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

Giải:

a) Đặt \(\frac{x}{10}=\frac{y}{6}=k\)

\(\Rightarrow x=10k,y=6k\)

Mà \(xy=60\)

\(\Rightarrow10k6k=60\)

\(\Rightarrow60k^2=60\)

\(\Rightarrow k^2=1\)

\(\Rightarrow k=\pm1\)

+) \(k=1\Rightarrow x=10;y=6\)

+) \(k=-1\Rightarrow x=-10;y=-6\)

Vậy cặp số \(\left(x;y\right)\) là \(\left(10;6\right);\left(-10;-6\right)\)

b) Hình như đề sai !!!

c) Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)

+) \(\frac{x^2}{9}=4\Rightarrow x^2=36\Rightarrow x=\pm6\)

+) \(\frac{y^2}{16}=4\Rightarrow y^2=64\Rightarrow y=\pm8\)

( x, y cùng dấu )

Vậy cặp số ( x; y ) là ( 6; 8 ) ; ( -6; -8 )
 

b) x-1/2=y-2/3=z-3/4 vã-2y+3z=16

Ta co:

\(\frac{x}{y}=\frac{17}{3}\Rightarrow\frac{x}{3}=\frac{y}{17}=\frac{x+y}{3+17}=3\)

\(\frac{x}{3}=3\Rightarrow x=9\)

\(\frac{y}{17}=3\Rightarrow y=51\)

b)Ta co:

\(\frac{x}{19}=\frac{y}{21}\Rightarrow\frac{2x}{38}=\frac{y}{21}=\frac{2x-y}{38-21}=2\)

\(\frac{2x}{38}=2\Rightarrow x=38\)

\(\frac{y}{21}=2\Rightarrow y=42\)

Ta co:

\(\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=4\)

\(\frac{x^2}{9}=4\Rightarrow x^2=36\Rightarrow x=6\)

\(\frac{y^2}{16}=4\Rightarrow y^2=64\Rightarrow y=8\)

g)\(3x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{3}\)

\(7y=5z\Leftrightarrow\frac{y}{5}=\frac{z}{7}\)

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

\(\frac{x}{10}=2\Rightarrow x=20;\frac{y}{15}=2\Rightarrow y=30;\frac{z}{21}=2\Rightarrow z=42\)

a) ADTCDTSBN

có: \(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3.\)

=> x/2 = 3 => x = 6

y/3 = 3 => y = 9

z/4 = 3 => z = 12

KL:...

b,c làm tương tự nha

d) ta có: \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{2x}{10}\)

ADTCDTSBN

có: \(\frac{2x}{10}=\frac{y}{-6}=\frac{z}{7}=\frac{2x+y-z}{10+\left(-6\right)-7}=\frac{49}{-3}\)

=>...

e) ADTCDTSBN

có: \(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}=\frac{x+1+y+2+z+3}{2+3+4}=\frac{\left(x+y+z\right)+\left(1+2+3\right)}{9}\)

\(=\frac{21+6}{9}=\frac{27}{9}=3\)

=>...

g) ta có: \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=4k\\y=3k\end{cases}}\)

mà xy = 12 => 4k.3k = 12

                          12.k² = 12

                              k² = 1

                        => k = 1 hoặc k = -1

=> x = 4.1 = 4

y = 3.1 = 3

x=4.(-1) = -4 

y=3.(-1) = -3

KL:...

h) ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)

ADTCDTSBN

có: \(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)

=>...

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

a) ) Ta có:\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\Rightarrow\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}\)

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

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

Suy ra: \(\frac{5x}{50}=2\Rightarrow5x=100\Rightarrow x=20\)

\(\frac{y}{6}=2\Rightarrow y=12\)

\(\frac{2z}{42}=2\Rightarrow2z=84\Rightarrow z=42\)

b) 3x=2y, 7y=5z \(\Rightarrow\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{21}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

Suy ra: \(\frac{x}{10}=2\Rightarrow x=20\)

\(\frac{y}{15}=2\Rightarrow y=30\)

\(\frac{z}{21}=2\Rightarrow z=42\)

c) \(\frac{x}{3}=\frac{y}{4};\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x}{9}=\frac{y}{12};\frac{y}{12}=\frac{z}{20}\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\Rightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}\) 

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

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

Suy ra: \(\frac{2x}{18}=3\Rightarrow2x=54\Rightarrow x=27\)

\(\frac{3y}{36}=3\Rightarrow3y=108\Rightarrow y=36\)

\(\frac{z}{20}=3\Rightarrow z=60\)