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Bài 1
a) (x + 3)(x + 2) = 0
x + 3 = 0 hoặc x + 2 = 0
*) x + 3 = 0
x = 0 - 3
x = -3 (nhận)
*) x + 2 = 0
x = 0 - 2
x = -2 (nhận)
Vậy x = -3; x = -2
b) (7 - x)³ = -8
(7 - x)³ = (-2)³
7 - x = -2
x = 7 + 2
x = 9 (nhận)
Vậy x = 9
Bài 9:
Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)
Vậy: (x,y,z,t)=(-10;6;34;-18)
Bài 11:
Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)
Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)
\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)
Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)
\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)
Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)
\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)
Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)
\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)
Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)
\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)
Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)
`-7/6=x/18`
`=>-21/18=x/18`
`=>x=-21(TM\ x in Z)`
`-7/6=-98/y`
`=>-98/84=-98/y`
`=>y=84(TM\ y in Z)`
`-7/6=-14/z`
`=>-14/12=-14/z`
`=>z=12(TM\ z in Z)`
`-7/6=t/102`
`=>-119/102=t/102`
`=>t=-119(TM\ t in Z)`
Vậy `(x,y,z,t)=(-21,84,12,-119)`
\(\frac{12}{-6}=\frac{x}{5}=\frac{-y}{3}=\frac{z}{-17}=\frac{-t}{-9}\)
\(-6x=12\cdot5=60\Rightarrow x=-10\)
\(-y\cdot\left(-6\right)=12\cdot3=36\Rightarrow y=6\)
\(-6z=-17\cdot12=>z=34\)
\(-t\cdot\left(-6\right)=-9\cdot12=>t=-18\)
=> 3x+3y+3z+3t = 1-10+6-3 = -6
=> x+y+z+t = -2
x = x+y+z+t-(y+z+xt) = -2 + 10 = 8
y = x+y+z+t-(x+z+t) = -2-6 = -8
z = x+y+z+t-(y+x+t) = -2+3 = 1
t = x+y+z -(x+y+z) = -2-1 = -3
k mk nha
Ta có:
x + y + z = 1 (1)
y + z + t = -10 (2)
x + z + t = 6 (3)
x + y + + t = -3 (4)
3(x + y + z + t) = -6
=> x + y + z + t = -2 (5)
Lấy (5) - (1) = t => t = -2 - 1 = -3
x = (5) - (2) = -2 + 10 = 8
y = (5) - (3) = -2 - 6 = -8
z = (5) - (4) = -2 + 3 = 1
=> 3x+3y+3z+3t = 1-10+6-3 = -6
=> x+y+z+t = -2
x = x+y+z+t-(y+z+xt) = -2 + 10 = 8
y = x+y+z+t-(x+z+t) = -2-6 = -8
z = x+y+z+t-(y+x+t) = -2+3 = 1
t = x+y+z -(x+y+z) = -2-1 = -3