tìm x,y,z biết
a)\(\frac{x}{y}=\frac{7}{3}và5x-2y=87\)
b)\(\frac{x}{19}=\frac{y}{21}và2x-y=34\)
c)\(\left(\frac{-2}{3}\right).x=\left(\frac{-2}{3}\right)^5\)
d)\(\left(\frac{-1}{3}\right)^3.x=\frac{1}{81}\)
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TA
22 tháng 8 2017
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
28 tháng 11 2016
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 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 \(x=58\)
*Bài làm:
a, Ta có: \(\frac{x}{y}\) = \(\frac{7}{3}\) (theo đề bài).
⇒ \(\frac{x}{7}\) = \(\frac{y}{3}\)
⇒ \(\frac{5x}{35}\) = \(\frac{2y}{6}\) . Mà \(5x-2y\) = \(87\) .
Áp dụng tính chất dãy tỉ số bằng nhau , ta được:
\(\frac{5x}{35}\) = \(\frac{2y}{6}\) = \(\frac{5x-2y}{35-6}\) = \(\frac{87}{29}\) = \(3\) .
⇒ \(\left\{{}\begin{matrix}\frac{5x}{35}=3\\\frac{2y}{6}=3\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}5x=3.35=105\\2y=3.6=18\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=105\div5=21\\y=18\div2=9\end{matrix}\right.\)
➤ Vậy: \(\left(x;y\right)=\left(21;9\right)\) .
b, Ta có: \(\frac{x}{19}\) = \(\frac{y}{21}\)
⇒ \(\frac{2x}{38}\) = \(\frac{y}{21}\) . Mà \(2x-y\) = \(34\) .
Áp dụng tính chất dãy tỉ số bằng nhau , ta được:
\(\frac{2x}{38}\) = \(\frac{y}{21}\) = \(\frac{2x-y}{38-21}\) = \(\frac{34}{17}\) = \(2\) .
⇒ \(\left\{{}\begin{matrix}\frac{2x}{38}=2\\\frac{y}{21}=2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}2x=2.38=76\\y=2.21=42\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=76\div2=38\\y=42\end{matrix}\right.\)
➤ Vậy: \(\left(x;y\right)=\left(38;42\right)\) .
c, Ta có: \(\left(\frac{-2}{3}\right)\) . \(x\) = \(\left(\frac{-2}{3}\right)^5\)
⇒ \(x\) = \(\left(\frac{-2}{3}\right)^5\) \(\div\) \(\left(\frac{-2}{3}\right)\)
⇒ \(x\) = \(\left(\frac{-2}{3}\right)^4\)
⇒ \(x\) = \(\frac{\left(-2\right)^4}{3^4}\)
⇒ \(x\) = \(\frac{16}{81}\)
➤ Vậy: \(x\) = \(\frac{16}{81}\) .
d, Ta có: \(\left(\frac{-1}{3}\right)^3\) . \(x\) = \(\frac{1}{81}\)
⇒ \(\frac{\left(-1\right)^3}{3^3}\) . \(x\) = \(\frac{1}{81}\)
⇒ \(\frac{-1}{27}\) . \(x\) = \(\frac{1}{81}\)
⇒ \(x\) = \(\frac{1}{81}\) \(\div\) \(\frac{-1}{27}\)
⇒ \(x\) = \(\frac{-1}{3}\)
➤ Vậy: \(x\) = \(\frac{-1}{3}\) .
☛ Chúc bạn học tốt!
c) \(\left(-\frac{2}{3}\right).x=\left(-\frac{2}{3}\right)^5\)
=> \(x=\left(-\frac{2}{3}\right)^5:\left(-\frac{2}{3}\right)\)
=> \(x=\left(-\frac{2}{3}\right)^4\)
=> \(x=\frac{16}{81}\)
Vậy \(x=\frac{16}{81}.\)
d) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
=> \(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
=> \(x=\frac{1}{81}:\left(-\frac{1}{27}\right)\)
=> \(x=-\frac{1}{3}\)
Vậy \(x=-\frac{1}{3}.\)
Chúc bạn học tốt!