a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

b: =(1-2x)(1+2x)

c: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)

d: =(x+3)^3

e: \(=\left(2x-y\right)^3\)

f: =(x+2y)(x^2-2xy+4y^2)

a) \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)

b) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)

c) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4}{25}y^2\)

d) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot\left(y^2\right)^2+\left(y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)

e) \(\left(3x^2-2y\right)^2=\left(3x^2\right)^2-2\cdot3x^2\cdot2y+\left(2y\right)^2=9x^4-12x^2y+4y^2\)

f) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)

g) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)

\(\)a: \(\left(x-2y\right)^3\)

\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)

\(=x^3-6x^2y+12xy^2-8y^3\)

b: \(\left(2x+y\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)

\(=8x^3+12x^2y+6xy^2+y^3\)

c: \(\left(\dfrac{1}{3}x-1\right)^3=\left(\dfrac{1}{3}x\right)^3-3\cdot\left(\dfrac{1}{3}x\right)^2\cdot1+3\cdot\dfrac{1}{3}x\cdot1^2-1^3\)

\(=\dfrac{1}{27}x^3-\dfrac{1}{3}x^2+x-1\)

d: \(\left(x+\dfrac{1}{3}y\right)^3\)

\(=x^3+3\cdot x^2\cdot\dfrac{1}{3}y+3\cdot x\cdot\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)

\(=x^3+x^2y+\dfrac{1}{3}xy^2+\dfrac{1}{27}y^3\)

e: (2x-3y)³

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot3y+3\cdot2x\cdot\left(3y\right)^2-\left(3y\right)^3\)

\(=8x^3-36x^2y+54xy^2-27y^3\)

f: \(\left(x^2-2y\right)^3\)

\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2y+3\cdot x^2\cdot\left(2y\right)^2-\left(2y\right)^3\)

\(=x^6-6x^4y+12x^2y^2-8y^3\)

g: \(\left(\dfrac{1}{2}x-y\right)^3=\left(\dfrac{1}{2}x\right)^3-3\cdot\left(\dfrac{1}{2}x\right)^2\cdot y+3\cdot\dfrac{1}{2}x\cdot y^2-y^3\)

\(=\dfrac{1}{8}x^3-\dfrac{3}{4}x^2y+\dfrac{3}{2}xy^2-y^3\)

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

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x+2y}{7+2\cdot5}=\dfrac{51}{17}=3\)

Do đó: x=21; y=15

a) \(\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{2y}{10}=\dfrac{x+2y}{7+10}=\dfrac{51}{17}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.7=21\\y=3.5=15\end{matrix}\right.\)

b) \(\dfrac{x}{5}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{9}\\y^2=\dfrac{16}{9}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{4}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\end{matrix}\right.\)

c) \(\dfrac{x}{y}=\dfrac{2}{5}\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

\(\Rightarrow xy=10k^2=40\Rightarrow k=\pm2\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-10\end{matrix}\right.\end{matrix}\right.\)

\(a,5x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x+2y}{7+10}=\dfrac{51}{17}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x=21\\y=15\end{matrix}\right.\)

\(b,\) Áp dụng t/c dtsbn:

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{1}{9}\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{9}\\y=\dfrac{4}{9}\end{matrix}\right.\)

\(c,\dfrac{x}{y}=\dfrac{2}{5}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\Leftrightarrow x=2k;y=5k\)

\(xy=40\Leftrightarrow10k^2=40\Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4;y=10\\x=-4;y=-10\end{matrix}\right.\)

Bài `10`

`a,` Ta có : `x/2=y/3=>(4x)/8 =(3y)/9`

ADTC dãy tỉ số bằng nhau ta có :

`(4x)/8 =(3y)/9=(4x-3y)/(8-9)=(-2)/(-1)=2`

`=> x/2=2=>x=2.2=4`

`=>y/3=2=>y=2.3=6`

`b,` Ta có : `2x=5y=>x/5=y/2`

ADTC dãy tỉ số bằng nhau ta có :

`x/5=y/2=(x+y)/(5+2)=-42/7=-6`

`=>x/5=-6=>x=-6.5=-30`

`=>y/2=-6=>y=-6.2=-12`

Bài `11`

`a,` Ta có : `x/3=y/4=z/6=>x/3=(2y)/8 =(3z)/18`

ADTC dãy tỉ số bằng nhau ta có :

`x/3=(2y)/8=(3z)/18=(x+2y-3z)/(3+8-18)=(-14)/(-7)=2`

`=>x/3=2=>x=2.3=6`

`=>y/4=2=>y=2.4=8`

`=>z/6=2=>z=2.6=12`

Bạn đăng lại `2` câu sau nhe , mình ko hiểu `x=y-z` với `15x-5y=3x=45`

`d,` Ta có :

`x/2=y/3=>x/4=y/6`

`y/2=z/3=>y/6=z/9`

`-> x/4=y/6=z/9=>x/4=(2y)/12 =(3z)/27`

ADTC dãy tỉ số bằng nhau ta có :

`x/4=(2y)/12=(3z)/27=(x-2y+3z)/(4-12+27)=19/19=1`

`=>x/4=1=>x=1.4=4`

`=>y/6=1=>y=1.6=6`

`=>z/9=1=>z=1.9=9`

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé