a) Ta có \(x:2=y:-5.\)

=> \(\frac{x}{2}=\frac{y}{-5}\) và \(x-y=14.\)

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

\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{14}{7}=2.\)

\(\left\{{}\begin{matrix}\frac{x}{2}=2=>x=2.2=4\\\frac{y}{-5}=2=>y=2.\left(-5\right)=-10\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(4;-10\right).\)

k) Ta có \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}.\)

\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{20}=\frac{z}{28}.\)

=> \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

=> \(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}\) và \(2x+3y-z=186.\)

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

\(\frac{2x}{30}=\frac{3y}{60}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3.\)

\(\left\{{}\begin{matrix}\frac{x}{15}=3=>x=3.15=45\\\frac{y}{20}=3=>y=3.20=60\\\frac{z}{28}=3=>z=3.28=84\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(45;60;84\right).\)

Mình chỉ làm 2 câu thôi nhé.

Chúc bạn học tốt!

Bạn này riết quá, mình cũng đang bận nữa :(

b) \(21x=19y\Leftrightarrow\frac{x}{19}=\frac{y}{21}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{x}{19}=\frac{y}{21}=\frac{x-y}{19-21}=\frac{14}{-2}=-7\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-38\\y=-42\end{matrix}\right.\)

Vậy...

c) Xem lại đề nhé.

d) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\Leftrightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{x^2+y^2-z^2}{4+9-25}=\frac{-12}{-12}=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4\\y^2=9\\z^2=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm2\\y=\pm3\\z=\pm5\end{matrix}\right.\)

Vậy...

e) \(5x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{5}\)(1)

\(3y=5z\Leftrightarrow\frac{y}{5}=\frac{z}{3}\)(2)

Từ (1) và (2) suy ra \(\frac{x}{2}=\frac{y}{5}=\frac{z}{3}=\frac{x+y+z}{2+5+3}=\frac{-720}{10}=-72\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-144\\y=-360\\z=-216\end{matrix}\right.\)

Vậy...

f) \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=12\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)

g) Áp dụng TCDTSBN:

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

\(=\frac{2x-2+3y-6-z+3}{9}=\frac{45}{9}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)

Vậy...

h) \(\frac{y-z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y-z+1+x+z+2+x+y-3}{x+y+z}=\frac{2x+2y}{x+y+z}\)

Suy ra \(\frac{2x+2y}{x+y+z}=\frac{1}{x+y+z}\Leftrightarrow2x+2y=1\Leftrightarrow x+y=\frac{1}{2}\)

\(\Leftrightarrow\frac{\frac{1}{2}-3}{z}=\frac{1}{\frac{1}{2}+z}\Leftrightarrow z=\frac{5}{6}\)

Từ đó suy ra : \(\frac{y-z+1}{x}=\frac{x+z+2}{y}=-3\)

Ta có hệ :

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

\(\Leftrightarrow\left\{{}\begin{matrix}y-\frac{5}{6}+1=-3x\\x+\frac{5}{6}+2=-3y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+\frac{1}{6}=-3x\\x+\frac{17}{6}=-3y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-3x-\frac{1}{6}\\x+\frac{17}{6}=-3\left(-3x-\frac{1}{6}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{7}{24}\\y=\frac{-25}{24}\end{matrix}\right.\)

Vậy...

a/

Theo đề,ta có:

+/ \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\)

+/\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)\(\left(2\right)\)

Từ (1) và (2), ta có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)

Do đó:

+/ \(\dfrac{x}{8}=\dfrac{28}{-19}\Rightarrow x=-\dfrac{224}{19}\)

+/\(\dfrac{y}{12}=\dfrac{28}{-19}\Rightarrow y=-\dfrac{336}{19}\)

+/\(\dfrac{z}{15}=\dfrac{28}{-19}\Rightarrow z=-\dfrac{420}{19}\)

Vậy: + \(x=-\dfrac{224}{19}\)

+ \(y=-\dfrac{336}{19}\)

+ \(z=-\dfrac{420}{19}\)

a,$\frac{x}{2}=\frac{y}{3},\frac{y}{4}=\frac{z}{5}$và x-y-z=28

Có \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)

=>\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng tính chất DTSBN có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)=\(\dfrac{x-y-z}{8-12-15}=\dfrac{-28}{19}\)

=> x=\(\dfrac{-224}{19}\)

y=\(\dfrac{-336}{19}\)

z=\(\dfrac{-420}{19}\)

e, Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\left(k\in Z\right)\)

\(\Leftrightarrow x=4k,y=5k\) (1)

Theo bài ra ta có: xy = 80

Từ (1) \(\Rightarrow4k.5k=80\Rightarrow20.k^2=80\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

+ Với k = 2 \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)

+ Với k = -2 \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)

Vậy \(\left(x,y\right)\in\left\{\left(8,10\right);\left(-8,-10\right)\right\}\)

a) \(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=\dfrac{-16}{4}=-4\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{z}{-2}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)

mình ko đọc được, chữ nhỏ quá

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

\(\dfrac{x}{6}=\dfrac{y}{10}=\dfrac{z}{21}=\dfrac{5x+y-2z}{6\cdot5+10-2\cdot21}=\dfrac{28}{-2}=-14\)

\(\Rightarrow x=\left(-14\right)6=-84;y=\left(-14\right)10=-140;z=\left(-14\right)21=-294\)

