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Ta có: \(\frac{a-x}{b-y}=\frac{a}{b}\Rightarrow\left(a-x\right)b=\left(b-y\right)a\)
\(\Rightarrow ab-bx=ab-ay\Rightarrow bx=ay\)
\(\Rightarrow\frac{x}{y}=\frac{a}{b}\left(ĐPCM\right)\)
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...
a) \(x\)=1 \(y\)= 12
b)\(x\)=4 \(y\)= 14
hoặc \(x\)= 6 \(y \)=21
...
a) \(\dfrac{x}{2}+\dfrac{y}{3}=\dfrac{x+y}{2+3}\)
\(\dfrac{x}{2}=\dfrac{x+y}{2+3}-\dfrac{y}{3}\)
\(\dfrac{x}{2}=\dfrac{x+y}{5}-\dfrac{y}{3}\)
\(\dfrac{x}{2}=\dfrac{3\left(x+y\right)}{15}-\dfrac{5y}{15}\)
\(\dfrac{x}{2}=\dfrac{3x-2y}{15}\)
\(\Rightarrow15x=2\left(3x-2y\right)\)
\(15x=6x-4y\)
\(15x-6x=4y\)
\(9x=4y\)
(CÒN LẠI MÌNH KHÔNG BIẾT LÀM)
b) \(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{4}{y}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{20}{5y}\)
\(\dfrac{x}{3}=\dfrac{1+4}{y+1}\)
\(\Rightarrow x\left(y+1\right)=15\)
(CÒN NHIÊU TỰ LÀM NHÉ)
\(\dfrac{x-y}{x+y}=\dfrac{3}{7}\)
\(\Leftrightarrow7x-7y=3x+3y\)
=>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: \(k=-4\sqrt{10}\)
\(x=-20\sqrt{10};y=-8\sqrt{10}\)
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
b/ Có \(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)
nên \(6.\left(x-7\right)=7.\left(y-6\right)\)
\(\rightarrow\) \(6.x-6.7=7.y-7.6\)
\(\Rightarrow\) \(6x=7y\). Mà \(x-y=-4\) nên \(6x-6y=-24\)
\(\rightarrow\) \(7y-6x=-24\)
\(\rightarrow1y=-24\)
Và \(x-y=-4\) \(\Rightarrow\) \(x=\left(-4\right)+y\) \(=\left(-4\right)+\left(-24\right)\)\(=-28\)
Vậy \(x=-28\) \(;\) \(y=-24\)
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)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
Biến đổi ta có :
\(\dfrac{x}{-5}=\dfrac{y}{11}=\dfrac{x-y}{-5-11}=\dfrac{-32}{16}=-2\)
\(\dfrac{x}{-5}=-2=>x=10\)
\(\dfrac{y}{11}=-2=>y=-22\)