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a: x/3=z/8
nên x/9=z/24
-6y=7z
nên \(\dfrac{y}{-7}=\dfrac{z}{6}\)
=>y/-28=z/24
=>x/9=y/-28=z/24
Áp dụng tính chất của dãytỉ số bằng nhau, ta được:
\(\dfrac{x}{9}=\dfrac{y}{-28}=\dfrac{z}{24}=\dfrac{2x-9y}{2\cdot9-9\cdot\left(-28\right)}=\dfrac{2}{270}=\dfrac{1}{135}\)
Do đó: x=1/15; y=-28/135; z=8/45
c: \(\Leftrightarrow\left(5x-3\right)^{2013}\cdot\left[\left(5x-3\right)^2-1\right]=0\)
=>(5x-3)(5x-4)(5x-2)=0
hay \(x\in\left\{\dfrac{3}{5};\dfrac{4}{5};\dfrac{2}{5}\right\}\)
\(a)\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+x+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Lại có: \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
\(\Rightarrow2=\dfrac{1}{x+y+z}\Rightarrow2\left(x+y+z\right)=1\Rightarrow x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+x+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1+\dfrac{1}{2}}{3}\\y=\dfrac{\dfrac{1}{2}+2}{3}\\z=\dfrac{\dfrac{1}{2}-3}{3}\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)
Chúc bạn học tốt!
a)
Ta có: \(9x=5y=15z\Rightarrow\dfrac{9x}{45}=\dfrac{5y}{45}=\dfrac{15z}{45}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{z}{3}\Rightarrow\dfrac{-x}{-5}=\dfrac{y}{9}=\dfrac{z}{3}_{\left(1\right)}\)
và \(-x+y-z=11_{\left(2\right)}.\)
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\), kết hợp tính chất dãy tỉ só bằng nhau có:
\(\dfrac{-x}{-5}=\dfrac{y}{9}=\dfrac{z}{3}=\dfrac{-x+y-z}{-5+9-3}=\dfrac{11}{1}=11.\)
Từ đó: \(\left\{{}\begin{matrix}\dfrac{-x}{-5}=11\Rightarrow-x=-55\Rightarrow x=55.\\\dfrac{y}{9}=11\Rightarrow y=99.\\\dfrac{z}{3}=11\Rightarrow z=33.\end{matrix}\right.\)
Vậy.....
b); c); d); e) làm tương tự.
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
\(1.\)
\(a.\)
\(\dfrac{x}{-150}=-\dfrac{6}{x}\)
\(\Rightarrow x^2=\left(-6\right)\left(-150\right)\)
\(\Rightarrow x^2=900\)
\(\Rightarrow x=\pm30\)
\(2.\)
\(a.\) \(2x=3y;5y=7z\) và \(3x-7y+5z=30\)
Ta có : \(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\) \(\left(1\right)\)
\(5y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\) \(\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
\(\Rightarrow\dfrac{x}{21}=2\Rightarrow x=42\)
\(\dfrac{y}{14}=2\Rightarrow y=28\)
\(\dfrac{z}{10}=2\Rightarrow z=20\)
Vậy : ..................
a: 3x=2y nên x/2=y/3
7y=5z nên y/5=z/7
=>x/10=y/15=z/21
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)
=>x=20; y=30; z=42
b: 2x=3y=5z
nên x/15=y/10=z/6
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
=>x=75; y=50; z=30
d: Đặt x/3=y/4=z/5=k
=>x=3k; y=4k; z=5k
2x^2+2y^2-3z^2=-100
=>18k^2+32k^2-3*25k^2=-100
=>25k^2=100
=>k^2=4
TH1: k=2
=>x=6; y=8; z=10
TH2: k=-2
=>x=-6; y=-8; z=-10
1.
Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=\left(5k\right)^2-\left(4k\right)^2=25k^2-16k^2=9k^2=4\)
\(\Rightarrow k^2=\dfrac{4}{9}\Rightarrow k=\pm\dfrac{2}{3}\)
\(\circledast k=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{8}{3}\end{matrix}\right.\)
\(\circledast k=-\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=-\dfrac{8}{3}\end{matrix}\right.\)
2.
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\Rightarrow y=\dfrac{\dfrac{2\cdot2+1}{5}\cdot7+2}{3}=3\)
3.
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-\left(z-3\right)}{4+9-4}=\dfrac{95-8+3}{9}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot4+2}{2}=21\\y=\dfrac{10\cdot9+6}{3}=32\\z=10\cdot4+3=43\end{matrix}\right.\)