Đề bài yêu cầu gì vậy em.

a) Tìm được A = (x- y)(x + 5y).

Thay x = 4 và y = -4 vào A tìm được A = -128.

b) Tìm được B = 9 ( x   - 1 ) 2 .

Thay x = - 4 vào B tìm được B = 81 4 .  

c) Tìm được C = (x - y)(y - z)(x - z).

Thay x = 6,y = 5 và z = 4 vào C tìm được C = 2.

d) Thay 10 = x +1 vào D và biến đổi ta được D = -1.

Bài 2:

\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)

\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)

bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc

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

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

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

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

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

\(=\left(z-x\right)\left[y^2\left(z^2+zx+x^2\right)-y^3\left(z+x\right)-z^2x^2\right]\)

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

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

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

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

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

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

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

\(=\left(z-x\right)\left(z-y\right)\left[\left(y^2z-x^2z\right)-\left(x^2y-xy^2\right)\right]\)

\(=\left(z-x\right)\left(z-y\right)\left[z\left(y^2-x^2\right)-xy\left(x-y\right)\right]\)

\(=\left(z-x\right)\left(z-y\right)\left[z\left(y-x\right)\left(y+x\right)+xy\left(y-x\right)\right]\)

\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left[z\left(y+x\right)+xy\right]\)

\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left(yz+xz+xy\right)\)

2) \(xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)

\(=xyz-xy-yz-xz+x+y+z-1\)

\(=\left(xyz-xy\right)-\left(yz-y\right)-\left(xz-x\right)+\left(z-1\right)\)

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

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

\(=\left(z-1\right)\left[\left(xy-y\right)-\left(x-1\right)\right]\)

\(=\left(z-1\right)\left[y\left(x-1\right)-\left(x-1\right)\right]\)

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

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x.y}{2.3}=\dfrac{54}{6}=9\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=81\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm9\end{matrix}\right.\)

b) \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{3}\right)^2=\dfrac{x^2-y^2}{5^2-3^2}=\dfrac{4}{16}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{4}\\y^2=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{5}{2}\\y=\pm\dfrac{3}{2}\end{matrix}\right.\)

c: Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)

nên \(\dfrac{x}{10}=\dfrac{y}{15}\)

Ta có: \(\dfrac{y}{5}=\dfrac{z}{7}\)

nên \(\dfrac{y}{15}=\dfrac{z}{21}\)

mà \(\dfrac{x}{10}=\dfrac{y}{15}\)

nên \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{92}{46}=2\)

Do đó: x=20; y=30; z=42

a) Ta có: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)

nên \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

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

Do đó:

\(\left\{{}\begin{matrix}\dfrac{2x}{3}=12\\\dfrac{3y}{4}=12\\\dfrac{4z}{5}=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=36\\3y=48\\4z=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=20\end{matrix}\right.\)

Vậy: (x,y,z)=(18;16;20)

b) Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=3k\end{matrix}\right.\)

Ta có: \(x^2-y^2=4\)

\(\Leftrightarrow\left(5k\right)^2-\left(3k\right)^2=4\)

\(\Leftrightarrow16k^2=4\)

\(\Leftrightarrow k\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

Trường hợp 1: \(k=\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=5\cdot\dfrac{1}{2}=\dfrac{5}{2}\\y=3k=3\cdot\dfrac{1}{2}=\dfrac{3}{2}\end{matrix}\right.\)

Trường hợp 2: \(k=-\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=5\cdot\dfrac{-1}{2}=\dfrac{-5}{2}\\y=3k=3\cdot\dfrac{-1}{2}=\dfrac{-3}{2}\end{matrix}\right.\)

Vậy: \(\left(x,y\right)\in\left\{\left(\dfrac{5}{2};\dfrac{3}{2}\right);\left(-\dfrac{5}{2};-\dfrac{3}{2}\right)\right\}\)

a)

Theo tính chất của dãy tỉ số bằng nhau, ta có : 

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

Suy ra : 

\(x=\dfrac{12.3}{2}=18\\ y=\dfrac{12.4}{3}=16\\ z=\dfrac{12.5}{4}=15\)

b)

\(x=\dfrac{y}{3}.5=\dfrac{5y}{3}\\ x^2-y^2=4\\ \Leftrightarrow\left(\dfrac{5y}{3}\right)^2-y^2=4\\ \Leftrightarrow\dfrac{16y^2}{9}=4\Leftrightarrow y=\pm\dfrac{3}{2} \)

Với $y = \dfrac{3}{2}$ thì $x = \dfrac{5}{2}$

Với $y = \dfrac{-3}{2}$ thì $x = \dfrac{-5}{2}$

c)

\(\dfrac{x}{y+z+1}=\dfrac{y}{z+x+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2x+2y+2z}=\dfrac{1}{2}\)

Suy ra : 

\(2x=y+z+1\Leftrightarrow y+z=2x-1\)

Mặt khác : 

\(x+y+z=\dfrac{1}{2}\Leftrightarrow x+2x-1=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(2y=x+z+1=z+\dfrac{3}{2}\)

Mà \(y+z=0\Leftrightarrow z=-y\)

nên suy ra:  \(y=\dfrac{1}{2};z=-\dfrac{1}{2}\)

bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

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