giải giúp ạ, siêu siêu gấp ạ
1.a
`x:7=5:3`
`x=5.3:7`
`x=15/7`
b.
\(\dfrac{x+2}{1,2}=\dfrac{3,6}{0,6}\)
\(\dfrac{x+2}{1,2}=6\)
\(x+2=6.1,2\)
\(x+2=7,2\)
`x=7,2-2`
`x=5,2`
1c.
\(\dfrac{x-3}{x-6}=\dfrac{8}{9}\)
`9.(x-3)=8.(x-6)`
`9x-27=8x-48`
`9x-8x=-48+27`
`x=-21`
d.
\(\dfrac{x^2}{6}=\dfrac{24}{25}\)
\(x^2=\dfrac{24}{25}.6\)
\(x^2=\dfrac{144}{25}\)
\(x^2=\left(\dfrac{12}{5}\right)^2\)
\(x=\dfrac{12}{5}\) hoặc \(x=-\dfrac{12}{5}\)
1e.
\(\dfrac{x+2}{5}=\dfrac{1}{x-2}\)
`(x+2)(x-2)=1.5`
`x^2-4=5`
`x^2=5+4`
`x^2=9`
`x=3` hoặc `x=-3` (thỏa mãn)
f.
\(\dfrac{x+2}{x+6}=\dfrac{3}{x+1}\)
`(x+2)(x+1)=3.(x+6)`
`x^2+3x+2=3x+18`
`x^2+3x-3x=18-2`
`x^2=16`
`x^2=4^2`
`x=4` hoặc `x=-4` (thỏa mãn)
2a.
\(\dfrac{x}{y}=\dfrac{9}{16}\Rightarrow\dfrac{x}{9}=\dfrac{y}{16}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{9}=\dfrac{y}{16}=\dfrac{x+y}{9+16}=\dfrac{75}{25}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.9=27\\y=3.16=48\end{matrix}\right.\)
b.
Áo dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{-5x}{-25}=\dfrac{7y}{42}=\dfrac{-5x+7y}{-25+42}=\dfrac{34}{17}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.5=10\\y=2.6=12\end{matrix}\right.\)
2c.
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{4x}{9}=\dfrac{5y}{11}=\dfrac{x}{9:4}=\dfrac{y}{11:5}=\dfrac{x+y}{9:4+11:5}=\dfrac{89}{\dfrac{89}{20}}=20\)
\(\Rightarrow\left\{{}\begin{matrix}x=20.9:4=45\\y=20.11:5=44\end{matrix}\right.\)
d.
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{4}=\dfrac{y}{-7}=\dfrac{x-y}{4-\left(-7\right)}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.4=12\\y=3.\left(-7\right)=-21\end{matrix}\right.\)
2e.
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{3x}{15}=\dfrac{-2y}{-4}=\dfrac{3x-2y}{15-4}=\dfrac{44}{11}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.5=20\\y=4.2=8\end{matrix}\right.\)
f.
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{-2}=\dfrac{y}{-3}=\dfrac{4x}{-8}=\dfrac{3y}{-9}=\dfrac{4x-3y}{-8-\left(-9\right)}=\dfrac{9}{1}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=9.\left(-2\right)=-18\\y=9.\left(-3\right)=-27\end{matrix}\right.\)
3a.
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)
Thay vào \(x^2+y^2=100\)
\(\left(3k\right)^2+\left(4k\right)^2=100\)
\(9k^2+16k^2=100\)
\(25k^2=100\)
\(k^2=100:25=4\)
\(k=\pm2\)
- Với \(k=2\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=2.4=8\end{matrix}\right.\)
- Với \(k=-2\Rightarrow\left\{{}\begin{matrix}x=-2.3=-6\\y=-2.4=-8\end{matrix}\right.\)
3b.
Đặt \(\dfrac{x}{4}=\dfrac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=3k\end{matrix}\right.\)
Thay vào `xy=12`
`4k.3k=12`
`12k^2=12`
`k^2=1`
\(k=\pm1\)
- Với \(k=1\Rightarrow\left\{{}\begin{matrix}x=4.1=4\\y=3.1=3\end{matrix}\right.\)
- Với \(k=-1\Rightarrow\left\{{}\begin{matrix}x=4.\left(-1\right)=-4\\y=3.\left(-1\right)=-3\end{matrix}\right.\)
3c.
