https://dethi.violet.vn/present/showprint/entry_id/11072330

bạn vào link trên sẽ có full đề và đáp án 

p/s: nhớ k cho mình nha <3

\(\frac{x-2}{4}=-\frac{16}{2-x}\)

\(\Leftrightarrow\frac{x-2}{4}=\frac{16}{x-2}\)

\(\Leftrightarrow\left(x-2\right)^2=4.16=64\)

\(\Leftrightarrow\left(x-2\right)^2=8^2\)

\(\Leftrightarrow\left(x-2-8\right)\left(x-2+8\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+6\right)=0\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)

        2^x+1 . 3^y = 12^x

\(\Rightarrow\)2^x+1 . 3^y =( 2^2 . 3 )

\(\Rightarrow\)2^x+1 . 3^y =2^2x . 3^x

\(\Rightarrow\)\(\frac{2^{2x}}{2^{x+1}}\)= \(\frac{3^y}{3^x}\)

\(\Rightarrow\)2^2x - (x+1) = 3^y - x

\(\Rightarrow\)2^x+1         = 3^y - x

Ta có : 3^y - x lẻ

\(\Rightarrow\)2^x - 1 lẻ

\(\Rightarrow\)x- 1 = 0 \(\Rightarrow\)x=1

3^y - 2 = 2^0 = 1

\(\Rightarrow\)y - x = 0

=> x = y = 1 

Lại gặp bn ha.Chúc bn hc tốt

a) ADTCDTSBN

có: \(\frac{x}{12}=\frac{y}{13}=\frac{z}{15}=\frac{x+y+z}{12+13+15}=\frac{160}{40}=4\)

=> x/12 = 4 => x = 48

...

b) ta có: \(x=\frac{y}{6}=\frac{z}{3}=\frac{2x}{2}=\frac{3y}{18}=\frac{4z}{12}\)

ADTCDTSBN

có: \(\frac{2x}{2}=\frac{3y}{18}=\frac{4z}{12}=\frac{2x-3y+4z}{2-18+12}=\frac{16}{-4}=-4\)

=>...

c) ta có: \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{3}=\frac{2x}{4}=\frac{3y}{-9}=\frac{2z}{8}\)

ADTCTDBN

có: \(\frac{2x}{4}=\frac{3y}{-9}=\frac{2z}{8}=\frac{2x+3y+2z}{4-9+8}=\frac{1}{3}\)

=>...

+) nếu x; y đều dương => x + y = 10

+) nếu x; y đều âm => - x + (- y) = 10 <=> x + y = -10

hc tốt

Xét 2 TH

TH1: /x/+/y/= 10

\(\Rightarrow\)x+y=10

TH2 : /x/+/y/=10

\(\Rightarrow\)-x  + (-y) =10

\(\Rightarrow\)-x - -y = 10

=> - ( x+y ) = 10

=> x+y = -10

 Vậy: x+y= -10 hoặc 10

a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)

b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)

\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)

\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)

a) �2=�5=�7;�+�+�=562x​=5y​=7z​;x+y+z=56

�2=�5=�7=�+�+�2+5+7=5614=42x​=5y​=7z​=2+5+7x+y+z​=1456​=4

⇒{�=4.2=8�=4.5=20�=4.7=28⇒⎩⎨⎧​x=4.2=8y=4.5=20z=4.7=28​

b) �1,1=�1,3=�1,4(1);2�−�=5,51,1x​=1,3y​=1,4z​(1);2x−y=5,5

(1)⇒2�−�1,1.2−1,3=5,50,9(1)⇒1,1.2−1,32x−y​=0,95,5​
    

⇒⎩⎨⎧​x=1,1.0,95,5​=0,96,05​y=1,3.0,95,5​=0,97,15​z=1,11,4​.x=1,11,4​.0,96,05​=0,998,47​​

d) �2=�3=�5;���=−302x​=3x​=5z​;xyz=−30

�2=�3=�5=���2.3.5=−3030=−12x​=3x​=5z​=2.3.5xyz​=30−30​=−1

⇒{�=2.(−1)=−2�=3.(−1)=−3�=5.(−1)=−5⇒⎩⎨⎧​x=2.(−1)=−2y=3.(−1)=−3z=5.(−1)=−5​
 

chết rùi nó bị lỗi

xin lỗi nha

tự làm đi bạn

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

áp dụng t.c dãy tỉ số bằng nhau ta có:

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

\(\frac{x-1}{2}=1\Rightarrow x-1=2\Rightarrow x=3\)

\(\frac{y-2}{3}=1\Rightarrow y-2=3\Rightarrow y=5\)

\(\frac{z-3}{4}=1\Rightarrow z-3=4\Rightarrow z=7\)

Vậy x=3,y=5,z=7

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

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{x-1}{2}=\frac{\left(y-2\right)\cdot2}{3\cdot2}=\frac{\left(z-3\right)\cdot3}{4\cdot3}\)\(=\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{\left(x-1\right)-\left(y-4\right)+\left(3z-9\right)}{2-6+12}\)\(=\frac{x-1-2y+4+3z-9}{8}=\frac{\left(x-2y+3z\right)-\left(1-4+9\right)}{8}\)\(=\frac{14-6}{8}=\frac{8}{8}=1\)

\(\frac{x-1}{2}=1\Rightarrow x=2\cdot1+1=3\)

\(\frac{y-2}{3}=1\Rightarrow y=1\cdot3+2=5\)

\(\frac{z-3}{4}=1\Rightarrow z=1\cdot4+3=7\)

Vậy \(x=3,y=5,z=7\)

Gook luck for you !!!