Giải:

Ta có: \(3\left(x-1\right)=2\left(y-2\right)\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}\)

\(4\left(y-2\right)=3\left(z-3\right)\Rightarrow\frac{y-2}{3}=\frac{z-3}{4}\)

\(\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6+z-3}{4+9+4}=\frac{\left(2x+3y+z\right)-\left(2+6+3\right)}{17}\)

\(=\frac{50-11}{17}=\frac{39}{17}\)

+) \(\frac{x-1}{2}=\frac{39}{17}\Rightarrow x-1=\frac{78}{17}\Rightarrow x=\frac{95}{17}\)

+) \(\frac{y-2}{3}=\frac{39}{17}\Rightarrow y-2=\frac{117}{17}\Rightarrow y=\frac{151}{17}\)

+) \(\frac{z-3}{4}=\frac{39}{17}\Rightarrow z-3=\frac{156}{17}\Rightarrow z=\frac{207}{17}\)

Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{95}{17};\frac{151}{17};\frac{207}{17}\right)\)

a) \(x^3-2x^2+x-xy^2\)

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

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

b) \(x^3-4x^2-12x+27\)

\(=x^3+27-4x^2-12x\)

\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

c) \(x^2-2xy+y^2+1\)

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

d) \(2x^2+4x+2-2y^2\)

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

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

e) \(2xy-x^2-y^2+16\)

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

\(=-\left(x-y-4\right)\left(x-y+4\right)\)

đăng 3 câu /1 lần hỏi thôi

Ta có: x/2=y/3 =>x/8=y/12  (1)

          y/4=z/5 =>y/12=z/15  (2)

Từ 1 và 2 => x/8=y/12=z/15

         => (x/8)²=(y/12)²=z/15

      hay  x²/64=y²/144=z/15

Áp dụng t/c của dãy tỉ số bằng nhau,có

 x²/64=y²/144=z/15=(x² - y²)/(64 - 144)= -16/-80=1/5

Khi đó: x²/64=1/5 => x²=1/5 . 64=64/5

                           =>x=\(\sqrt{\frac{64}{5}}\)

            y²/144=1/5 => y²=144 . 1/5=144/5

                             =>y=\(\sqrt{\frac{144}{5}}\)

            z/15 = 1/5 => z =15 . 1/5=3

  mk lm sai thì thôi nha ^-^

a) ADTCDTSBN

có: \(\frac{x}{12}=\frac{y}{13}=\frac{z}{15}=\frac{x+y+z}{12+13+15}=\frac{160}{40}=4\)

=> x/12 = 4 => x = 48

...

b) ta có: \(x=\frac{y}{6}=\frac{z}{3}=\frac{2x}{2}=\frac{3y}{18}=\frac{4z}{12}\)

ADTCDTSBN

có: \(\frac{2x}{2}=\frac{3y}{18}=\frac{4z}{12}=\frac{2x-3y+4z}{2-18+12}=\frac{16}{-4}=-4\)

=>...

c) ta có: \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{3}=\frac{2x}{4}=\frac{3y}{-9}=\frac{2z}{8}\)

ADTCTDBN

có: \(\frac{2x}{4}=\frac{3y}{-9}=\frac{2z}{8}=\frac{2x+3y+2z}{4-9+8}=\frac{1}{3}\)

=>...

a) Ta có: \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)

\(=\left[\left(3-xy^2\right)-\left(2+xy^2\right)\right]\cdot\left[\left(3-xy^2\right)+\left(2+xy^2\right)\right]\)

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

\(=5\cdot\left(1-2xy^2\right)\)

\(=5-10xy^2\)

b) Ta có: \(9x^2-\left(3x-4\right)^2\)

\(=\left[3x-\left(3x-4\right)\right]\left[3x+\left(3x-4\right)\right]\)

\(=\left(3x-3x+4\right)\cdot\left(3x+3x-4\right)\)

\(=4\cdot\left(6x-4\right)\)

\(=24x-16\)

c) Ta có: \(\left(a-b^2\right)\left(a+b^2\right)\)

\(=a^2-b^4\)

d) Ta có: \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)

\(=\left(a^2+2a\right)^2-9\)

\(=a^4+4a^3+4a^2-9\)

e) Ta có: \(\left(x-y+6\right)\left(x+y-6\right)\)

\(=x^2+xy-6x-yx-y^2+6y+6x+6y-36\)

\(=x^2-y^2+12y-36\)

f) Ta có: \(\left(y+2z-3\right)\left(y-2z-3\right)\)

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

\(=y^2-6y+9-4z^2\)

g) Ta có: \(\left(2y-5\right)\left(4y^2+10y+25\right)\)

\(=\left(2y\right)^3-5^3\)

\(=8y^3-125\)

h) Ta có: \(\left(3y+4\right)\left(9y^2-12y+16\right)\)

\(=\left(3y\right)^3+4^3\)

\(=27y^3+64\)

i) Ta có: \(\left(x-3\right)^3+\left(2-x\right)^3\)

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

\(=x^3-9x^2+27x-27-\left(x^3-6x^2+12x-8\right)\)

\(=x^3-9x^2+27x-27-x^3+6x^2-12x+8\)

\(=-3x^2+15x-19\)

j) Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=\left[\left(x+y\right)-\left(x-y\right)\right]\cdot\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

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

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

\(=6x^2y+2y^3\)