A= x+7- (-12 )voi |x|=42
B= 27-x+y biết |x|=9 y=14
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a) (-12) . (x - 14) + 7 . (3 - x) = 12.15
=> -12x + 168 + 21 - 7x = 180
=> -19x = -9
=> x = \(\frac{9}{19}\)
b) (2x - 8) . ( y - 2) = 0
=> \(\left\{\begin{matrix}2x-8=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=4\\y=2\end{matrix}\right.\)
c) x^2 - 75 = (-50)
=> \(x^2\)=25
=> \(\left[\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
d) (2x - 5)^2 - 1 = 0
=> ( 2x - 5 ) ^2 = 1
=> \(\left[\begin{matrix}2x-5=1\\2x-5=-1\end{matrix}\right.\)
=>\(\left[\begin{matrix}x=3\\x=2\end{matrix}\right.\)
e) (3y + 6)^3 + 27 = 0
=> \(\left(3y+6\right)^3=-27\)
=> 3y + 6 = -3
=> y = -3
( 2 x y + 2/15 ) x 3 = 4/5
( 2 x y + 2/15 ) = 4/5 : 3
( 2 x y + 2/15 ) = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 :2
y = 1/15
(2 x y + 2/15) x 3 = 4/5
2 x y + 2/15) = 4/5 : 3
2 x y + 2/15 = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 : 2
y = 1/15
7/9 x (2 - 1/3 x y) = 14/15
(2 - 1/3 x y) = 14/15 : 7/9
(2 - 1/3 x y) = 6/5
2 - y = 6/5 x 1/3
2 - y = 2/5
y = 2/5 + 2
y = 12/5
4/21 + 5 x y - 8/7 = 1/3
4/21 + 5 x y = 1/3 + 8/7
4/21 + 5 x y = 31/21
5 x y = 31/21 - 4/21
5 x y = 9/7
y = 9/7 : 5
y = 9/35
7/12 x y - 3/12 x y = 5
y x (7/12 - 3/12) = 5
y x 1/3 = 5
y = 5 : 1/3
y = 15
a) y : 4/11 x 3/5 = 27/4
y : 4/11 = 27/4 : 3/5
y : 4/11 = 45/4
y = 45/4 x 4/11
y = 45/11
b) y + 7/2 - 5/9 = 14/3
y + 7/2 = 14/3 + 5/9
y + 7/2 = 47/9
y = 47/9 - 7/2
y = 31/18
a) y : 4/11 x 3/5 = 27/4
y : 4/11 = 27/4 : 3/5
y : 4/11 = 45/4
y = 45/4 x 4/11
y = 45/11
b) y + 7/2 - 5/9 = 14/3
y + 7/2 = 14/3 + 5/9
y + 7/2 = 47/9
y = 47/9 - 7/2
y = 31/18
a, \(x+y+z-x-y+z=2z\)
Thay z=-10 Thì \(2z=-20\)
b, \(a-b-c+a+b=2a-c=2\cdot\left(-12\right)-\left(-27\right)=3\)
Lời giải:
a. $(-12)-x=(-12)-(-28)=-12+28=28-12=16$
b. $(23+x)-(56-x)=(23+7)-(56-7)=30-49=-19$
c. $25-x-(29+y-8)=25-13 -(29+11-8)=12-32=-20$
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)
a) Ta có: \(\left|x\right|=42\Rightarrow x=\pm42\)
Nếu x = 42 => A = 42 + 7 + 12 = 61
Nếu x = -42 => A = -42 + 7 + 12 = -23
b) \(\left|x\right|=9\Rightarrow x=\pm9\)
Nếu x = 9 và y = 14 => B = 27 - 9 + 14 = 32
Nếu x = -9 và y = 14 => B = 27 + 9 + 14 = 50
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