ta có: \(\left(x-y+z\right)^2-\left(2x-y+1\right)^2-\left(y-z+2\right)^2\) \(+\left(2x-1\right)^2\) \(-3\left(2y-3z\right)^2\)

= \(x^2+y^2+z^2-2xy-2yz+2xz-4x^2-y^2-1+4xy\) \(+2y-4x\) \(-y^2-z^2-4+2yz+4z-4y\) \(+4x^2-4x+1\) \(-12y^2+36yz-27z^2\)

= \(x^2-13y^2-27z^2+2xz+2xy+36yz+4z-2y\) \(-8x-4\)

Đặt \(\hept{\begin{cases}\frac{1}{x^2}=a\\\frac{1}{y^2}=b\\\frac{1}{z^2}=c\end{cases}}\Rightarrow abc=1\) và ta cần chứng minh 

\(\frac{1}{2a+b+3}+\frac{1}{2b+c+3}+\frac{1}{2c+a+3}\le\frac{1}{2}\left(1\right)\)

Áp dụng BĐT AM-GM ta có: 

\(2a+b+3=\left(a+b\right)+\left(a+1\right)+2\ge2\left(\sqrt{ab}+\sqrt{a}+2\right)\)

\(\Rightarrow\frac{1}{2a+b+3}\le\frac{1}{2\left(\sqrt{ab}+\sqrt{a}+1\right)}=\frac{1}{2}\cdot\frac{1}{\sqrt{ab}+\sqrt{a}+1}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\frac{1}{2b+c+3}\le\frac{1}{2}\cdot\frac{1}{\sqrt{bc}+\sqrt{b}+1};\frac{1}{2c+a+3}\le\frac{1}{2}\cdot\frac{1}{\sqrt{ac}+\sqrt{c}+1}\)

Cộng theo vế 3 BĐT trên ta có: 

\(VT_{\left(1\right)}\le\frac{1}{2}\left(\frac{1}{\sqrt{ab}+\sqrt{a}+1}+\frac{1}{\sqrt{b}+\sqrt{bc}+1}+\frac{1}{\sqrt{c}+\sqrt{ac}+1}\right)\le\frac{1}{2}=VP_{\left(2\right)}\left(abc=1\right)\)

t nghĩ ôg có chút nhầm lẫn , phải là sigma (1/2b+a+3) </ 1/2 

1) ADTCDTSBN, ta có:

 \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)= \(\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}\)= 4

* \(\frac{x}{3}=4\)=> x = 3 . 4 = 12

- \(\frac{y}{4}=4\)=> y = 4 . 4 = 16

* \(\frac{z}{5}=4\)=> z = 5 . 4 = 20

Vậy x = 12

       y = 16

       z = 20

x=12

y=16

z=20

đừng nên dựa vào trang này quá 

bài trên thuộc dạng SGK , SBT mà không làm được à

a, Theo đề bài ta có :

a, Theo đề bài ta có :\(\frac{x}{3}\)=\(\frac{y}{5}\)=\(\frac{z}{\left(-2\right)}\)

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

\(\frac{x}{3}\)=\(\frac{y}{5}\)=\(\frac{z}{\left(-2\right)}\)=\(\frac{5x}{15}\)=\(\frac{3z}{\left(-6\right)}\)=\(\frac{5x-y+3z}{15-5+\left(-6\right)}\)=\(\frac{124}{4}\)= 31 (Vì \(5x-y+3z=124\))

Suy ra : \(x=31\times3=93\)

 \(y=31\times5=155\)

\(z=31\times\left(-2\right)=-62\)

Vậy .................

\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)

\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)

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

\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

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

a: \(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\left(3x-y\right)\cdot\left(3x+y\right)=9x^2-y^2\)

b: \(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\cdot\dfrac{1}{2}\)

\(=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\)

\(=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

c: \(\left(5y-3x\right)\cdot\dfrac{1}{4}\cdot\left(12x+20y\right)\)

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

\(=25y^2-9x^2\)

d: \(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(\dfrac{3}{2}y+x\right)\cdot2\)

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

\(=\dfrac{9}{4}y^2-x^2\)

e: \(\left(a+b+c\right)\left(a+b-c\right)\)

\(=\left(a+b\right)^2-c^2\)

\(=a^2+2ab+b^2-c^2\)

\(x-\frac{1}{2}=y-\frac{2}{3}=z-\frac{3}{4}\)va \(x-2y+3z=14\)

\(\frac{\Rightarrow\left(x-1\right)}{2}=\frac{\left(-2y+4\right)}{-6}=\frac{\left(3z-9\right)}{12}\)

\(=\frac{\left(x-1-2y+4+3z-9\right)}{\left(2-6+12\right)}\)

\(\Rightarrow-\frac{16}{8}=-2\)

\(\frac{\Rightarrow\left(y-2\right)}{2}=-2\Leftrightarrow x-1=-4\Leftrightarrow x=-3\)

\(\Rightarrow\frac{\left(y-2\right)}{3}=-2\Leftrightarrow x-1=-4\Leftrightarrow x=-3\)

\(\Rightarrow\frac{\left(x-3\right)}{4}=-2\Leftrightarrow z-3=-8\Leftrightarrow z=-5\)

\(b)\)

Theo đề ra:

\(x:y:z=3:4:5\)

\(2x^2+2y^2-3z^2=-100\)

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

\(\Leftrightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}\)

\(\Leftrightarrow\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}\)

Áp dụng tính chất dãy tỷ số bằng nhau:

\(\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}=\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}=4\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x}{3}=4\Leftrightarrow x=12\\\frac{y}{4}=4\Leftrightarrow y=16\\\frac{z}{5}=4\Leftrightarrow z=20\end{cases}}\)

a) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)và \(2x-y+3z=28\)

Theo tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x}{5}=\dfrac{2x-y+3z}{4-3+15}=\dfrac{28}{16}=1,75\)

\(\Rightarrow\dfrac{x}{2}=1,75\Rightarrow x=3,5\)

\(\dfrac{y}{3}=1,75\Rightarrow y=5,25\)

\(\dfrac{z}{5}=1,75\Rightarrow z=8,75\)