1:

Biểu thức toán học: \(\frac{a+b}{a-b}\)

Biểu thức pascal: (a+b)/(a-b)

2:

Biểu thức toán học: \(S=pi.r^2\)

Biểu thức pascal: S=pi*sqr(r)

3:

Biểu thức toán học: \(V=\sqrt{2}GH\)

Biểu thức pascal: V=sqrt(2)*g*h

4:

Biểu thức toán học: \(\frac{\frac{4x^2+2y}{2-3a}}{4a+b}\)

Biểu thức pascal: (\(4\cdot x^2+2\cdot y\))/(2-3*a)/(4*a+b)

5:

Biểu thức toán học: \(\sqrt{3a+b}>5\left(a+b\right)^2\)

Biểu thức pascal:\(\sqrt{3\cdot a+b}>5\cdot\left(a+b\right)^2\)

6:

Biểu thức toán học: \(\frac{\frac{5a^2+b}{6-5a}}{6a+b}\)

Biểu thức pascal: (5*sqr(a)+b)/(6-5*a)/(6*a+b)

7:

Biểu thức toán học: \(\left|a+b\right|>0\)

Biểu thức pascal: abs(a+b)>0

8:

Biểu thức toán học: \(sin\left(x^2\right)+cos\left(x^2\right)=1\)

Biểu thức pascal: sin(sqr(x))+cos(sqr(x))=1

9:

Biểu thức toán học: \(\frac{x+y}{2z}\)

Biểu thức pascal: (x+y)/(2*z)

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

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

3, \(4x\left(a-b\right)+6xy\left(b-a\right)=\left(4x-6xy\right)\left(a-b\right)=2x\left(2-3y\right)\left(a-b\right)\)

4, \(y\left(a-b\right)-x\left(b-a\right)=\left(x+y\right)\left(a-b\right)\)

5, \(6x\left(x-y\right)+8y\left(y-x\right)=\left(x-y\right)\left(6x-8y\right)=2\left(3x-4y\right)\left(x-y\right)\)

6, \(4\left(x-3\right)^2-2x\left(x-3\right)=\left(x-3\right)\left[4\left(x-3\right)-2x\right]=2\left(x-3\right)\left(x-6\right)\)

Câu hỏi là jv bn

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

2.\(a^2\left(x-5\right)-3\left(x-5\right)=\left(x-5\right)\left(a^2-3\right)\)

3.\(4x\left(a-b\right)+6xy\left(b-a\right)=4x\left(a-b\right)-6xy\left(a-b\right)=2x\left(a-b\right)\left(2-3y\right)\)

4.\(y\left(a-b\right)-x\left(b-a\right)=y\left(a-b\right)+x\left(a-b\right)=\left(c-b\right)\left(x+y\right)\)

5.\(6x\left(x-y\right)+8y\left(y-x\right)=6x\left(x-y\right)-8y\left(x-y\right)=2\left(x-y\right)\left(3x-4y\right)\)

6.\(4\left(x-3\right)^2-2x\left(x-3\right)=\left(x-3\right)\left[4\left(x-3\right)-2x\right]=\left(x-3\right)\left(4x-12-2x\right)\)

\(=\left(x-3\right)\left(2x-12\right)=2\left(x-3\right)\left(x-6\right)\)

\(\dfrac{4}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)

\(\Rightarrow\dfrac{4}{x}-\dfrac{2y}{6}=\dfrac{1}{6}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{1+2y}{6}\)

\(\Rightarrow24=x\left(1+2y\right)\)

\(\Rightarrow x;1+2y\inƯ\left(24\right)\)

\(Ư\left(24\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm8;\pm12;\pm24\right\}\)

Mà 1+2y lẻ nên:

\(\left\{{}\begin{matrix}1+2y=1\Rightarrow2y=0\Rightarrow y=0\\x=24\\1+2y=-1\Rightarrow2y=-2\Rightarrow y=-1\\x=-24\end{matrix}\right.\)

\(\left\{{}\begin{matrix}1+2y=3\Rightarrow2y=2\Rightarrow y=1\\x=8\\1+2y=-3\Rightarrow2y=-4\Rightarrow y=-2\\x=-8\end{matrix}\right.\)

1)\(4\left(a^4-1\right)x=5\left(a-1\right)\)

<=>x=\(\frac{5\left(a-1\right)}{a^4-1}\)

<=>x=\(\frac{5\left(a-1\right)}{\left(a-1\right)\left(a+1\right)\left(a^2+1\right)}=\frac{5}{\left(a+1\right)\left(a^2+1\right)}\)

Tương tự ta tính được y=\(\frac{4a^6+4}{5a^4-5a^2+5}\)

