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8 tháng 10 2020

a, Trừ vế theo vế hai phương trình ta được

\(x^2+6y-y^2-6x=0\)

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

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

Nếu \(x=y,pt\left(1\right)\Leftrightarrow x^2+x=5x+3\)

\(\Leftrightarrow x^2-4x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y=2+\sqrt{7}\\x=y=2-\sqrt{7}\end{matrix}\right.\)

Nếu \(x=6-y,pt\left(2\right)\Leftrightarrow y^2+6-y=5y+3\)

\(\Leftrightarrow y^2-6y+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=3+\sqrt{6}\\y=3-\sqrt{6}\end{matrix}\right.\)

\(y=3+\sqrt{6}\Rightarrow x=3-\sqrt{6}\)

\(y=3-\sqrt{6}\Rightarrow x=3+\sqrt{6}\)

8 tháng 10 2020

b, Trừ vế theo vế hai phương trình

\(3x^3-3y^3=y^2-x^2\)

\(\Leftrightarrow3\left(x-y\right)\left(x^2+xy+y^2+x+y\right)=0\)

Từ \(pt\left(1\right)\) \(3x^3=y^2+2>0\Rightarrow x>0\)

Tương tự \(y>0\)

\(\Rightarrow x^2+xy+y^2+x+y>0,\forall x;y\)

\(\Rightarrow x=y\)

\(pt\left(1\right)\Leftrightarrow3x^3=x^2+2\)

\(\Leftrightarrow3x^3-x^2-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x^2+2x+2\right)=0\)

\(\Leftrightarrow x=y=1\left(\text{vì }3x^2+2x+2=2x^2+\left(x+1\right)^2+1>0\right)\)

22 tháng 12 2023

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

\(=3xy\left(x^2y-2\right)\)

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

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

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

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

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

\(=\left(2x-1\right)^2-\left(3y\right)^2\)

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

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-2\right)\)

7 tháng 11 2017

1)

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)

d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)

e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)

f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)

\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)

2)

a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)

b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)

c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )

d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)

\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)

e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)

\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)

25 tháng 7 2021

a, mình nghĩ đề là cm đẳng thức nhé 

\(VT=\left(5x^4-3x^3+x^2\right):3x^2=\frac{5x^4}{3x^2}-\frac{3x^3}{3x^2}+\frac{x^2}{3x^2}=\frac{5}{3}x^2-x+\frac{1}{3}=VP\)

Vậy ta có đpcm 

b, \(VT=\left(5xy^2+9xy-x^2y^2\right):\left(-xy\right)=\frac{5xy^2}{-xy}+\frac{9xy}{-xy}-\frac{x^2y^2}{-xy}\)

\(=-5y-9+xy=VP\)

Vậy ta có đpcm 

c, \(VT=\left(x^3y^3-x^2y^3-x^3y^2\right):x^2y^2=\frac{x^3y^3}{x^2y^2}-\frac{x^2y^3}{x^2y^2}-\frac{x^3y^2}{x^2y^2}=xy-y-x=VP\)

Vậy ta có đpcm 

20 tháng 3 2019

b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

20 tháng 3 2019

caau a) binh phuong len ra no x=y tuong tu

11 tháng 7 2019

\(a,3x^3y^3-15x^2y^2=3x^2y^2\left(xy-5\right)\)

\(b,5x^3y^2-25x^2y^3+40xy^4\)

\(=5xy^2\left(x^2-5xy+8y^2\right)\)

\(c,-4x^3y^2+6x^2y^2-8x^4y^3\)

\(=-2x^2y^2\left(2x-3+4x^2y\right)\)

\(d,a^3x^2y-\frac{5}{2}a^3x^4+\frac{2}{3}a^4x^2y\)

\(=a^3x^2\left(y-\frac{5}{2}x^2+\frac{2}{3}ay\right)\)

\(e,a\left(x+1\right)-b\left(x+1\right)=\left(x+1\right)\left(a-b\right)\)

\(f,2x\left(x-5y\right)+8y\left(5y-x\right)\)

\(=2x\left(x-5y\right)-8y\left(x-5y\right)=\left(x-5y\right)\left(2x-8y\right)\)

\(g,a\left(x^2+1\right)+b\left(-1-x^2\right)-c\left(x^2+1\right)\)

\(=\left(x^2+1\right)\left(a-b-c\right)\)

\(h,9\left(x-y\right)^2-27\left(y-x\right)^3\)

\(=9\left(x-y\right)^2+27\left(x-y\right)^3\)

\(=9\left(x-y\right)^2\left(1+3x-3y\right)\)

11 tháng 7 2019

a,3x3y3−15x2y2=3x2y2(xy−5)

b,5x3y2−25x2y3+40xy4

=5xy2(x2−5xy+8y2)

c,−4x3y2+6x2y2−8x4y3

=−2x2y2(2x−3+4x2y)

d,a3x2y−52a3x4+23a4x2y

=a3x2(y−52x2+23ay)

e,a(x+1)−b(x+1)=(x+1)(a−b)

f,2x(x−5y)+8y(5y−x)

=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)

g,a(x2+1)+b(−1−x2)−c(x2+1)

=(x2+1)(a−b−c)

h,9(x−y)2−27(y−x)3

18 tháng 8 2016

bài này đặt ẩn đi nhìn hệ to quá cx ngại

18 tháng 8 2016

dung ham dac trung do'