Bài 3:

a, (\(x\)+y+z)²

=((\(x\)+y) +z)²

= (\(x\) + y)² + 2(\(x\) + y)z + z²

= \(x^2\) + 2\(xy\) + y² + 2\(xz\) + 2yz + z²

=\(x^2\) + y² + z² + 2\(xy\) + 2\(xz\) + 2yz

b, (\(x-y\))(\(x^2\) + y² + z² - \(xy\) - yz - \(xz\))

= \(x^3\) + \(xy^2\) + \(xz^2\) - \(x^2\)y - \(xyz\) - \(x^2\)z - y³ 

Đến dây ta thấy xuất hiện \(x^3\) - y³ khác với đề bài, em xem lại đề bài nhé

c,

(\(x\) + y + z)³ 

=(\(x\) + y)³ + 3(\(x\) + y)²z + 3(\(x\)+y)z² + z³

= \(x^3\) + 3\(x^2\)y + 3\(xy^{2^{ }}\) + y³ +  3(\(x\)+y)z(\(x\) + y + z) + z³

= \(x^3\) + y³ + z³ + 3\(xy\)(\(x\) + y) + 3(\(x+y\))z(\(x+y+z\))

= \(x^3\) + y³ + z³ + 3(\(x\) + y)( \(xy\) + z\(x\) + yz + z²)

= \(x^3\) + y³ + z³ + 3(\(x\) + y){(\(xy+xz\)) + (yz + z²)}

= \(x^3\) + y³ + z³ + 3(\(x\) + y){ \(x\)( y +z) + z(y+z)}

= \(x^3\) + y³ + z³ + 3(\(x\) + y)(y+z)(\(x+z\)) (đpcm)

Đề bài yêu cầu gì vậy em.

Biến đổi tương đương nhé bạn.

a: Ta có: \(\left(x+y\right)^2\)

\(=x^2+2xy+y^2\)

\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)

a: \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)

=>\(\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{4}\right)^3=\left(\dfrac{z}{6}\right)^3\)

=>\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)

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

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

=>x=k; y=2k; z=3k

\(x^2+y^2+z^2=14\)

=>\(k^2+4k^2+9k^2=14\)

=>\(14k^2=14\)

=>\(k^2=1\)

=>k=1 hoặc k=-1

TH1: k=1

=>\(x=k=1;y=2k=2\cdot1=2;z=3k=3\cdot1=3\)

TH2: k=-1

=>\(x=k=-1;y=2k=2\cdot\left(-1\right)=-2;z=3k=3\cdot\left(-1\right)=-3\)

b: \(\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\)

=>\(\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{3}\right)^3=\left(\dfrac{z}{4}\right)^3\)

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

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

=>x=2k; y=3k; z=4k

\(x^2+2y^2-3z^2=-650\)

=>\(\left(2k\right)^2+2\cdot\left(3k\right)^2-3\cdot\left(4k\right)^2=-650\)

=>\(4k^2+18k^2-3\cdot16k^2=-650\)

=>\(-26\cdot k^2=-650\)

=>\(k^2=25\)

=>\(\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)

TH1: k=5

=>\(x=2\cdot5=10;y=3\cdot5=15;z=4\cdot5=20\)

TH2: k=-5

=>\(x=2\cdot\left(-5\right)=-10;y=3\cdot\left(-5\right)=-15;z=4\cdot\left(-5\right)=-20\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)

Ta có: \(x^2-y^2+2z^2=108\)

\(\Leftrightarrow\left(2k\right)^2-\left(3k\right)^2+2\cdot\left(4k\right)^2=108\)

\(\Leftrightarrow4k^2-9k^2+2\cdot16k^2=108\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot2=4\\y=3k=3\cdot2=6\\z=4k=4\cdot2=8\end{matrix}\right.\)

Trường hợp 2: k=-2

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot\left(-2\right)=-4\\y=3k=3\cdot\left(-2\right)=-6\\z=4k=4\cdot\left(-2\right)=-8\end{matrix}\right.\)

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Đề bài yêu cầu gì vậy bạn?