Tìm x , y , z
b,\(\frac{b}{3}=\frac{c}{\frac{2}{3}}\left(b+c=11\right)\)
c, \(\frac{x}{2}=\frac{8}{x}\)
e,\(\frac{x}{3}=\frac{y}{7}=\frac{z}{2}\left(2x^2+y^2+3z^2=316\right)\)
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5 tháng 7 2017
b. Áp dụng t/c dãy tỉ số = nhau:
\(\frac{x}{2}=\frac{y}{5}=\frac{x-y}{2-5}=-\frac{7}{3}\)
\(\Rightarrow\frac{x}{2}=-\frac{7}{3}\Leftrightarrow x=-\frac{7}{3}.2=-\frac{14}{3}\)
\(\Rightarrow\frac{y}{5}=-\frac{7}{3}\Leftrightarrow y=-\frac{7}{3}.5=-\frac{35}{3}\)
Vậy \(\hept{\begin{cases}x=-\frac{14}{3}\\y=-\frac{35}{3}\end{cases}}\)
c, Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\Rightarrow x=2k;y=3k;z=4k\)
Ta có: \(xyz=192\Leftrightarrow2k.3k.4k=192\)
\(\Leftrightarrow24k^3=192\)
\(\Leftrightarrow k^3=8\)
\(\Leftrightarrow k=2\)
\(\Rightarrow x=2.2=4\)
\(y=2.3=6\)
\(z=2.4=8\)
e, Ta có: \(x=\frac{y}{2}=\frac{z}{3}=\frac{2x}{2}=\frac{3z}{9}\)
Áp dụng t/c dãy tỉ số = nhau:
\(\frac{2x}{2}=\frac{y}{2}=\frac{3z}{9}=\frac{2x-y+3z}{2-2+9}=\frac{10}{9}\)
\(\Rightarrow x=\frac{10}{9}\)
\(y=\frac{10}{9}.2=\frac{20}{9}\)
\(z=\frac{10}{9}.3=\frac{10}{3}\)
5 tháng 7 2017
b,\(\frac{x}{2}=\frac{y}{5}=\frac{x-y}{2-5}=\frac{7}{-3}.\)
=>x= \(\frac{7}{-3}.2=-4\frac{2}{3}\)
y, \(\frac{7}{-3}.5=-11\frac{2}{3}\)
VM
12 tháng 10 2019
Bài 1:
\(A=\frac{a+b}{b+c}.\)
Ta có:
\(\frac{b}{a}=2\Rightarrow\frac{b}{2}=\frac{a}{1}\) (1)
\(\frac{c}{b}=3\Rightarrow\frac{c}{3}=\frac{b}{1}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{b}{2}=\frac{c}{6}.\)
\(\Rightarrow\frac{a}{1}=\frac{b}{2}=\frac{c}{6}=\frac{a+b}{3}=\frac{b+c}{8}.\)
\(\Rightarrow A=\frac{a+b}{b+c}=\frac{3}{8}\)
Vậy \(A=\frac{a+b}{b+c}=\frac{3}{8}.\)
Bài 2:
a) \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow648+280=7x+9x\)
\(\Rightarrow928=16x\)
\(\Rightarrow x=928:16\)
\(\Rightarrow x=58\)
Vậy \(x=58.\)
b) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=5.20\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=100\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10.\)
\(\Rightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-4\\x=\left(-10\right)-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
Vậy \(x\in\left\{6;-14\right\}.\)
Chúc bạn học tốt!
DH
12 tháng 10 2019
Bài 2:
a, \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow9.72-9.x=7.x-7.40\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow-9x-7x=-280-648\)
\(\Rightarrow-16x=-648\)
\(\Rightarrow x=58\)
Vậy \(x=58\)
TA
22 tháng 8 2017
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
ta có \(\frac{b}{3}=\frac{c}{\frac{2}{3}}\)=>\(\frac{b+c=11}{3+\frac{2}{3}=\frac{11}{3}}\)=3
Từ \(\frac{b}{3}=3\)=> b=3.3=9
\(\frac{c}{\frac{2}{3}}\)=3;=> c=3.3/2=9/2