e) y x 9,2 + y : 0,25 - y x 3,2 = \(\frac{3}{4}x\left(1\frac{1}{2}-\frac{6}{8}\right)\) g) 71 + 65 x 4 = \(\frac{X+140}{X}+260\)
\(y.9,2+y:0,25-y.3,2=\frac{3}{4}.\left(1\frac{1}{2}-\frac{6}{8}\right)\)
\(y.9,2+y:0,25-y.3,2=\frac{3}{4}.\left(1\frac{1}{2}-\frac{6}{8}\right)\)
\(y.9,2+y.4-y.3,2=\frac{3}{4}.\frac{3}{4}\)
\(y.\left(9,2+4-3,2\right)=\frac{9}{16}\)
\(y.10=\frac{9}{16}\)
\(y=\frac{9}{16}:10\)
\(y=\frac{9}{160}\)
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Bài 3 : Tìm y ,x
a) 71 + 65 x 4 = \(\frac{X+140}{X}+260\)
b) y x 15/2 - 1/3 x ( 1/4 + y ) = \(96\frac{2}{3}\)
\(a,\)\(71+65\times4=\frac{x+140}{x}+260\)
\(\Rightarrow71+260=\frac{x-140}{x}+260\)
\(\Rightarrow71=\frac{x-140}{x}\)
\(\Rightarrow71x=x-140\)
\(\Rightarrow71x-x=-140\)
\(\Rightarrow70x=-140\)
\(\Rightarrow x=-2\)
\(b,\)\(y\times\frac{15}{2}-\frac{1}{3}\times\left(\frac{1}{4}+y\right)=90\frac{2}{3}\)
\(\Rightarrow\frac{15y}{2}-\frac{1}{12}-\frac{y}{3}=\frac{272}{3}\)
\(\Rightarrow\frac{90y}{12}-\frac{1}{12}-\frac{4y}{12}=\frac{1088}{12}\)
\(\Rightarrow90y-1-4y=1088\)
\(\Rightarrow86y=1089\)
\(\Rightarrow y=\frac{1089}{86}\)
Tìm x :
a) \(\times.\left(0,25+1999\right).2000=\left(53.1999\right).2000\)
b) 71 + 65 x 4 = \(\frac{x+140}{x}+260\)
hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
a)\(3,6-\left|x-0,4\right|=0\)
b)\(\frac{x}{2}=y=\frac{z}{3}\) và \(x-2y+z=210\)
c)\(\left|x+0,25\right|-4=\frac{1}{4}\)
d)\(x:\left(0,25\right)^4=\left(0,5\right)^2\)
e)\(3^{x-1}+5.3^{x-1}=162\)
f)\(\frac{x}{-25}=\frac{2}{5}\)
g)\(\left|x+\frac{3}{4}\right|-\frac{3}{4}=\sqrt[]{\frac{1}{9}}\)
a) \(3,6-\left|x-0,4\right|=0\)
\(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
Vậy \(x\in\left\{4;-3,2\right\}\)
b) Ta có:
\(\frac{x}{2}=y=\frac{z}{3}=\frac{2y}{2}=\frac{x-2y+z}{2-2+3}=\frac{210}{3}=70\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=70\\y=70\\\frac{z}{3}=70\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=140\\y=70\\z=210\end{matrix}\right.\)
Vậy \(x=140\); \(y=70\); \(z=210\)
c)\(\left|x+0,25\right|-4=\frac{1}{4}\)
\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=\frac{-17}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{-9}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{4;\frac{-9}{2}\right\}\)
d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,25\right)^4.\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,5\right)^8.\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,5\right)^{10}=\left(\frac{1}{2}\right)^{10}=\frac{1}{2^{10}}=\frac{1}{1024}\)
Vậy \(x=\frac{1}{1024}\)
e) \(3^{x-1}+5.3^{x-1}=162\)
\(\Leftrightarrow6.3^{x-1}=162\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
f) \(\frac{x}{-25}=\frac{2}{5}\)
\(\Leftrightarrow x=\left(-25\right).\frac{2}{5}=-10\)
Vậy \(x=-10\)
g) \(\left|x+\frac{3}{4}\right|-\frac{3}{4}=\sqrt{\frac{1}{9}}\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|-\frac{3}{4}=\frac{1}{3}\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{13}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{13}{12}\\x+\frac{3}{4}=-\frac{13}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-\frac{11}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};-\frac{11}{6}\right\}\)
a) \(3,6-\left|x-0,4\right|=0\)
\(\Rightarrow\left|x-0,4\right|=3,6-0\)
\(\Rightarrow\left|x-0,4\right|=3,6.\)
\(\Rightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,6+0,4\\x=\left(-3,6\right)+0,4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
Vậy \(x\in\left\{4;-3,2\right\}.\)
c) \(\left|x+0,25\right|-4=\frac{1}{4}\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{1}{4}+4\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=-\frac{17}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{17}{4}-\frac{1}{4}\\x=\left(-\frac{17}{4}\right)-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{9}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\frac{9}{2}\right\}.\)
d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)
\(\Rightarrow x:\left(0,25\right)^4=0,25\)
\(\Rightarrow x=\left(0,25\right).\left(0,25\right)^4\)
\(\Rightarrow x=\left(0,25\right)^5\)
\(\Rightarrow x=\frac{1}{1024}\)
Vậy \(x=\frac{1}{1024}.\)
Chúc bạn học tốt!
CMR: \(\frac{1}{\left(x+y\right)^3}.\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}.\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}.\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x^3y^3}\)
rút gọn biểu thức
\(A_8=\left(1-\frac{1}{x+2}\right):\left(\frac{4-x^2}{x-6}-\frac{x-2}{3-x}-\frac{x-3}{x+2}\right)\)
\(A=\frac{y-x}{xy}:\left[\frac{y^2}{\left(x-y\right)^2\left(x+y\right)}-\frac{2x^2y}{x^4-2x^2y^2+y^4}+\frac{x^2}{\left(y^2-x^2\right)\left(x+y\right)}\right]\)
\(\frac{1}{\left(x+y\right)^2}\cdot\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^{\text{4}}}\cdot\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\)
Giúp vs cần gấp
Thiếu điều kiện xy = 1; x+y khác 0 nhá bn
Bài này tương tự câu 1 ở đây