Suy ra x.y=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\cdot\left(a^6+1\right)}{5\left(a^4-a^2+1\right)}\)=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\left(a^2+1\right)\left(a^4-a^2+1\right)}{5\left(a^4-a^2+1\right)}\)

=\(\frac{5}{a+1}\)

Tương tự với x:y

\(A=\frac{4.6}{4.2}:\left(\frac{8.10}{6.8}.\frac{12.14}{10.12}.\frac{16.18}{14.16}...\frac{54.56}{54.53}\right)=\frac{6}{2}:\frac{56}{6}=\)

2b/\(\frac{5\left(a-b\right)^2}{2\left(a^2-ab+b^2\right)}:\frac{8\left(a-b\right)}{10\left(a+b\right)\left(a^2-ab+b^2\right)}=x\)

\(\Leftrightarrow x=\frac{50\left(a-b\right)^2\left(a+b\right)\left(a^2-ab+b^2\right)}{16\left(a^2-ab+b^2\right)\left(a-b\right)}\)

\(\Leftrightarrow x=\frac{25\left(a-b\right)}{8}\)

1, 5a²xy-10a³x-15ay = 5a( axy - 2a\(^2\)x - 3y )

2, mxy-m²x+my = m( xy - mx + y )

3, 2mx-4m²xy+6mx = 2mx( 1 - 2my + 3 ) = 2mx( 4 - 2my )

4, a²b-2ab²+ab = ab( a - 2b + 1 )

5, 5a²b-2ab²+ab = ab( 5a - 2b +1 )

6, -3x²y³-6x³y²-x²y² = -3x\(^2\)y\(^2\) ( y + 2x + 1 )

7, 5x²y⁴-10x⁴y²+5x²y² = 5x\(^{^2y^2}\)( y\(^2\) - 2x\(^2\) + 1 )

8, -2x³y⁴-4x⁴y³+2x³y³ = 2\(x^3y^3\) ( -y - 2x + 1 )

9, 4x³y²-8x³y+16xy²-24 = 4( x\(^3\)y\(^2\) - 2x\(^3\)y + 4 xy\(^2\) - 6 )

10, 12x³y-6xy+3x = 3x( 4x\(^2\)y - 2y + 1 )

11, 2(x-y)-a(x-y) = ( 2 - a ) ( x - y )

12, a(x-y)+b(x-y)= ( a + b ) ( x - y )

13, m(x+y)-n(x+y) = ( m - n ) ( x + y )

14, 2a(x+y)-4(x+y) = ( 2a - 4 )( x + y ) = 2( a - 2 ) ( x + y )

15, 3a(x+y)-6ab(x+y) = ( 3a - 6ab )( x + y ) = 3a( 1 - 2b ) ( x + y )

16, 5a²(x-y)+10a(x-y) = ( 5a\(^2\)+10a )( x - y ) = 5a( a + 2 ) ( x - y )

17, -2ab(x-y)-4a(x-y) = ( -2ab - 4a )( x - y ) = -2a( b + 2 )( x - y )

18, 3a(x-y)+2(x-y) = ( 3a + 2 ) ( x - y )

19, m(a-b)-m²(a-b) = ( m - m\(^2\) ) ( a - b ) = m( 1 - m ) ( a - b )

20, mx(a+b)-m(a+b) = ( mx - m ) ( a + b ) = m( x - 1 )( a + b )

21, x(a-b)-y(b-a) = x( a - b ) + y( a - b ) = ( x + y ) ( a - b )

22, ab(x-5)-a²(5-x) = ab( x - 5 ) + a\(^2\)( x - 5 ) = ( ab + a\(^2\) ) ( x - 5 ) = a( b + a )( x - 5 )

23, 2a²(x-y)-4a(y-x)= 2a\(^2\)( x - y ) + 4a( x - y )=( 2a\(^2\) + 4a ) ( x - y )= 2a( a + 2 )( x - y )

Đăng ít thôi =))

a. \(5a^2xy-10a^3x-15ay=5a\left(axy-2a^2x-3y\right)\)

b. \(mxy-m^2x+my=m\left(xy-mx+y\right)\)

c. \(2mx-4m^2xy+6mx=2mx\left(1-2my+3\right)=2mx\left(-2my+4\right)\)

d. \(a^2b-2ab^2+ab=ab\left(a-2b+1\right)\)

e. \(5a^2b-2ab^2+ab=ab\left(5a-2b+1\right)\)

g.

Mệt :v

\(a.A=\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x+y\right)\sqrt{3}}{2}=\dfrac{\sqrt{3}}{x-y}\) ( x # y )

\(b.\dfrac{1}{2x-1}.\sqrt{5a^4\left(1-4x+4a^2\right)}=\dfrac{1}{2a-1}.\left(2a-1\right)a^2\sqrt{5}=a^2\sqrt{5}\) ( a # \(\dfrac{1}{2}\) )