Tìm x,y,z nguyên biết
1) x/5 = -12/10
2) (x + 1)/2 = x/3
3) (1 - 2x)/9 = 1/ (1- 2x)
4) 4/5 = 12/x = y/20 = 8.(y -x)/z
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\(x+\frac{1}{2}=\frac{x}{3}\)
\(\Rightarrow3x+\frac{3}{2}=x\)
\(\Rightarrow x-3x=\frac{3}{2}\)
\(\Rightarrow-2x=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}:-2\)
\(\Rightarrow x=\frac{3}{2}.-\frac{1}{2}\)
\(\Rightarrow x=-\frac{3}{4}\)
Vậy \(x=-\frac{3}{4}\)
\(x+\frac{1}{2}=\frac{x}{3}\)
\(x=\frac{x}{3}-\frac{1}{2}\)
\(\frac{6x}{6}=\frac{2x}{6}-\frac{3}{6}\)
\(\Rightarrow6x=2x-3\)
\(6x-2x=3\)
\(4x=3\)
\(x=\frac{3}{4}\)
\(\frac{1-2x}{9}=\frac{1}{1-2x}\)
\(\Rightarrow\left(1-2x\right).\left(1-2x\right)=1.9\)
\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}1-2x=3\\1-2x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
\(\frac{4}{5}=\frac{12}{x}=\frac{y}{20}=\frac{8.\left(y-x\right)}{z}\)
\(\Rightarrow\frac{4}{5}=\frac{12}{x}\)
\(4x=60\)
\(x=15\)
\(\frac{4}{5}=\frac{y}{20}\)
\(y5=80\)
\(y=16\)
\(\frac{4}{5}=\frac{8\left(y-x\right)}{z}\Leftrightarrow\frac{4}{5}=\frac{8.\left(80-15\right)}{z}\Leftrightarrow\frac{4}{5}=\frac{520}{z}\)
\(\Leftrightarrow4z=2600\)
\(z=650\)
Vậy ... ( bạn tự kết luận nhé )
chúc bạn học tốt !!!
1, \(\frac{x+1}{2}=\frac{x}{3}\)
<=> 3(x+1)=2x
<=> 3x+3-2x=0
<=> x+3=0
<=> x=-3
2, \(\frac{1-2x}{9}=\frac{1}{1-2x}\left(x\ne\frac{1}{2}\right)\)
<=> (1-2x)2=9
<=> \(\orbr{\begin{cases}1-2x=3\\1-2x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\left(tmđk\right)}\)
Vậy x={-1;2}
a: =>-2x=90/91
hay x=-45/91
b: =>2x=-7
hay x=-7/2
c: ->-3x=-12
hay x=4
Bài 1:
a: =>2x-9=10/91
=>2x=829/91
hay x=829/182
b: =>2x=-7
hay x=-7/2
c: =>-3x=-12
hay x=4
Bài 2:
\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)
\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
Mọi người giúp em với ạ
Bài tập này em chưa hiểu nên hỏi
Mong mọi người giúp:((
1) x/5 = -12/10
\(x.10=5.\left(-12\right)\)
\(10x=\left(-60\right)\)
\(x=-6\)
2) (x + 1)/2 = x/3
\(3\left(x+1\right)=2x\)
\(3x+3=2x\)
\(3x-2x=3\)
\(x=3\)
3) (1 - 2x)/9 = 1/ (1- 2x)
\(\left(1-2x\right).\left(1-2x\right)=9.1\)
\(\left(1-2x\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}1-2x=3\\1-2x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
4) 4/5 = 12/x = y/20 = 8.(y -x)/z
\(\frac{4}{5}=\frac{12}{x}\Leftrightarrow4x=60\Leftrightarrow x=15\)
\(\frac{4}{5}=\frac{y}{20}\Leftrightarrow y5=80\Leftrightarrow y=16\)
\(\frac{4}{5}=\frac{8.\left(y-x\right)}{z}\Leftrightarrow\frac{4}{5}=\frac{8}{z}\Leftrightarrow4z=40\Leftrightarrow z=10\)
Vậy \(x=15;y=16;z=10\)
chúc bạn học tốt