b) \(\dfrac{7x-21}{14x-42}=\dfrac{2}{4}\)

\(\Leftrightarrow\dfrac{7\left(x-3\right)}{14\left(x-3\right)}=\dfrac{2}{4}\)

Ở tử và mẫu đều có chung x-3 nên loại

\(\Rightarrow\dfrac{7}{14}=\dfrac{2}{4}\Leftrightarrow\dfrac{2}{4}=\dfrac{2}{4}\) (đpcm)

c) \(\dfrac{9x-18}{18y-54}=\dfrac{2x-4}{4y-12}\)

\(\Leftrightarrow\dfrac{9\left(x-2\right)}{18\left(y-3\right)}=\dfrac{2\left(x-2\right)}{4\left(y-3\right)}\)

Ở tử VT và VP đều có tử là x-2 và mẫu là y-3 nên loại

\(\Leftrightarrow\dfrac{9}{18}=\dfrac{2}{4}\Leftrightarrow\dfrac{1}{2}=\dfrac{1}{2}\) (đpcm)

a: \(\dfrac{\overline{ab}}{\overline{abab}}=\dfrac{10a+b}{1000a+100b+10a+b}=\dfrac{10a+b}{1010a+101b}\)

\(=\dfrac{10a+b}{101\left(10a+b\right)}=\dfrac{1}{101}=\dfrac{3}{303}\)

b: \(\dfrac{7x-21}{14x-42}=\dfrac{7\left(x-3\right)}{14\left(x-3\right)}=\dfrac{7}{14}=\dfrac{2}{4}\)

c: \(\dfrac{9x-18}{18y-54}=\dfrac{9\left(x-2\right)}{18\left(y-3\right)}=\dfrac{x-2}{2\left(y-3\right)}=\dfrac{2\left(x-2\right)}{4\left(y-3\right)}=\dfrac{2x-4}{4y-12}\)

d: Đề sai rồi bạn

1. = 2(a+b)

2. =2(a-b)

3.=2(a+2b-3c)

4.=3(a-2b-3c)

5.=-4(a+2b+3c)

6.=-5(x+2xy+3y)

7.=-7(a+2ab+3b)

8.=6(xy-2x-3y)

9.=8(xy-3y+2x)

10.=9(ab-2a+1)

Lần sau bn cần ghi đề rõ ràng hơn

11.=x(y-1)

12.=a(x+1)

13.=m(x+y+1)

14.=-a(x+y+1)

15.=-a(x²+x+1)

16.=-2a(x+2y)

17.=2a(x-y+1)

18.=2(2ax-ay-2)

19.=5a(1-2x-3)

20.=-2ab(a+2b+3)

1, 2a+2b = 2( a + b )

2, 2a-2b = 2( a - b )

3, 2a+4b-6c = 2( a + 2b - 3c )

4, 3a-6b-9c = 3( a - 2b - 3c )

5, xy-x = x( y - 1 )

6, -5x-10xy-15y = -5( x + 2xy + 3y )

7, -7a-14ab-21b = -7( a + 2ab + 3b )

8, 6xy-12x-18y = 6 ( xy - 2x - 3y )

9, 8xy-24y+16x = 8 ( xy - 3y + 2x )

10, 9ab-18a+9 = 9( ab - 2a + 1 )

11, ax+a = a( x + 1 )

12, mx+my+m = m( x + y + 1 )

13, -ax-ay-a = -a( x + y +1 )

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

15, 2ax-2ay+2a = 2a( x - y + 1 )

16, 4ax-2ay-2 = 2( 2ax - a - 1 )

17, 5a-10ax-15a = 5a( 1 - 2x - 3 ) = 5a( -2x - 2 )

18, 5ax-15ay+20a = 5a( x - 3y + 4 )

19, 3a²x-6a²y+12a = 3a( ax - 2ay + 4 )

20, 2axy-4a²xy²+6a³x\(^2\) = 2ax( y - 2ay\(^2\) + 3a\(^2\)x )

\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)

\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)

\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)

Bài 2;

\(a)x^4-16x=0\Rightarrow x^4=16x\Leftrightarrow x^3=16\Leftrightarrow x=\sqrt[3]{16}\)

\(c)4x^2-\frac{1}{4}=0\Leftrightarrow4x^2=\frac{1}{4}\Leftrightarrow x^2=\frac{1}{16}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)

\(x.\left(x-3\right)-x+3=0\)

\(x.\left(x-3\right)-\left(x-3\right)=0\)

\(\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

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

\(\left(x-1\right)^3=0\)( hằng đẳng thức số 5 )

\(\Rightarrow x=1\)

Vậy \(x=1\)

1)\(21x^2y-12xy^2=xy.\left(21x-12y\right)\)

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

3)\(3x.\left(x-1\right)+7x^2\left(x-1\right)=\left(x-1\right).\left(3x+7x^2\right)=x.\left(x-1\right)\left(3+7x\right)\)

15)\(\left(2a+3\right)^2-\left(2a+1\right)^2=\left(2a+3-2a-1\right)\left(2a+3+2a+1\right)=2.\left(4a+4\right)=8\left(a+1\right)\)

14) \(-4y^2+4y-1=-\left[\left(2y\right)^2-2.2y.1+1^2\right]=-\left(2y-1\right)^2\)

13) \(x^6+1=\left(x^2\right)^3+1=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

12) \(\left(x+1\right)^2-\left(y+6\right)^2=\left(x+1-y-6\right)\left(x+1+y+6\right)=\left(x-y-5\right)\left(x+y+7\right)\)

4) \(3x\left(x-a\right)+4a\left(a-x\right)=3x.\left(x-a\right)-4a\left(x-a\right)=\left(x-a\right)\left(3x-4a\right)\)

Sao nhiều thế!

Đúng là nhiều thật , dù sao cx cảm ơn bn nhìn nha!!!