Vậy \(x=-84;y=-140;z=-294\)

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

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{2\cdot15+3\cdot20-28}=\dfrac{124}{62}=2\)

\(x=2\cdot15=30;y=2\cdot20=40;z=2\cdot28=56\)

Vậy \(x=30;y=40;z=56\)

c. Ta có: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\)

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

\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\dfrac{12x+12y+12z}{18+16+15}=\dfrac{12\left(x+y+z\right)}{49}=\dfrac{12\cdot49}{49}=12\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{12x}{18}=12\\\dfrac{12y}{16}=12\\\dfrac{12z}{15}=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}12x=216\\12y=192\\12z=180\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)

Vậy \(x=18;y=16;z=15\)

d. Ta có:

\(3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\)

\(7y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{15}=\dfrac{z}{21}\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)

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

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

\(\Rightarrow x=2\cdot10=20;y=2\cdot15=30;z=2\cdot21=42\)

Vậy \(x=20;y=30;z=42\)

a) \(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}\Leftrightarrow\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{42}\)\(=\dfrac{5x+y-2z}{50+6-42}=\dfrac{28}{14}=2\)

\(\Rightarrow\dfrac{5x}{50}=2\Rightarrow5x=100\Rightarrow x=20\)

\(\Rightarrow\dfrac{y}{6}=2\Rightarrow y=2.6\Rightarrow y=12\)

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

Vậy \(x=20;y=12\) và \(z=42\)

Cậu không làm được hay cần gấp con nào nhỉ ?

Bài 1:

a: \(\Leftrightarrow\dfrac{x+2}{2}=x-5\)

=>2x-10=x+2

=>x=12

b: \(\Leftrightarrow\left(x+2\right)^2=100\)

=>x+2=10 hoặc x+2=-10

=>x=-12 hoặc x=8

c: \(\Leftrightarrow\left(2x-5\right)^3=27\)

=>2x-5=3

=>2x=8

=>x=4

\(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\) va \(x+y-z=69\)

Ta co: \(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\) ; \(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)

➤ \(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\) ➤ \(\dfrac{x+y-z}{20+24-21}\)

➤ \(\dfrac{69}{23}=3\) ➤ \(x=20.3=60\)

\(y=24.3=72\)

\(z=21.3=63\)

\(Vay\) \(x=60;y=72;z=63\)

\(2a=3b;5b=7c\) va \(3a+5c-7c=30\)

Ta co: \(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\)

\(5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\)

⇒ \(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\) ⇒ \(\dfrac{3a}{63}=\dfrac{5c}{50}=\dfrac{7b}{98}\) ⇒ \(\dfrac{3a+5c-7b}{63+50-98}\)

⇒ \(\dfrac{30}{15}=2\) ➤ \(3a=63.2=126\) ➤ \(a=126:3=42\)

\(5c=50.2=100\) \(c=100:5=20\)

\(7b=98.2=196\) \(b=196:7=28\)

Vay \(a=42;c=20;b=28\)

\(x\div y\div z=3\div8\div5\) va \(3x+y-2z=14\)

Ta co: \(x\div y\div z=3\div8\div5\Rightarrow\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\)

⇒ \(\dfrac{3x}{9}=\dfrac{y}{8}=\dfrac{2z}{10}\) ⇒ \(\dfrac{3x+y-2z}{9+8-10}\)

⇒ \(\dfrac{14}{7}=2\) ➤ \(3x=9.2=18\) ➤ \(x=18:3=6\)

\(y=8.2\) \(y=16\)

\(2z=10.2=20\) \(z=20:2=10\)

Vay \(x=6;y=16;z=10\)

Chuc ban hoc tot

ko ai trả lời hẳn một đống cho cậu đâu chi

k cần trả lời hết cũng đc

nhưng có trả lời là đc rùi

1.

\(\left(\dfrac{-2}{3}\right).0,75+1\dfrac{2}{3}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)^2\)

\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(\dfrac{9}{-4}\right)+\dfrac{1}{4}\)

\(=-\dfrac{1}{2}+\dfrac{45}{-12}+\dfrac{1}{4}\)

\(=-\dfrac{6}{12}+\dfrac{-45}{12}+\dfrac{3}{4}\)

\(=\dfrac{-48}{12}\)

\(=-4\)

2.

a) \(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{10}{20}\)

\(\Leftrightarrow x=\dfrac{-11}{20}\)

b) \(\left|x-\dfrac{2}{5}\right|+\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{11}{4}-\dfrac{3}{4}\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=-2\Rightarrow x=-2+\dfrac{2}{5}=\dfrac{-8}{5}\\x-\dfrac{2}{5}=2\Rightarrow x=2+\dfrac{2}{5}=\dfrac{12}{5}\end{matrix}\right.\)

3.

a) \(\dfrac{16}{2^n}=2\)

\(\Leftrightarrow2^n=16:2\)

\(\Leftrightarrow2^n=8\)

\(\Leftrightarrow2^n=2^3\)

\(\Leftrightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{81}=-27\)

\(\Leftrightarrow\left(-3\right)^n=\left(-27\right).81\)

\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)

\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)

\(\Leftrightarrow n=7\)

4. Ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)

\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)

Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

Vì \(x-y+x=-49\) ta có:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)

Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)