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a) Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}=\dfrac{x-3y+4z}{4-3.3+4.9}=\dfrac{63}{31}=2\)
\(\Rightarrow x=8\)
\(\Rightarrow y=6\)
\(\Rightarrow z=18\)
b. c. Xem lại đề.
Bài 1:
a: =>3x-3-4=0
=>3x=7
hay x=7/3
b: =>2x-2+3x+6=0
=>5x+4=0
hay x=-4/5
c: =>\(4x^2+4x-1=0\)
hay \(x\in\left\{\dfrac{-1+\sqrt{2}}{2};\dfrac{-1-\sqrt{2}}{2}\right\}\)
d: \(\Leftrightarrow3x-3+2x-4+6=0\)
=>5x+1=0
hay x=-1/5
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
Bài 2: a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow7x-5x=21+25\)
\(\Leftrightarrow2x=46\)
\(\Rightarrow x=46:2=23\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Rightarrow x^2=\left(\pm8\right)^2\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
2)a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(7x-21=5x+25\)
\(7x-5x+25=21\)
\(2x+25=21\)
\(2x=-4\Rightarrow x=-2\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x+1\right)\left(x-1\right)\)
\(63=x\left(x-1\right)+1\left(x-1\right)\)
\(63=x^2-x+x-1\)
\(x^2=63+1=64\)
\(x=\left\{\pm8\right\}\)
c) \(\dfrac{x+4}{20}=\dfrac{2}{x+4}\)
\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=2.20=40\)
\(x\left(x+4\right)+4\left(x+4\right)=40\)
\(x^2+4x+4x+16=40\)
\(x^2+8x=40-16=24\)
\(x\left(x+8\right)=24\)
\(x\in\left\{\varnothing\right\}\)
d) \(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+3\right)\)
\(x\left(x-2\right)+2\left(x-2\right)=x\left(x+3\right)-1\left(x+3\right)\)
\(x^2-2x+2x-4=x^2+3x-x-3\)
\(\)\(x^2-4=x^2+2x-3\)
\(\Leftrightarrow x^2-x^2-2x+3=4\)
\(-2x+3=4\)
\(-2x=1\)
\(x=-\dfrac{1}{2}\)
a. \(\Rightarrow\left\{\begin{matrix}\dfrac{-10}{15}=\dfrac{x}{-9}\\\dfrac{-10}{15}=\dfrac{-8}{y}\\\dfrac{-10}{15}=\dfrac{z}{-21}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b. \(\Rightarrow\left\{\begin{matrix}\dfrac{-7}{6}=\dfrac{x}{18}\\\dfrac{-7}{6}=\dfrac{-98}{y}\\\dfrac{-7}{6}=\dfrac{-14}{z}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-21\\y=84\\z=-12\end{matrix}\right.\)
a) Ta có: \(\dfrac{-10}{15}=\dfrac{x}{-9}\)
\(\Rightarrow15x=-10.\left(-9\right)\)
\(\Rightarrow15x=90\)
\(\Rightarrow x=6\)
Khi đó: \(\dfrac{6}{-9}=\dfrac{-8}{y}=\dfrac{z}{-21}\)
\(\Rightarrow y=\dfrac{-8\left(-9\right)}{6}=12\)
và \(z=\dfrac{-8\left(-21\right)}{12}\) \(=14\)
Vậy \(\left[{}\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b) Lại có: \(\dfrac{-7}{6}=\dfrac{x}{18}\)
\(\Rightarrow6x=-7.18\)
\(\Rightarrow6x=-126\)
\(\Rightarrow x=-21\)
Khi đó \(\dfrac{-21}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}\)
\(\Rightarrow y=\dfrac{-98.18}{-21}=84\)
và \(z=\dfrac{-14.84}{-98}=12\)
Vậy \(\left[{}\begin{matrix}x=-21\\y=84\\z=12\end{matrix}\right.\)
x=\(\dfrac{-4.\left(-10\right)}{8}=5\).
y=\(\dfrac{-10.\left(-7\right)}{5}=14.\)
z=\(\dfrac{-7.\left(-24\right)}{14}=12.\)
\(\dfrac{4x}{2x+9}=8\)
=>16x+72=4x
=>12x=-72
=>x=-6
\(\dfrac{9^{x+9}}{3^{5y}}=243\)
\(\Leftrightarrow\dfrac{9^{-6+9}}{3^{5y}}=243\)
\(\Leftrightarrow3^{5y}=\dfrac{9^3}{243}=3\)
=>5y=1
hay y=1/5
=>xy=-6/5
(không cần "z" nha bn)
ta có : \(\dfrac{x-3}{8}=\dfrac{y-1}{3}\Leftrightarrow3\left(x-3\right)=8\left(y-1\right)\Leftrightarrow3x-9=8y-8\)
\(\Leftrightarrow3x-8y=-8+9\Leftrightarrow3x-8y=1\)
\(\Rightarrow\) ta có hệ phương trình \(\left\{{}\begin{matrix}4x-5y=10\\3x-8y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x-15y=30\\12x-32y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}17y=26\\3x-8y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{26}{17}\\3x-8.\dfrac{26}{17}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{26}{17}\\3x=1+8.\dfrac{26}{17}=\dfrac{225}{17}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{26}{17}\\x=\dfrac{225}{17.3}=\dfrac{75}{17}\end{matrix}\right.\) vậy \(y=\dfrac{26}{17};x=\dfrac{75}{17}\)
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