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b: 2x^3-1=15
=>2x^3=16
=>x=2
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
=>y-25=32; z+9=50
=>y=57; z=41
d: 3/5x=2/3y
=>9x=10y
=>x/10=y/9=k
=>x=10k; y=9k
x^2-y^2=38
=>100k^2-81k^2=38
=>19k^2=38
=>k^2=2
TH1: k=căn 2
=>\(x=10\sqrt{2};y=9\sqrt{2}\)
TH2: k=-căn 2
=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)
Ta có: \(\dfrac{3}{5}x=\dfrac{2}{3}y\)
\(\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{5}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{5}}=\dfrac{x^2-y^2}{\dfrac{4}{9}-\dfrac{9}{25}}=\dfrac{38}{\dfrac{19}{225}}=450\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}.450=300\\y=\dfrac{3}{5}.450=270\end{matrix}\right.\)
Vậy \(x=300;y=270\)
Ta có:
3/5x=2/3y(1) , x^2-y^2=38(2)
(1)=> (9/10) x . thay vào (1) =>x^2 -((9/10x))^2=38<=>x^2-(81/100)x^2=38
<=> (19/100)x^2=38<=>x^2=(38/19)*100=200
<=> x=10 can 2, y=(9/10)x=9can2
<=> X=-10can2, y=(9/10)x=-9can2
\(\dfrac{3}{5}x=\dfrac{2}{3}y\Rightarrow\dfrac{x}{\dfrac{5}{3}}=\dfrac{y}{\dfrac{3}{2}}\Rightarrow\left(\dfrac{x}{\dfrac{5}{3}}\right)^2=\left(\dfrac{y}{\dfrac{3}{2}}\right)^2\)
\(\Rightarrow\dfrac{x^2}{\dfrac{25}{9}}=\dfrac{y^2}{\dfrac{9}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{\dfrac{25}{9}}=\dfrac{y^2}{\dfrac{9}{4}}=\dfrac{x^2-y^2}{\dfrac{25}{9}-\dfrac{9}{4}}=\dfrac{28}{\dfrac{19}{36}}\)
Áp dụng tính
Ta có:\(\frac{3}{5}x=\frac{2}{3}y\)
\(\Leftrightarrow\left(\frac{3}{5}x\right)^2=\left(\frac{2}{3}y\right)^2\)
\(\Leftrightarrow\frac{9}{25}x^2=\frac{4}{9}y^2\left(2\right)\)
Mà \(x^2-y^2=38\Rightarrow x^2=38+y^2\left(1\right)\)
Lấy (1) thay vào (2) ta đc:\(\frac{9}{25}\left(38+y^2\right)=\frac{4}{9}y^2\)
\(\Leftrightarrow\frac{342}{25}+\frac{9}{25}y^2-\frac{4}{9}y^2=0\)
\(\Leftrightarrow\frac{19}{225}y^2=\frac{342}{25}\)
\(\Leftrightarrow\)\(y=\sqrt{162}\)
Ko bt có đúng ko mong bn kiểm tra lại rồi nói với mk
2. Tham khảo thêm tại đây nha bạn
https://hoc24.vn/hoi-dap/question/417550.html
a,\(x-\dfrac{3}{5}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}+\dfrac{3}{5}\)
\(x=\dfrac{6}{5}\)
b,\(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)
\(\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)
\(\left|x\right|=\dfrac{6}{5}\)
\(\Rightarrow x=\pm\dfrac{6}{5}\)
c,\(\dfrac{x}{-5}=\dfrac{24}{15}\)
\(x=\dfrac{-5.24}{15}\)
\(x=\dfrac{-24}{5}\)
d,Áp dụng tc dãy TSBN, ta có:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)
+\(\dfrac{x}{4}=-21\Rightarrow x=-21.4=-84\)
+\(\dfrac{y}{5}=-21\Rightarrow y=-21.5=-105\)
Vậy x=-84 ; y=-105
a/ \(x-\dfrac{3}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{3}{5}+\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
Vậy...
b/ \(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{6}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-\dfrac{6}{5}\end{matrix}\right.\)
Vậy...
c/ \(\dfrac{x}{-5}=\dfrac{24}{15}\)
\(\Leftrightarrow15x=-120\)
\(\Leftrightarrow x=-8\)
Vậy...
c/ Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=-21\\\dfrac{y}{5}=-21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-84\\y=-105\end{matrix}\right.\)
Vậy..
Giải:
Áp dụng tính chất của dãy tỉ số bằng nhau,
Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x^2-y^2+2z^2}{2^2-3^2+2.4^2}=\dfrac{108}{27}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{y}{3}=4\\\dfrac{z}{4}=4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=2.4\\y=3.4\\z=4.4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=8\\y=12\\z=16\end{matrix}\right.\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\left(\text{*}\right)\)
Thay \(\left(\text{*}\right)\) vào \(xy=10\)
\(\text{Ta được : }2k\cdot3k=10\\ \Leftrightarrow6k^2=10\\ \Leftrightarrow k^2=\dfrac{5}{3}\\ \Leftrightarrow k=\sqrt{\dfrac{5}{3}}\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\sqrt{\dfrac{5}{3}}\\y=3\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)
Vậy \(x=2\sqrt{\dfrac{5}{3}};y=3\sqrt{\dfrac{5}{3}}\)
\(\dfrac{3x}{5}=\dfrac{2y}{3}\Leftrightarrow\dfrac{3x}{5}.\dfrac{1}{6}=\dfrac{2y}{3}.\dfrac{1}{6}\Leftrightarrow\dfrac{3x}{30}=\dfrac{2y}{18}\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{9}\Leftrightarrow\dfrac{x^2}{100}=\dfrac{y^2}{81}\)Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{100}=\dfrac{y^2}{81}=\dfrac{x^2-y^2}{100-81}=\dfrac{38}{19}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=2.100=200\\y^2=2.81=162\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{200}\\y=\pm\sqrt{162}\end{matrix}\right.\)
\(\dfrac{3}{5}x=\dfrac{2}{3}y\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{5}}\) và \(x^2-y^2=38\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{5}}=\dfrac{x^2}{\dfrac{4}{6}}=\dfrac{y^2}{\dfrac{6}{10}}=\dfrac{x^2+y^2}{\dfrac{4}{6}+\dfrac{6}{10}}=\dfrac{38}{\dfrac{19}{15}}=30\)
\(\dfrac{x}{\dfrac{2}{3}}=30\Rightarrow x=30.\dfrac{2}{3}=20\)
\(\dfrac{y}{\dfrac{3}{5}}=30\Rightarrow y=30.\dfrac{3}{5}=18\)
Vậy x=20 ; y=18