a)\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Leftrightarrow\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}\)

Áp dụng tc dãy tỉ 

\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)

Với \(\frac{x}{\frac{3}{2}}=12\Rightarrow x=18\)

Với \(\frac{y}{\frac{4}{3}}=12\Rightarrow y=16\)

Với \(\frac{z}{\frac{5}{4}}=12\Rightarrow z=15\)

b)\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Leftrightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}\)

Áp dụng tc dãy tỉ

\(\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4\)

Với \(\frac{a^2}{4}=4\Rightarrow a=4\)

Với \(\frac{b^2}{9}=4\Rightarrow b=6\)

Với \(\frac{2c^2}{32}=4\Rightarrow c=8\)

x=18 

y=16 

z = 15

a)\(\left(x+y\right)^2:\left(x+y\right)=x+y\)

b)\(\left(x-y\right)^5:\left(y-x\right)^4=\left(x-y\right)^5:\left(x-y\right)^4=x-y\)

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

d)\(\left(x^3y^3-\frac{1}{2}x^2y^3+x^3y^2\right):\frac{1}{2}x^2y^2=2xy-y+x\)

1 a

2c

3b

4d

5c

6c

a, \(x^2\) + 6x + 5 = 0
=>\(x^2\) + x + 5x +5 = 0
=>x(x + 1) + 5(x + 1) = 0
=>(x + 1)(x + 5) = 0
=> x + 1 =0 hoặc x + 5 =0
=> x = -1 hoặc x = -5

c) \(\dfrac{x+3}{x-1}+\dfrac{2x+5}{x-1}+\dfrac{14-3x}{1-x}\)

\(=\dfrac{x+3}{x-1}+\dfrac{2x+5}{x-1}-\dfrac{14-3x}{x-1}\)

\(=\dfrac{x+3+2x+5-14+3x}{x-1}\)

\(=\dfrac{6x-6}{x-1}\)

\(=\dfrac{6\left(x-1\right)}{x-1}\)

\(=6.\)

d) \(\dfrac{3x}{2y-2x}+\dfrac{3y}{x+y}+\dfrac{3y\left(3y-x\right)}{2\left(x^2-y^2\right)}\)

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

\(=-\dfrac{3x\left(x+y\right)+6y\left(x-y\right)+3y\left(3y-x\right)}{2\left(x-y\right)\left(x+y\right)}\)

\(=-\dfrac{3x^2+3xy+6xy-6y^2+9y^2-3xy}{2\left(x-y\right)\left(x+y\right)}\)

\(=-\dfrac{3x^2+6xy+3y^2}{2\left(x-y\right)\left(x+y\right)}\)

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

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

\(=-\dfrac{3\left(x+y\right)}{2\left(x-y\right)}\).

f: \(=x^2-9-x^2+6x-9=6x-18\)

           Bài làm :

 \(\text{a)}9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

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

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

\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)

 \(\text{b)}3x^4y^2+3x^3y^2+3xy^2+3y^2\)

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

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

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

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

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

 \(\text{c)}\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)

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

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

\(d ) x^3+3x^2+3x+1-27z^3\)

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

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

A = 2x² - 6xy - 3xy - 6y - 2x² + 8xy + 6y

   = - xy

  = \(\frac{2}{3}\)\(x\)\(\frac{3}{4}\)

  = \(\frac{1}{2}\)

mk đang bận mấy câu kia tương tự nha