+y^2+4-4xy+8x-4y\)

a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2

b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y

=>A-B=12xy^2-14x^2y

c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2

=>A-B=-5x^2y^3-x^3y^2

d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2

Ta có : xy - 2x + 3y - 5 = 0 

<=> x(y - 2) + 3y - 6 + 1 = 0

<=> x(y - 2) + 3(y - 2) + 1 = 0

=> (y - 2) (x + 3) = -1

Suy ra :  (y - 2) (x + 3) thuộc Ư(-1) = {-1;1}

Th1 : nếu y - 2 = -1 thì x + 3 = -1 => y = 1 ; x = -4

Th2 : nếu y - 2 = 1 thì x + 3 = 1 => y = 3 , x = -2 

what the hell???

avatar mèo đen

a: (3x^2-4)(x+3y)

=3x^2*x+3x^2*3y-4x-4*3y

=3x^3+9x^2y-4x-12y

b: (c+3)(x^2+3x)

=c*x^2+c*3x+3x^2+9x

=cx^2+3cx+3x^2+9x

c: (xy-1)(xy+5)

=xy*xy+5xy-xy-5

=x^2y^2+4xy-5

d: (3x+5y)(2x-7y)

=3x*2x-3x*7y+5y*2x-5y*7y

=6x^2-21xy+10xy-35y^2

=6x^2-11xy-35y^2

e: -(x-1)(-x^2+2y)

=(x-1)(x^2-2y)

=x^3-2xy-x^2+2y

f: (-x^2+2y)(x^2+2y)

=(2y)^2-x^4

=4y^2-x^4

a.

\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

TH1:

\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

c.

\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)

Xét pt:

\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)

TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)

\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)

\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)

TH2: \(y=6-x\) thế vào...

\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)

\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)