tìm x biết
a)3x-1+5.3x-1=162
b)\(\frac{x}{3}=\frac{y}{-9}=\frac{-16}{48}=\frac{-8}{z}\left(tim:x;y;z\right)\)
Tìm min,max của P=xyz biết A= \(\frac{8-x^2}{16+x^4}+\frac{8-y^2}{16+y^4}+\frac{8-z^2}{16+z^4}\ge0.\)
Cho a;b;c >0 thỏa mã \(a+b+c\le3\)Tìm min P \(=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)\)
Tìm x \(\in\) Z biết:
a) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\) b) \(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)\)Ta có: 1/2 - (1/3 + 1/4) = 1/2 - 7/12 = -1/12 ;
1/48 - (1/16 - 1/6) = 1/48 + 5/48 = 1/8
Vì \(-\frac{1}{12}< x< \frac{1}{8}\) nên x = 0
b) \(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)\)Ta có :
\(4\frac{5}{9}:2\frac{5}{18}-7=2-7=-5\)
\(\left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)=\left(1+\frac{38}{5}\right):\left(-21\frac{2}{3}\right)=\frac{43}{5}:\frac{-65}{3}=-\frac{129}{325}\)
Vì \(-5< x< -\frac{129}{325}\) nên \(x\in\left\{-4;-3;-2;-1\right\}\)
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\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
a)\(\left\{{}\begin{matrix}\frac{x-12}{4}=\frac{y-9}{3}=z-1\\3x+5y-z=2\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{a+b}{6}=\frac{b+c}{7}\frac{a+c}{8}\\a+b+c=14\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x+y+z=9\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\zy+yz+zx=27\end{matrix}\right.\)
a.
\(\frac{3x-36}{12}=\frac{5y-45}{15}=\frac{z-1}{1}=\frac{3x+5y-z-50}{26}=\frac{-48}{26}\)
\(\Rightarrow\frac{x-12}{4}=\frac{-48}{26}\Rightarrow x=...\)
Tương tự với y, z, nhưng chắc bạn nhầm đề, nếu pt bên dưới là -2 thì nó ra \(\frac{-52}{26}=-2\) kết quả đẹp hơn nhiều
b. Không rõ đề
c.
\(x+y+z=9\Rightarrow\left(x+y+z\right)^2=81=3.27=3\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\Rightarrow\frac{3}{x}=1\Rightarrow x=y=z=3\)
Tìm x \(\in\)Z, biết:
a) \(\frac{1}{2}-\)\(\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
b) \(4\frac{5}{9}:2\frac{5}{18}-7\le x< \frac{-2}{3}\cdot\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
a)Ta có: 1/2-(1/3+1/4)= -1/12
1/48-(1/16-1/6)=1/8
suy ra: -1/12<x<1/8
<=> -2/24<x<3/24
=>x thuộc:(-1/24 ;0 ;1/24 ;2/24 ;3/24)
a)\(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
\(\Rightarrow\frac{1}{2}-\frac{7}{12}< x< \frac{1}{48}-\frac{-5}{48}\)
\(\Rightarrow\frac{-1}{12}< x< \frac{1}{8}\)
Mà \(x\in Z\Rightarrow x=0\)
b)\(4\frac{5}{9}:2\frac{5}{18}-7\le x< \frac{-2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow\frac{41}{9}:\frac{41}{18}-7\le x< \frac{-2}{3}.\frac{-11}{12}\)
\(\Rightarrow2-7\le x< \frac{11}{18}\)
\(\Rightarrow-5\le x< \frac{11}{18}\)
Mà \(x\in Z\Rightarrow x\in\){-5;-4;-3;-2;-1;0}
Chúc bạn học tốt
a)\(\frac{3x-2}{16}=\frac{-4}{2x-3}\)
b)2/x-1/-3x=7
c)/5x-3/=/7-x/
d)\(\frac{x}{y}=\frac{9}{7};\frac{y}{z}=\frac{7}{3};x+y-z=-15\)
e)\(\frac{x}{-3}=\frac{y}{-8};x^2-y^2=\frac{-44}{5}\)
GIẢI DẦN NHA CÁC BẠN
Cho x,y,,z >0 và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=16\). C/m\(\frac{1}{3x+2y+z}+\frac{1}{x+3y+2z}+\frac{1}{y+3z+2x}\le\frac{8}{3}\)
\(\frac{1}{3x+2y+z}=\frac{1}{x+x+x+y+y+z}\le\frac{1}{6^2}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=\frac{1}{36}\left(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}\right)\)
Tương tự thì ta có:
\(\frac{1}{3x+2y+z}+\frac{1}{x+3y+2z}+\frac{1}{y+3z+2x}\)
\(\le\frac{1}{36}\left(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}\right)+\frac{1}{36}\left(\frac{1}{x}+\frac{3}{y}+\frac{2}{z}\right)+\frac{1}{36}\left(\frac{1}{y}+\frac{3}{z}+\frac{2}{x}\right)\)
\(=\frac{6}{36}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{16}{6}=\frac{8}{3}\)
Dấu "=" xảy ra <=> x = y = z = 3/16
1/ Tìm x, y, z khi
a/ 3x=y ; 5y=4z và 6X+7Y+8Z= 456
b/ \(\frac{4-z}{1}=\frac{y+z}{2}=\frac{x+y}{3}=\frac{y+8}{5}\)
C/ \(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
\(3x=y\)=> \(\frac{x}{1}=\frac{y}{3}\)
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
Cho x , y , z \(\in Z\)thỏa : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\). Hãy tính giá trị biểu thức :
\(M=\frac{3}{4}+\left(x^8-y^8\right)\left(y^9+z^9\right)\left(z^{10}-x^{10}\right)\)
tự suy nghĩ nhé bn o0o I am a studious person o0o,chỉ cần ngồi quy đồng sau làm từng bước là được nhaaaaaaaaa :)) ^_^