Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a. Theo t/c dãy tỉ số = nhau:
\(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{42}{7}=6\)
=>\(\frac{x}{2}=6\Rightarrow x=6.2=12\)
=>\(\frac{y}{5}=6\Rightarrow y=6.5=30\)
Vậy x=12; y=30.
b. \(\left|x-0,25\right|-\frac{5}{6}=1\frac{2}{3}\)
=> \(\left|x-0,25\right|=1\frac{2}{3}+\frac{5}{6}\)
=> \(\left|x-0,25\right|=\frac{5}{2}=2,5\)
+) x-0,25=2,5
=> x=2,5+0,25
=> x=2,75
+) x-0,25=-2,5
=> x=-2,5+0,25
=> x=-2,25
Vậy x \(\in\){-2,25; 2,75}.
c. y=kx
=> -17=k.8
=> k=-17/8
Vậy hệ số tỉ lệ là -17/8.
a) \(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{42}{7}=6\)
=> x=12 ; y = 30
b) \(\left|x-0,25\right|-\frac{5}{6}=1\frac{2}{3}=>\left|x-0,25\right|=\frac{5}{3}+\frac{5}{6}=\frac{5}{2}=2,5\)
=> x-0,25 = 2,5 hoac: -2,5
=> x = 2,75 hoac x= -2,25
Vay: x la { 2,75 ; -2,25 }
c) Ti le gi vay ban.
Neu thuan thi he so ti le la: \(-\frac{17}{8}\)
Neu nghich thi he so ti le la : -136
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{2}{3a}-\frac{3}{a}=\frac{2}{3a}-\frac{9}{3a}=\frac{-7}{3a}=\frac{7}{15}\Leftrightarrow-3a=15\Leftrightarrow a=-5\)
b)\(2x^3-1=15\Leftrightarrow2x^3=16\Leftrightarrow x^3=8\Leftrightarrow x=2\)
\(\Rightarrow\frac{2+16}{9}=\frac{y-15}{16}=2\Leftrightarrow y-15=32\Leftrightarrow y=47\)
c) \(\left|x\right|=3\Rightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\) rồi xét 2 trường hợp để tính A nhé :)
Bài 1: ĐK của a: \(a\ne0\)
Quy đồng VT ta có: \(\frac{2a-9a}{3a^2}=\frac{7}{15}\)
\(\Leftrightarrow\frac{-7a}{3a^2}=\frac{7}{15}\)
\(\Leftrightarrow-7a.15=3a^2.7\)
\(\Leftrightarrow-105a=21a^2\)
\(\Leftrightarrow-105a-21a^2=0\)
\(\Leftrightarrow a\left(-105-21a\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}a=0\left(l\right)\\-105-21a=0\end{cases}\Leftrightarrow a=-5\left(n\right)}\)
Vậy:..
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(B=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{x}.\left(1+2+3+...+x\right)\)
\(B=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{x}.\frac{x.\left(x+1\right)}{2}\)
\(B=1+\frac{3}{2}+\frac{4}{2}+...+\frac{x+1}{2}\)
\(B=\frac{2+3+4+...+\left(x+1\right)}{2}\)
để B = 115 thì \(\frac{2+3+4+...+\left(x+1\right)}{2}=115\)
\(\Rightarrow\)\(\left(x+3\right)x=115.2.2\)
\(\Rightarrow\)\(\left(x+3\right)x=23.20\)
\(\Rightarrow\)x = 20
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\frac{x}{4}=\frac{y}{3}=\frac{z}{9}=\frac{3y}{9}=\frac{4z}{36}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{4}=\frac{3y}{9}=\frac{4z}{36}=\frac{x-3y+4z}{4-9+36}=\frac{62}{31}=2\)
=> x=2.4=8
3y=2.9=18 => y=6
4z=2.36=72 => z=18
Vậy x=8; y=6; z=18
b) Đặt \(\frac{x}{3}=\frac{y}{4}=k\)
=> x=3k; y=4k
Mà: xy=192
=> 3k.4k=192
=> 12k2=192
=> k2=16
=> k=\(\pm\)4
TH1: k=4
=> x=4.3=12; y=4.4=16
TH2: k=-4
=> x= -4.3=-12; y=-4,3.4=-16
Vậy (x;y) thõa mãn là (12;16);(-12;-16)
a) Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{4}=\frac{y}{3}=\frac{z}{9}=\frac{x}{4}=\frac{3y}{9}=\frac{4z}{36}=\frac{62}{4-9+36}=\frac{62}{31}=2\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=2.4\\y=2.3\\z=2.9\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=8\\y=6\\z=18\end{array}\right.\)
Vậy x = 8 ; y = 6 ; z = 18
b) Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{4}=\frac{xy}{3y}=\frac{192}{3y}\)
\(\Rightarrow\frac{y}{4}=\frac{192}{3y}\Rightarrow y.3y=192.4\)
\(\Rightarrow y^2.3=768\Rightarrow y^2=\frac{768}{3}=256\)
\(\Rightarrow y=\sqrt{256}=16;y=-\sqrt{256}=-16\)
Với y = 16 => x = \(\frac{192}{16}=12\)
Với y = -16 => x = \(\frac{192}{-16}=-12\)
Vậy x = 12 ; y = 16
hoặc x = -12 ; y = -16
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{x-y}{x+2y}=\frac{3}{4}\)
\(\Rightarrow\left(x-y\right).4=\left(x+2y\right).3\)
\(\Rightarrow4x-4y=3x+6y\)
\(\Rightarrow4x=3x+10y\)
\(\Rightarrow x=10y\)
Thay \(x=10y\) vào \(\frac{x-y}{x+2y}=\frac{3}{4}\), ta có:
\(\frac{10y-y}{10y+2y}=\frac{3}{4}\)
\(\Rightarrow\frac{9y}{12y}=\frac{3}{4}\)
êk? thôi chắc chịu, pai pai, cứ để hiện lên cho oách
2, Đặt \(\frac{x}{3}=\frac{y}{2}=\frac{z}{6}=a\)
\(\Rightarrow x=3a;y=2a;z=6a\)
\(5x^2+y^2-z^2=117\Rightarrow5.\left(3a\right)^2+\left(2a\right)^2-\left(6a\right)^2=117\)
\(\Rightarrow13a^2=117\Rightarrow a^2=9\)\(\Rightarrow a=3\) hoặc \(a=-3\)
+ Với \(a=3\) thì \(x=3.3=9;y=3.2=6;z=3.6=18\)
+Với \(a=-3\) thì \(x=-9;y=-6;z=-18\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow\frac{x+1+y+2+z+3}{3+4+5}\)
\(\Rightarrow\frac{24}{12}=2\)
\(\frac{x+1}{3}=2\Rightarrow x=5\)
\(\frac{y+2}{4}=2\Rightarrow y=6\)
\(\frac{z+3}{5}=2\Rightarrow z=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng t/c dãy tỉ số bằng nhau ta có:
1+3y/12=1+7y/4x=2+10y/12+4x=2(1+5y)/2(6+2x)
=1+5y/6+2x
do đó : 1+5y/6+2x=1+5y/5x<=>6+2x=5x<=>6=5x-2x
<=>3x=6=>x=2
Vậy x=2. chúc bạn học tốt
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
\(\Leftrightarrow x+116=0\Leftrightarrow x=-116\)
\(\frac{x-1}{117}+\frac{x-2}{118}+\frac{x-3}{119}=\frac{x-4}{120}+\frac{x-5}{121}+\frac{x-6}{122}\)
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}+1=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
Vì \(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\ne0\)
Nên x + 116 = 0
<=> x = -116