Bài 1:

Giải:

Ta có: \(\left\{{}\begin{matrix}3x=4y\\5y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=\dfrac{y}{3}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)

Mà \(xyz=30\)

\(\Rightarrow240k^3=30\)

\(\Rightarrow k^3=\dfrac{1}{8}\)

\(\Rightarrow k=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=3\\z=2,5\end{matrix}\right.\)

Vậy...

Bài 2: sai đề

Bài 3:

Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\Rightarrow\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)

Ta có: \(x+2y+3z=38\)

\(\Rightarrow2k+1+8k-6+18k+15=38\)

\(\Rightarrow28k=28\)

\(\Rightarrow k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=1\\z=11\end{matrix}\right.\)

Vậy...

1) Ta có :

\(3x=4y\Rightarrow\dfrac{3x}{12}=\dfrac{4y}{12}\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\) <=> \(\dfrac{x}{8}=\dfrac{y}{6}\)

\(5y=6z\Rightarrow\dfrac{5y}{30}=\dfrac{6z}{30}\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\)

=> \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)

Thay vào đẳng thức xyz = 30

=> 8k.6k.5k = 30

<=> 240k³ = 30

<=> k³ = 8

<=> k = 2

\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=6.2=12\\z=5.2=10\end{matrix}\right.\)

b) Câu này cũng tương tự câu 1 nha ! Đặt k luôn , còn không bình phương lên rồi dùng tính chất dãy tỉ số bằng nhau .

c) Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\)

=> \(\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)

Thay vào đẳng thức , ta có :

x + 2y + 3z = 2k + 1 + 2(4k - 3) + 3(6k + 5) = 38

=> 28k = 38

=> k = \(\dfrac{19}{14}\)

Vậy .....

a: 3x=2y nên x/2=y/3

7y=5z nên y/5=z/7

=>x/10=y/15=z/21

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)

=>x=20; y=30; z=42

b: 2x=3y=5z

nên x/15=y/10=z/6

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

=>x=75; y=50; z=30

d: Đặt x/3=y/4=z/5=k

=>x=3k; y=4k; z=5k

2x^2+2y^2-3z^2=-100

=>18k^2+32k^2-3*25k^2=-100

=>25k^2=100

=>k^2=4

TH1: k=2

=>x=6; y=8; z=10

TH2: k=-2

=>x=-6; y=-8; z=-10

Bài 1:

a) ta có: \(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}=\frac{2y-4}{6}\)

ADTCDTSBN

có: \(\frac{x-1}{5}=\frac{2y-4}{6}=\frac{z-2}{2}=\frac{x-1+2y-4-z+2}{5+6-2}\)\(=\frac{\left(x+2y-z\right)-\left(1+4-2\right)}{9}=\frac{6-3}{9}=\frac{3}{9}=\frac{1}{3}\)

=>...

bn tự tính típ nhé!

b) ta có: \(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}\)

ADTCDTSBN

có: \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{x^2+y^2}{4+9}=\frac{52}{13}=4\)

=>...

Bài 2:

a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)

\(\Rightarrow\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a+b}{b}=\frac{c+d}{b}\left(đpcm\right)\)

b) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\) (*)

mà \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)

Từ (*) \(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)

\(A=2x+2y+3xy\left(x+y\right)+5\left(x^3y^2+x^2y^3\right)\)

\(\Rightarrow A=2\left(x+y\right)+3xy\left(x+y\right)+5x^2y^2\left(x+y\right)\)

\(\Rightarrow A=0\) ( do x+y = 0 )

Đề thiếu r

sao bạn ?

Bài2:

Vì x:y:z tỉ lệ với 4:5:6 =>\(\dfrac{x}{4}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\) mà \(x^2\)-\(2y^2\)+\(z^2\)= 18

Ta có:

\(\dfrac{x}{4}\)=\(\dfrac{x^2}{16}\)

\(\dfrac{y}{5}\)=\(\dfrac{2y}{5}\)=\(\dfrac{2y^2}{10}\)

\(\dfrac{z}{6}\)=\(\dfrac{z^2}{36}\)

Áp dụng tính chất dãy tỉ số= nhau,ta có:

\(\dfrac{x^2}{16}\)=\(\dfrac{2y^2}{10}\)=\(\dfrac{z^2}{36}\)=\(\dfrac{x^2-2y^2+z^2}{16-10+36}\)=\(\dfrac{18}{42}\)=\(\dfrac{3}{7}\)

\(\dfrac{x^2}{16}\)=\(\dfrac{3}{7}\)

=> \(x^2\)=\(\dfrac{48}{7}\)

=> x=\(\sqrt{\dfrac{48}{7}}\)

\(\dfrac{2y^2}{10}\)=\(\dfrac{3}{7}\)

=> \(2y^2\)=\(\dfrac{30}{7}\)

2y=\(\sqrt{\dfrac{30}{7}}\)

y=\(\sqrt{\dfrac{30}{7}}\):2

y= 1,035098339.....

\(\dfrac{z^2}{36}\)=\(\dfrac{3}{7}\)

=> \(z^2\)=\(\dfrac{108}{7}\)

z= \(\sqrt{\dfrac{108}{7}}\)

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}\)

\(\Leftrightarrow\dfrac{x^2}{25}=\dfrac{y^2}{47}=\dfrac{z^2}{9}\)

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

\(\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2}{25}=9\\\dfrac{y^2}{49}=9\\\dfrac{z^2}{9}=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=15\\x=-15\end{matrix}\right.\\\left[{}\begin{matrix}y=21\\y=-21\end{matrix}\right.\\\left[{}\begin{matrix}z=9\\z=-9\end{matrix}\right.\end{matrix}\right.\)

Vậy ..

Ta có:

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}\Rightarrow\left(\dfrac{x}{5}\right)^2=\left(\dfrac{y}{7}\right)^2=\left(\dfrac{z}{3}\right)^2\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)

Theo tính chất của dãy các tỉ số bằng nhau ta có:

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{25+49-9}=\dfrac{585}{65}=9\)

Vậy:

\(\left(\dfrac{x}{5}\right)^2=3^2\Rightarrow\dfrac{x}{5}=3\) hoặc \(\dfrac{x}{5}=-3\)

\(\left(\dfrac{y}{7}\right)^2=3^2\Rightarrow\dfrac{y}{7}=3\) hoặc \(\dfrac{y}{7}=-3\)

\(\left(\dfrac{z}{3}\right)^2=3^2\Rightarrow\dfrac{z}{3}=3\) hoặc \(\dfrac{z}{3}=-3\)

Do đó:

x =15 x = -15

y =21 hoặc y = -21

z = 9 z = -9

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\Rightarrow\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{16}\)\(=\dfrac{x^2-2y^2+z^2}{4-18+16}=\dfrac{8}{2}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{4}=4\\\dfrac{y^2}{9}=4\\\dfrac{z^2}{16}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=16\\y^2=36\\z^2=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=8\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-4\\y=-6\\z=-8\end{matrix}\right.\)