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\end{matrix}\right.\)
Thay vào \(x^2-y^2=16\)
`(5k)^2 - (3k)^2=16`
`25k^2 -9k^2=16`
`16k^2 =16`
`k^2=1`
\(k=\pm1\)
- Với \(k=1\Rightarrow\left\{{}\begin{matrix}x=5.1=5\\y=3.1=3\end{matrix}\right.\)
- Với \(k=-1\Rightarrow\left\{{}\begin{matrix}x=5.\left(-1\right)=-5\\y=3.\left(-1\right)=-3\end{matrix}\right.\)
3d.
\(2x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}\)
Đặt \(\dfrac{x}{5}=\dfrac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)
Thay vào \(x^3+y^3=133\)
`(5k)^3 + (2k)^3 =133`
`125k^3 +8k^3 =133`
`133k^3=133`
`k^3=1`
`k=1`
\(\Rightarrow\left\{{}\begin{matrix}x=5.1=5\\y=2.1=2\end{matrix}\right.\)
3e.
Do \(5x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)
Thay vào \(x^3y^2=200\)
`(2k)^3 .(5k)^2=200`
`8k^3 .25k^2=200`
`200.k^5 =200`
`k^5=1`
`k=1`
\(\Rightarrow\left\{{}\begin{matrix}x=2.1=2\\y=5.1=5\end{matrix}\right.\)
3f.
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)
Thay vào `2x^2 -3y^2 =-1200`
`2.(3k)^2 -3.(4k)^2 =-1200`
`2.9k^2 -3.16k^2 =-1200`
`18k^2 -48k^2 =-1200`
`-30k^2 =-1200`
`k^2=-1200:(-30)`
`k^2=40`
\(k=\pm2\sqrt{10}\)
- Với \(k=2\sqrt{10}\Rightarrow\left\{{}\begin{matrix}x=3.2\sqrt{10}=6\sqrt{10}\\y=4.2\sqrt{10}=8\sqrt{10}\end{matrix}\right.\)
- Với \(k=-2\sqrt{10}\Rightarrow\left\{{}\begin{matrix}x=3.\left(-2\sqrt{10}\right)=-6\sqrt{10}\\y=4.\left(-2\sqrt{10}\right)=-8\sqrt{10}\end{matrix}\right.\)
4a.
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{81}{9}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.9=18\\y=3.9=27\\z=4.9=36\end{matrix}\right.\)
b.
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{5}=\dfrac{2x}{4}=\dfrac{3y}{-9}=\dfrac{5z}{25}=\dfrac{2x+3y+5z}{4-9+25}=\dfrac{6}{20}=\dfrac{3}{10}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{10}.2=\dfrac{3}{5}\\y=\dfrac{3}{10}.\left(-3\right)=-\dfrac{9}{10}\\z=\dfrac{3}{10}.5=\dfrac{3}{2}\end{matrix}\right.\)
4c.
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{-10}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x}{-20}=\dfrac{3y}{18}=\dfrac{2z}{6}=\dfrac{2x+3y-2z}{-20+18-6}=\dfrac{16}{-8}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2.\left(-10\right)=20\\y=-2.6=-12\\z=-2.3=-6\end{matrix}\right.\)
d.
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{0,3}=\dfrac{y}{0,7}=\dfrac{z}{1}=\dfrac{3x}{0,9}=\dfrac{z-3x}{1-0,9}=\dfrac{1}{0,1}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=10.0,3=3\\y=10.0,7=7\\z=10.1=10\end{matrix}\right.\)
5.
Ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}\) (1)
\(\dfrac{y}{2}=\dfrac{z}{5}\Rightarrow\dfrac{y}{6}=\dfrac{z}{15}\) (2)
Từ (1);(2) \(\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{15}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{15}=\dfrac{x+y+z}{4+6+15}=\dfrac{50}{25}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.4=8\\y=2.6=12\\z=2.15=30\end{matrix}\right.\)
6.
ta có:
\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Rightarrow\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{4}\right)^3=\left(\dfrac{z}{6}\right)^3\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=4k\\z=6k\end{matrix}\right.\)
Thay vào \(x^2+y^2+z^2=14\)
`(2k)^2 +(4k)^2 +(6k)^2 =14`
`4k^2 +16k^2 +36k^2 =14`
`56k^2=14`
\(k^2=\dfrac{1}{4}\)
\(k=\pm\dfrac{1}{2}\)
- Với \(k=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{1}{2}=1\\y=4.\dfrac{1}{2}=2\\z=6.\dfrac{1}{2}=3\end{matrix}\right.\)
- Với \(k=-\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}x=2.\left(-\dfrac{1}{2}\right)=-1\\y=4.\left(-\dfrac{1}{2}\right)=-2\\z=6.\left(-\dfrac{1}{2}\right)=-3\end{matrix}\right.\)
