1D  2C

Câu 1: D

Câu 2: C

Ta có: \(\left\{{}\begin{matrix}x\left(x+2y+3z\right)=-5\\y\left(x+2y+3z\right)=27\\z\left(x+2y+3z\right)=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-5}=x+2y+3z\\\dfrac{y}{27}=x+2y+3z\\\dfrac{z}{5}=x+2y+3z\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{-5}=\dfrac{y}{27}=\dfrac{z}{5}\Rightarrow\left\{{}\begin{matrix}y=\dfrac{-27}{5}x\\z=-x\end{matrix}\right.\)

Ta có: \(x\left(x+2y+3z\right)=-5\Rightarrow x\left(x+2.\dfrac{-27}{5}x-3x\right)=-5\)

\(\Rightarrow\dfrac{-64}{5}x^2=-5\Rightarrow x^2=\dfrac{25}{64}\Rightarrow x=\dfrac{5}{8}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=-\dfrac{27}{5}x=-\dfrac{27}{8}\\z=-x=-\dfrac{5}{8}\end{matrix}\right.\)

B(2y + z) (4 x - 7 y)

B nha

B

\(A=2xy-x^2-2yz+4y^2=2\cdot2\cdot\dfrac{1}{2}-2^2-2\cdot\dfrac{1}{2}\cdot\left(-1\right)+4\cdot\left(\dfrac{1}{2}\right)^2\)

\(=2-4+1+4\cdot\dfrac{1}{4}=-2+1+1=0\)

A = 2 (2 . 1/2 - 2) - 2 . 1/2 (-1 - 2 . 1/2) 

A = 2 (1 - 2) - 1 (-1 - 1) 

A = 2 . (-1)  - 1 . (-2) 

A = -2 + 2

A = 0

@Liz.Ald2094

=2⋅2⋅12−22−2⋅12⋅(−1)+4⋅(12)2A=2xy−x2−2yz+4y2=2⋅2⋅12−22−2⋅12⋅(−1)+4⋅(12)2

? = 100

a ) x + x + x = 90

=> 3  . x = 90

x = 90 : 3

x = 30

b ) x + 2y + 2y = 230

hay : 30 + 4y = 230

4y = 230 - 30

4y = 200

y = 200 : 4

y = 50

c ) z + 2y + x = 210

hay : z + 2 . 50 + 30 = 210

z + 100 + 30 = 210

z = 210 - 30 - 100

z = 80

d ) y - x + z = 50 - 30 + 80 = 100

x+x+x=90

x.3=90

x=90:3

x=30

30+2y+2y=230

30+4y=230

4y=230-30

4y=200

y=200:4

y=50

z+2.50+30=210

z+100+30=210

z+100=210-30

z+100=180

z=180-100

z=80

50-30+80=100