Tìm x, biết a)2.(3-6x)+8.(x-5)= - 42 b)(-3).(x+4)-12(3-x)= - 21
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Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
[ ( 6x - 39 ) : 7 ] . 4 = 12
=> ( 6x - 39 ) : 7 = 12 : 4
=> ( 6x - 39 ) : 7 = 3
=> 6x - 39 = 3.7
=> 6x - 39 = 21
=> 6x = 21+39
=> 6x = 60
=> x = 10.
( x : 3 - 4 ) . 5 = 15
=> x : 3 - 4 = 15 : 5
=> x : 3 - 4 = 3
=> x : 3 = 7
=> x = 7.3
=> x = 21
x - [ 42 + (-28) ] = -8
=> x - 14 = -8
=> x = -8 + 14
=> x = 6.
[ (6x - 39): 7] . 4=12
(6x-39) : 7=12:4
(6x -39) :7=3
(6x-39) =3 . 7
6x-39 =21
6x =21 +39
6x=20
x=10
\(a.\dfrac{12+x}{42}=\dfrac{35}{42}\Leftrightarrow12+x=35\Leftrightarrow x=23\)
\(b.\dfrac{25-x}{40}=\dfrac{3}{8}\Leftrightarrow\dfrac{25-x}{40}=\dfrac{15}{40}\Leftrightarrow25-x=15\Leftrightarrow x=10\)
\(c.\dfrac{13+x}{20}=\dfrac{3}{4}\Leftrightarrow\dfrac{13+x}{20}=\dfrac{15}{20}\Leftrightarrow13+x=15\Leftrightarrow x=2\)
\(d.\dfrac{23-x}{25}=\dfrac{20}{25}\Leftrightarrow23-x=20\Leftrightarrow x=3\)
a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
a) 12 : x = 3
X = 12 : 3
X = 4
b) 21 : x = 7
X = 21 : 7
X = 3
c) 30 : x = 3
X = 30 : 3
X = 10
d) X : 7 = 4
X = 4 x 7
X = 28
e) 20 : x = 5
X = 20 : 5
X = 4
f) X x 6 = 42
X = 42 : 6
X = 7
Bài 8:
Ta có: \(A=-x^2+2x+4\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=1
a) Ta có:
2.(3-6x)+8.(x-5)=-42
2.3-2.6x+8x-8.5=-42
6-12x+8x-40=-42
=>-12x+8x=-42+40-6
-12x+8x=-2-6
-12x+8x=-8
x(-12+8)=-8
x(-4)=-8
Mà 2(-4)=-8
Nên x=2
Vậy x=2
câu b) giải tương tự nha