Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A) ta có \(\frac{X}{2}=\frac{Y}{3}\)=>\(\frac{X}{8}=\frac{Y}{12}\)(1)
\(\frac{Y}{4}=\frac{Z}{5}\)=>\(\frac{Y}{12}=\frac{Z}{15}\)(2)
Từ (1)và (2)=>\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\) và x-y-z=28
đến đây tự làm
c) \(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
\(\Rightarrow\left(x-\frac{1}{5}\right)^{2004}=0\) và \(\left(y+0,4\right)^{100}=0\) và \(\left(z-3\right)^{678}=0\)
+) \(\left(x-\frac{1}{5}\right)^{2004}=0\Rightarrow x-\frac{1}{5}=0\Rightarrow x=\frac{1}{5}\)
+) \(\left(y+0,4\right)^{100}=0\Rightarrow y+0,4=0\Rightarrow y=-0,4\)
+) \(\left(z-3\right)^{678}=0\Rightarrow z-3=0\Rightarrow z=3\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{1}{5};-0,4;3\right)\)
\(3x=y\)=> \(\frac{x}{1}=\frac{y}{3}\)
hay \(\frac{x}{4}=\frac{y}{12}\)
\(5y=4z\)=> \(\frac{y}{4}=\frac{z}{5}\)
hay \(\frac{y}{12}=\frac{z}{15}\)
suy ra: \(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)
đến đây bạn ADTCDTSBN nhé
a: \(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{9}=\dfrac{11}{18}\)
hay \(x=\dfrac{11}{18}:\dfrac{1}{4}=\dfrac{11}{18}\cdot4=\dfrac{44}{18}=\dfrac{22}{9}\)
d: =>x+1;x-2 khác dấu
Trường hợp 1: \(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Leftrightarrow-1< x< 2\)
Trường hợp 2: \(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Leftrightarrow2< x< -1\left(loại\right)\)
e: =>x-2>0 hoặc x+2/3<0
=>x>2 hoặc x<-2/3
Đặt \(\frac{3\left|x\right|+5}{3}=\frac{3\left|y\right|-1}{5}=\frac{3-z}{7}=k\)
\(\Rightarrow\left|x\right|=\frac{3k-5}{3}\Rightarrow2\left|x\right|=\frac{6k-10}{3}\)
\(\Rightarrow\left|y\right|=\frac{5k+1}{3}\Rightarrow7\left|y\right|=\frac{35k+7}{3}\)
\(\Rightarrow z=3-7k\Rightarrow3z=9-21k\)
Vì \(2\left|x\right|+7\left|y\right|+3z=-14\)\(\Rightarrow\frac{6k-10}{3}+\frac{35k+7}{3}+\left(9-21k\right)=-14\)
\(\Rightarrow\frac{\left(6k-10\right)+\left(35k+7\right)+\left(27-63k\right)}{3}=-14\)
\(\Rightarrow\frac{-22k+24}{3}=-14\)
\(\Rightarrow-22k+24=-42\)
\(\Rightarrow k=\frac{-42-24}{22}=3\)
\(\Rightarrow\left|x\right|=\frac{3.3-5}{3}=\frac{4}{3}\Rightarrow x=-\frac{4}{3};\frac{4}{3}\)
\(\Rightarrow\left|y\right|=\frac{5.3+1}{3}=\frac{16}{3}\Rightarrow y=-\frac{16}{3};\frac{16}{3}\)
\(\Rightarrow z=3-7.3=-18\)
a) Ta có : \(\orbr{\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}}\)
=> \(\left(x+20\right)^{100}+\left|y+4\right|\ge0\)
Do đó \(\left(x+20\right)^{100}=0\)=> \(x=-20\)
\(y+4=0\Rightarrow y=-4\)
Vậy x = -20 và y = -4
b) \(\left(x-\frac{2}{5}\right)\left(x+\frac{3}{7}\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{2}{5}=0\\x+\frac{3}{7}=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{3}{7}\end{cases}}\)
\(a;x^2+\left(9-\frac{1}{10}\right)^2=0\)
\(\Leftrightarrow x^2+\frac{89^2}{100}=0\)
\(\Leftrightarrow x^2=-\frac{7921}{100}\)
Mà\(x^2\ge0\Rightarrow x\in\varnothing\)
Bài 1:
a)Ta có:
\(\frac{4}{5}\left(\frac{7}{2}+\frac{1}{4}\right)^2=\frac{4}{5}\left(\frac{15}{4}\right)^2=\frac{4}{5}.\frac{15}{4}.\frac{15}{4}=\frac{45}{4}\)
b)Ta có:
\(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
Bài 2:
Ta có:
\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{10}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{20}{7}\\y=\frac{-50}{7}\end{matrix}\right.\)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
Bài giải
a, Đặt \(\frac{x}{2}=\frac{y}{5}=k\text{ }\Rightarrow\text{ }\hept{\begin{cases}x=2k\\y=5k\end{cases}}\text{ }\Rightarrow\text{ }x\cdot y=2k\cdot5k=10k^2=90\text{ }\Rightarrow\text{ }k^2=9\text{ }\Rightarrow\text{ }k=\pm3\)
\(\Rightarrow\text{ }\hept{\begin{cases}x=2\cdot\left(-3\right)=-6\\y=5\cdot\left(-3\right)=-15\end{cases}}\) hoặc \(\hept{\begin{cases}x=2\cdot3=6\\y=5\cdot3=15\end{cases}}\)
Vậy \(\left(x\text{ ; }y\right)=\left(-3\text{ ; }-15\right)\text{ ; }\left(6\text{ ; }15\right)\)
b, Do \(\hept{\begin{cases}\left(x-\frac{1}{5}\right)^{2004}\ge0\\\left(y+0,4\right)^{100}\ge0\\\left(z-3\right)^{678}\ge0\end{cases}}\text{ mà }\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-\frac{1}{5}\right)^{2004}\ge0\\\left(y+0,4\right)^{100}\ge0\\\left(z-3\right)^{678}\ge0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x-\frac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{cases}}\Rightarrow\hept{\begin{cases}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{5}\\y=-0,4\\z=3\end{cases}}\)
Vậy \(x=\frac{1}{5}\text{ , }y=-0,4\text{ , }z=3\)
a) ĐẶt \(\frac{x}{2}=\frac{y}{5}=k\)suy ra x=2k, y=5k
Mà x.y=90
suy ra 2k. 5k = 90 suy ra k2=9 suy ra k\(\in\){3;-3}
Với k=3 suy ra x=6, y=15
Với k = -3 suy ra x=-1; y=-15
b) Vì \(\left(x-\frac{1}{5}\right)^{2004}\ge0,\forall x\)
\(\left(y+0,4\right)^{100}\ge0,\forall y\)
\(\left(z-3\right)^{678}\ge0,\forall z\)
Suy ra \(\left(x-\frac{1}{5}\right)^{2004}\)+\(\left(y+0,4\right)^{100}\)+\(\left(z-3\right)^{678}\ge0;\forall x,y,z\)
suy ra \(\left(x-\frac{1}{5}\right)^{2004}=0\)và \(\left(y+0,4\right)^{100}=0\)và \(\left(z-3\right)^{678}=0\)
suy ra x=\(\frac{1}{5}\); y=-0,4 ; z=3