đề

Tìm x,y,z biết

#)Giải :

a) Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)

\(\left\{{}\begin{matrix}\frac{x}{10}=2\\\frac{y}{6}=2\\\frac{z}{21}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\y=12\\z=42\end{matrix}\right.\)

b) Ta có : \(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\)

\(7y=5z\Rightarrow\frac{y}{7}=\frac{z}{7}\Rightarrow\frac{y}{15}=\frac{z}{21}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)

\(\left\{{}\begin{matrix}\frac{x}{10}=2\\\frac{y}{15}=2\\\frac{z}{21}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\y=30\\z=42\end{matrix}\right.\)

c) Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\)

\(\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\)

\(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

\(\left\{{}\begin{matrix}\frac{x}{9}=3\\\frac{y}{12}=3\\\frac{z}{20}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)

d) \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{12x}{18}=\frac{12y}{6}=\frac{12z}{15}=\frac{12x+12y+12z}{18+16+5}=\frac{12\left(x+y+z\right)}{18+16+15}=\frac{12.49}{49}=12\)

\(\left\{{}\begin{matrix}\frac{12x}{18}=12\\\frac{12y}{16}=2\\\frac{12z}{15}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}12x=216\\12y=192\\12z=180\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

a) \(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x}{50}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)(vì \(5x+y-z=28\))

⇒ \(x=20;y=12;z=42\)

b) \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)(vì \(x-y+z=32\))

⇒ \(x=20;y=30;z=42\)

c) \(\frac{x}{9}=\frac{y}{12}=\frac{z}{15}=\frac{2x}{18}=\frac{3y}{36}=\frac{2x-3y+z}{18-36+15}=\frac{6}{-3}=-2\)

⇒ x= -18; y= -24; z= -30

d) \(\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{49}{49}=1\)

⇒ x=18; y=16; z=15

Ta có : \(3\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow x=y=z\)

Khi đó : \(3x^{2018}=27^{673}=\left(3^3\right)^{673}=3^{2019}\)

\(\Leftrightarrow x^{2018}=3^{2018}\)

\(\Leftrightarrow\orbr{\begin{cases}x=y=z=3\\x=y=z=-3\end{cases}}\)

Đến đây tự tính A nha!

à....cái đó thì mình chưa tính ra được

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

Suy ra: x=y=z

\(\Rightarrow3x^{2018}=3y^{2018}=3z^{2018}=27^{673}=3^{2019}\)

\(\Leftrightarrow x^{2018}=y^{2018}=z^{2018}=3^{2018}\)

\(\Rightarrow x,y,z=3\)

Dễ tính A

Cảm ơn bạn nhé ,,.... 

a) B= 2x²-3x+1

=(2x²-2x)-(x-1)

=2x(x-1)-(x-1)

=(2x-1)(x-1)

\(\left|x\right|=\frac{1}{2}\)nên ta có \(x=\frac{1}{2}\)hoặc\(x=\frac{-1}{2}\)

nếu \(x=\frac{1}{2}\)thì

B=(2*\(\frac{1}{2}\)-1)(\(\frac{1}{2}\)-1)

B=0

nếu x= -1/2

thì B= (2*(-1/2)-1)(-1/2-1)

B=(-2)*(-3/2)

B=3

giúp e câu b vs a Phong

a/\(\left(x-1\right)\left(x^2+x+1\right)=x^3+x^2+x-x^2-x-1=x^3-1\left(đpcm\right)\)

b/ \(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4-x^3y+x^3y-x^2y^2+x^2y^2-xy^3+xy^3-y^4=x^4-y^4\left(đpcm\right)\)

c/ \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+y^2+xy+yz+z^2+zx+yz=x^2+y^2+z^2+2xy+2yz+2zx\left(đpcm\right)\)

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

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

\(=x^3+3x^2y+3xy^2+y^3+3x^2z+6xyz+3y^2z+3z^2x+3yz^2+z^3\)

\(=x^3+y^3+z^3+3xyz+3x^2y+3xy^2+3x^2z+3y^2z+3y^2x+3yz^2+3xyz\)

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

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

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

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

a, Xét vế trái ta có:

(x-1)(x^2+ x+1)=x^3+ x^2+ x- x^2- x-1

=x^3+ (x^2- x^2)+(x-x)-1

=x^3-1

Vậy...

b,Xét vế trái ta có:(x^3+ x^2y+ xy^2+ y^3)(x-y)

=x^4- x^3y+ x^3y- x^2- y^2+ x^2y^2- xy^3+ xy^3- y^4

=x^4-y^4

Vậy ........

c, Xét vế trái ta có:

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

=x^2+ xy+ xz+ yx+y^2+ yz+ zx+ zy+ z^2

=x^2+ y^2+ z^2+ 2xy+ 2xz+ 2yz

Vậy...............

d, Xé vế trái ta có:

(x+y+x)^3=(x+y+z)(x+y+z)(x+y+z)(x+y+z)

=(x^2+y^2+z^2+2xy+2xz+2yz)(x+y+z)

=x^3+ xy^2+ xz^2+ 2x^2y+ 2xyz+ 2x^2z+ x^2y+ y^3+ yz^2+2xy^2+ 2y^2z+z^3+ 2xyz+ x^2z+ y^2z+2xyz+ 2yz^2+ 2xz^2

=x^3+ 3xy^2+ 6xy+ 3x^2y+3xz^2+ 3x^2z+ 3yz^2+ y^3z^3 (1)

Xét vế phải ta có:x^3+ y^3+ z^3+ 3(x+y)(x+y)(y+z)

=x^3+ y^3+ z^3+ 3(xy+ xz+ y^2+ yz)(z+x)

=x^3+ y^3+ z^3+ 3(xyz+ xz^2+ y^2z+ yz^2+ x^2y+ x^2z+ xy^2+xyz)

=x^2+ y^3+ z^3 +3(2xyz+ xz^2+ y^2z+ yz^2+x^2y+x^2z+ xy^2)

=x^3+ y^3+ z^3+6xyz+ 3xz^2+ 3y^2z+3yz^2+ 3x^2y+3x^2z+3xy^2(2)

Từ (1) và (2)=>.......