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Lời giải:
\(\frac{x-y}{x+y}=\frac{3}{7}\Rightarrow 7(x-y)=3(x+y)\)
\(\Leftrightarrow 4x=10y\Rightarrow y=0,4x\)
Lại có: \(x^3y^3=1000\Leftrightarrow (xy)^3=1000\Rightarrow xy=\sqrt[3]{1000}=10\)
Thay \(y=0,4x\) ta có:
\(x.0,4x=10\Leftrightarrow x^2=25\Rightarrow x=\pm 5\)
Nếu \(x=5\rightarrow y=0,4x=2\)
Nếu \(x=-5\rightarrow y=0,4x=-2\)
ta có x^3.y^3=(x.y)^3=1000
<=>(x.y)^3=10^3
<=>x.y=10
ta có (x-y)/(x+y)=3/7 <=> 7x-7y=3x+3y
<=> 4x=10y
<=>x=y.5/2
thay x= y.5/2 vào x.y=10 ta có:
y.5/2.y=10
<=>y^2=4
<=>y=2 hoặc y=-2
với y=2 ta có x=5, với y=-2 ta có x=-5
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}=\dfrac{\left(y^2-x^2\right)-\left(y^2+x^2\right)}{3+5}=\dfrac{\left(y^2-x^2\right)-\left(y^2-x^2\right)}{3-5}\Rightarrow\dfrac{2y^2}{8}=\dfrac{-2x^2}{-2}\Rightarrow\dfrac{y^2}{4}=x^2\Rightarrow y^2=4x^2\)
Ta có: \(x^{10}.y^{10}=x^{10}.\left(4x^2\right)^5=1024.x^{20}=1024\Rightarrow x^{20}=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(\Rightarrow y^2=4\Rightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)
Vậy \(x\in\left\{1;-1\right\}\) và \(y\in\left\{4;-4\right\}\)
\(\dfrac{y^2-x^2}{3}=\dfrac{y^2+x^2}{5}\)
\(\Leftrightarrow5\left(y^2-x^2\right)=3\left(y^2+x^2\right)\)
\(\Leftrightarrow5y^2-5x^2=3y^2+3x^2\)
\(\Leftrightarrow2y^2=8x^2\)
\(\Leftrightarrow y^2=4x^2\)
\(\Leftrightarrow y^{10}=1024.x^{10}\)
Mà \(x^{10}.y^{10}=1024\)
\(\Leftrightarrow x^{10}.1024x^{10}=1024\)
\(\Leftrightarrow x^{20}=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
+)Với \(x=1\Leftrightarrow y^{10}=1024\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
+) Với \(x=-1\Leftrightarrow y^{10}=1024\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy...
Lời giải:
Ta có: \(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}\Rightarrow 5(y^2-x^2)=3(y^2+x^2)\)
\(\Rightarrow 2y^2=8x^2\Rightarrow y^2=4x^2\)
\(\Rightarrow y^{10}=4^5x^{10}=(2x)^{10}\)
Do đó:
\(x^{10}y^{10}=x^{10}.(2x)^{10}=1024\)
\(\Leftrightarrow (2x^2)^{10}=1024=2^{10}=(-2)^{10}\)
\(\Rightarrow \left[\begin{matrix} 2x^2=2\\ 2x^2=-2(\text{vô lý})\end{matrix}\right.\)
\(\Rightarrow x^2=1\Rightarrow x=\pm 1\)
\(y^2=4x^2=4\Rightarrow y=\pm 2\)
Vậy \((x,y)=(1,-2); (1,2); (-1,2); (-1,-2)\)
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
a) \(x\)=1 \(y\)= 12
b)\(x\)=4 \(y\)= 14
hoặc \(x\)= 6 \(y \)=21
...
c, Vì \(\hept{\begin{cases}\left|x+3\right|\ge0\\\left|5y+20\right|\ge0\end{cases}\Rightarrow\left|x+3\right|+\left|5y+20\right|\ge0}\)
Mà |x+ 3| + |5y + 20| \(\le\) 0
\(\Rightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left|5y+20\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\y=-5\end{cases}}}\)
d, 5xy - 5x + y = 5
<=> 5x(y - 1) + (y - 1) = 5 - 1
<=> (5x + 1)(y - 1) = 4
=> 5x + 1 và y - 1 thuộc Ư(4) = {1;-1;2-2;4;-4}
Ta có bảng:
| 5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
| y-1 | 4 | -4 | 2 | -2 | 1 | -1 |
| x | 0 | -2/5 (loại) | 1/5 (loại) | -3/5 (loại) | 3/5 (loại) | -1 |
| y | 5 | -3 | 3 | -1 | 2 | 0 |
Vậy các cặp (x;y) là (0;5);(-1;0)
e, Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{cases}\Rightarrow\left(x+1\right)^2+\left(y-1\right)^2\ge0}\)
Mà (x+1)2+(y-1)2 \(\le\) 0
\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)
a) \(\dfrac{5}{7}-1\dfrac{4}{7}\left(450\%+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)
\(\dfrac{5}{7}-\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)
\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{5}{7}+\dfrac{1}{14}\)
\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{11}{14}\)
\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}.\dfrac{7}{11}\)
\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{1}{2}\)
\(\dfrac{2}{3}x=\dfrac{1}{2}-\dfrac{9}{2}=-4\)
\(x=-4:\dfrac{2}{3}=-4.\dfrac{3}{2}=-6\)
Vậy x = \(-6\)
b) \(100=6.7^{\left|x+2\right|}-194\)
\(100+194=6.7^{\left|x+2\right|}\)
\(294=6.7^{\left|x+2\right|}\)
\(294:6=49=7^{\left|x+2\right|}\)
\(\Rightarrow7^2=7^{\left|x+2\right|}\)
\(\Rightarrow2=\left|x+2\right|\Rightarrow\pm2=x+2\)
+ x + 2 = -2 \(\Rightarrow\) x = - 4
+ x + 2 = 2 \(\Rightarrow\) x = 0
Vậy x = - 4 hoặc 0
\(\dfrac{x-y}{x+y}=\dfrac{3}{7}\)
\(\Leftrightarrow7x-7y=3x+3y\)
=>4x=10y
=>2x=5y
hay x/5=y/2
Đặt x/5=y/2=k
=>x=5k; y=2k
\(x^2y^2=1600\)
\(\Leftrightarrow10k^2=1600\)
\(\Leftrightarrow k^2=160\)
TH1: \(k=4\sqrt{10}\)
\(x=20\sqrt{10};y=8\sqrt{10}\)
TH2: \(k=-4\sqrt{10}\)
\(x=-20\sqrt{10};y=-8\sqrt{10}\)