Lời giải:

$z=(x+y+z)-(x+y)=21-4=17$

$y=z-5=17-5=12$

$2k=z+x=(x+y+z)-y=21-12=9$

$k=\frac{9}{2}$

Không đáp án nào đúng.

Bài 9:

Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)

Vậy: (x,y,z,t)=(-10;6;34;-18)

Bài 11:

Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)

Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)

\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)

Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)

\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)

Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)

\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)

Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)

\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)

Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)

\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)

Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ

a) Với x = 11 <=> 12 = x+1

\(A\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-...+12x-1\)

\(A\left(x\right)=12x-11=12.11-1=120\)

b) \(B=6x-6y+10-3ax+3ay+15a\)

\(B=6\left(x-y\right)+10-3a\left(x-y\right)+15a\)

\(B=6.5+10-3.a.5+15a\)

\(B=40\)

c)\(C=\frac{x-y}{x+6}=\frac{x-y}{x+x-2y}=\frac{x-y}{2\left(x-y\right)}=\frac{1}{2}\left(x-2y=6\right)\)

\(C=\frac{2x+6}{3x-2y}+\frac{2y-6}{4y-x}\)

\(C=\frac{2x+1-2y}{3x-2y}+\frac{2y-x+2y}{4y-x}\)

\(C=1+1=2\)

d) ta có : x-y-x = 0

\(\Rightarrow\left\{{}\begin{matrix}x-z=y\\x-y=z\\x=y+z\end{matrix}\right.\).Thay vào B, ta có :

\(B=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}\)

\(B=\frac{y}{x}.\frac{\left(-z\right)}{y}.\frac{x}{z}\)

B= -1

A

A

\(\hept{\Rightarrow\begin{cases}\frac{x}{5}=2\Rightarrow x=10\\\frac{y}{3}=2\Rightarrow y=6\\\frac{z}{17}=2\Rightarrow z=34\end{cases}}\)

\(\frac{12}{6}=\frac{x}{5}=\frac{y}{3}=\frac{z}{17}\)

\(\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{z}{17}=2\)

\(\Rightarrow x=2.5=10\)

\(y=3.2=6\)

\(z=17.2=34\)

\(x.x+y.y+z.z=12\)

\(\Leftrightarrow\frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}=\frac{12}{3}=4\)

\(\Rightarrow x^2=1.4=4\Leftrightarrow x=2\)

\(y^2=1.4=4\Leftrightarrow y=2\)

\(z^2=1.4=4\Leftrightarrow z=2\)

Áp dụng BĐT Cauchy - schwarz:

\(x^2+y^2+z^2=\frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}\ge\frac{\left(x+y+z\right)^2}{1+1+1}=\frac{36}{3}=12\)

(Dấu "="\(\Leftrightarrow x=y=z\))

\(pt\Leftrightarrow3x^2=12\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

\(\Rightarrow\orbr{\begin{cases}x=y=z=2\\x=y=z=-2\left(L\right)\end{cases}}\)(Vì x + y + z = 6)

Vậy x = y = z = 2

\(\frac{12}{-6}=-2=\frac{-10}{5}=\frac{-6}{3}=\frac{34}{-17}=\frac{-18}{9}\)

Vậy...........

\(\frac{-24}{-6}=4=\frac{12}{3}=\frac{4}{\left(\pm1\right)^2}=\frac{\left(-2\right)^3}{-2}\)

Vậy......