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1: \(x=\dfrac{3}{4}-\dfrac{2}{4}=\dfrac{1}{4}\)

2: \(x=\dfrac{2}{11}+\dfrac{1}{5}=\dfrac{10+11}{55}=\dfrac{21}{55}\)

b4 

1, (x-y)³

3, [(x-y)(x+y)] - 4(x-y) 

= (x-y) [(x+y) - 4]

Bài 4: 

d: Ta có: \(x^2-y^2-2x-2y\)

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

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

e: Ta có: \(x^3-y^3-3x+3y\)

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

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

a. Xét \(\Delta HAC\) vuông tại H và \(\Delta ABC\) vuông tại A có: \(\widehat{C}\) là góc chung

\(\Rightarrow\)\(\Delta HAC\)\(\sim\)\(\Delta ABC\) (đpcm)

b. Xét \(\Delta ABH\) vuông tại H có HD là đường cao ứng với cạnh huyền

\(\Rightarrow AH^2=AB.AD\) (đpcm)

c. Tương tự xét \(\Delta AHC\) vuông tại H có HE là đường cao ứng với cạnh huyền

\(\Rightarrow AH^2=AC.AE\)

\(\Rightarrow AD.AB=AE.AC\) (đpcm)

2.

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

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

\(c,\left(3x-2y\right)^2-\left(2x-3y\right)^2\\ =\left(3x-2y-2x+3y\right)\left(3x-2y+2x-3y\right)\\ =\left(x+y\right)\left(5x-5y\right)=5\left(x-y\right)\left(x+y\right)\)

\(d,9\left(x-y\right)^2-4\left(x+y\right)^2\\ =\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\)

\(e,\left(4x^2-4x+1\right)-\left(x+1\right)^2\\ =\left(2x-1\right)^2-\left(x+1\right)^2\\ =\left(2x-1-x-1\right)\left(2x-1+x+1\right)\\ =3x\left(x-2\right)\)

\(f,x^3+27=\left(x+3\right)\left(x^2+3x+9\right)\)

\(g,27x^3-0,001=\left(3x-0,1\right)\left(9x^2+0,027x+0,01\right)\)

\(h,125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)

Bài 3 : 

a) \(x^4+2x^2+1=\left(x^2+1\right)^2\)

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

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

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

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

g) \(x^3+6x^2+12x+8=\left(x+2\right)^3\)

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

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

Bài 4: 

a: Ta có: \(4x^2-49=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)

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

b: Ta có: \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x-6=0\)

hay x=6

c: Ta có: \(\dfrac{1}{16}x^2-x+4=0\)

\(\Leftrightarrow\left(\dfrac{1}{4}x-2\right)^2=0\)

\(\Leftrightarrow x\cdot\dfrac{1}{4}-2=0\)

\(\Leftrightarrow x\cdot\dfrac{1}{4}=2\)

hay x=8

d: Ta có: \(x^3-3\cdot x^2\cdot\sqrt{3}+9x-3\sqrt{3}=0\)

\(\Leftrightarrow\left(x-\sqrt{3}\right)^3=0\)

\(\Leftrightarrow x-\sqrt{3}=0\)

hay \(x=\sqrt{3}\)

a: =1/3x7/5=7/15

b: =11/9x1/2=11/18

c: =1/3x3/4=1/4

d: =1/5x4/5=4/25

e: =2/3x1/4=2/12=1/6

f: =1/2x1/3=1/6

g: =1/3x1/2=1/6