Câu 17: Tìm x Î Z biết:
a) - 30 + (25 – x) = - 1;
b) -7(3x – 5) + 2(7x – 14) = 28
c) 125: (3x – 13) = 25
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a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
a) Áp dụng t/x dtsbn:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-5}=\dfrac{3x}{6}=\dfrac{2z}{-10}=\dfrac{3x-2z}{6+10}=\dfrac{48}{16}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=3.3=9\\z=3.\left(-5\right)=-15\end{matrix}\right.\)
b) \(\dfrac{x}{10}=\dfrac{y}{-13}=\dfrac{z}{17}=\dfrac{2y}{-26}=\dfrac{3z}{51}=\dfrac{2y-3z}{-26-51}=\dfrac{77}{-77}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=10.\left(-1\right)=-10\\y=\left(-13\right).\left(-1\right)=13\\z=17.\left(-1\right)=-17\end{matrix}\right.\)
a) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-5}\Rightarrow\dfrac{3x}{6}=\dfrac{y}{3}=\dfrac{2z}{-10}\)
Áp dụng t/c của DTSBN, ta có: \(\dfrac{3x-2z}{6-\left(-10\right)}=\dfrac{48}{16}=3\)
\(\dfrac{x}{2}=3\Rightarrow x=6\)
\(\dfrac{y}{3}=3\Rightarrow y=9\)
\(\dfrac{z}{-5}=3\Rightarrow z=-15\)
\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)
Từ đó bạn tìm những giá trị của x nha!
a: \(A=\left\{0;1;2;3;4;5\right\}\)
b: \(B=\left\{2;3;4;5\right\}\)
c: \(C=\left\{0;1;-1;2;-2;3;-3\right\}\)
`a)`
`x-15 = -21`
`<=> x = -21 + 15`
`<=> x = -6`
`b)`
`42 - x = -7`
`<=> x = 42 - (-7)`
`<=> x = 42 + 7`
`<=> x = 49`
`c)`
`12 - (30-x) = -23`
`30-x = 12 - (-23)`
`30-x = 35`
`x = 30-35`
`x = -5`
`d)`
`31 - (17+x)=18`
`<=> 17+x = 31-18`
`<=> 17+x =13`
`<=> x = 13-17`
`<=> x = -4`
a) \(x-15=-21\)
\(\Rightarrow x=-21+15=-6\)
b) \(42-x=-7\)
\(\Rightarrow x=42+7=49\)
c) \(12-\left(30-x\right)=-23\)
\(\Rightarrow30-x=12+23=35\)
\(\Rightarrow x=30+35=65\)
d) \(31-\left(17+x\right)=18\)
\(\Rightarrow17+x=31-18=13\)
\(\Rightarrow x=13-17=-4\)
a) x−15=−21x−15=−21
⇒x=−21+15=−6⇒x=−21+15=−6
b) 42−x=−742−x=−7
x=42+7=49⇒x=42+7=49
c) 12−(30−x)=−2312−(30−x)=−23
30−x=12+23=35⇒30−x=12+23=35
x=30+35=65⇒x=30+35=65
d) 31−(17+x)=1831−(17+x)=18
17+x=31−18=13⇒17+x=31−18=13
x=13−17=−4
Lời giải:
a.
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=60\\ y=45\\ z=40\end{matrix}\right.\)
b)
Từ đkđb suy ra \(\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{10x-5y+z}{1-\frac{1}{3}+\frac{1}{6}}=\frac{25}{\frac{5}{6}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=3\\ y=2\\ z=5\end{matrix}\right.\)
a) - 30 + (25 – x) = - 1
<=>25-x=(-30)-1
<=>25-x=-31
<=>x=25-(-31)
<=>x=56
Vậy x= 56
b) -7(3x – 5) + 2(7x – 14) = 28
<=>-21x+35+14x-28=28
<=>-21x+14x=28+28-35
<=>-7x=21
<=>x=21:(-7)
<=>x=-3
Vậy x=-3
c) 125: (3x – 13) = 25
<=>3x-13=125:25
<=>3x-13=5
<=>3x=5+13
<=>3x=18
<=>x=18:3
<=>x=6
Vậy x=6