c) (5+x).7=42
d)12:(x-4)=6
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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
a, 2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= 2 x 12 x 31 + 4 x 6 x 42 + 8 x 3 x 27
= 24 x 31 + 24 x 42 + 24 x 27
= 24 x ( 31 + 24 + 27 )
= 24 x 82
= 1968
b, 2 x 53 x 12 + 4 x 6 x 87 - 3 x 8 x 40
= 24 x 53 + 24 x 87 - 24 x 40
= 24 x ( 53 + 87 - 40 )
= 24 x 100
= 2400
c, Tương tự
a)=>x=13+3/5=68/5.
Vậy x = 68/5
b)=>x/6=5/7+-23/42
=>x/6=1/6
=>x=1
Vậy x=1
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a) x-3/5=13
suy ra x=13+3/5
suy ra x=68/5
suy ra x=13.6
b)x/6=5/7+23/-42
suy ra x/6=1/6
suy ra x=1
d) x+11/12=2/5
suy ra x=2/5-11/12
suy ra x=-31/60
\(\frac{x}{3,15}=\frac{0,15}{7,2}\)
\(\Rightarrow x=\frac{0,15.3,15}{7,2}\)
\(\Rightarrow x=0,065625\)
b) \(\frac{-2,6}{x}=\frac{-12}{42}\)
\(\Rightarrow x=\frac{-2,6.42}{-12}\)
\(\Rightarrow x=9,1\)
\(a.\dfrac{12+x}{42}=\dfrac{35}{42}\Leftrightarrow12+x=35\Leftrightarrow x=23\)
\(b.\dfrac{25-x}{40}=\dfrac{3}{8}\Leftrightarrow\dfrac{25-x}{40}=\dfrac{15}{40}\Leftrightarrow25-x=15\Leftrightarrow x=10\)
\(c.\dfrac{13+x}{20}=\dfrac{3}{4}\Leftrightarrow\dfrac{13+x}{20}=\dfrac{15}{20}\Leftrightarrow13+x=15\Leftrightarrow x=2\)
\(d.\dfrac{23-x}{25}=\dfrac{20}{25}\Leftrightarrow23-x=20\Leftrightarrow x=3\)
6 x 4 = 24 12 : 6 = 2 7 x 3 = 21 63 : 7 = 9
7 x 5 = 35 42 : 7 = 6 6 x 8 = 48 48 : 6 = 8
6 x 6 = 36 28 : 7 = 4 7 x 8 = 56 49 : 7 = 7
6 x 4 = 24 12 : 6 = 02 7 x 3 = 21 63 : 7 = 09 7 x 5 = 35 42 : 7 = 06
6 x 8 = 48 48 :6 = 8 6 x 6 = 36 28 : 7 = 04 7 x 8 = 56 49 : 7 = 07
a) \(⇔x^2-9x+20=12 \)
\(⇔x^2-9x+8=0\)
\(⇔x^2-x-8x+8=0\)
\(⇔(x-1)(x-8)=0\)
\(⇔\left[\begin{array}{} x-1=0\\ x-8=0 \end{array}\right.⇔\left[\begin{array}{} x=1\\ x=8 \end{array}\right.\)
b) \(⇔4x^2-12x+8=3\)
\(⇔4x^2-12x+5=0\)
\(⇔(2x-1)(2x-5)=0\)
\(⇔\left[\begin{array}{} 2x-1=0\\ 2x-5=0 \end{array}\right.⇔\left[\begin{array}{} x=\frac{1}{2}\\ x=\frac{5}{2} \end{array}\right.\)
c) \(⇔x^2+x-30=42\)
\(⇔x^2+x-72=0\)
\(⇔(x-9)(x+8)=0\)
\(⇔\left[\begin{array}{} x-9=0\\ x+8=0 \end{array}\right.⇔\left[\begin{array}{} x=9\\ x=-8 \end{array}\right.\)
d) \(⇔2x^2+5x-3=-6\)
\(⇔2x^2+5x+3=0\)
\(⇔(x+1)(2x+3)=0\)
\(⇔\left[\begin{array}{} x+1=0\\ 2x+3=0 \end{array}\right.⇔\left[\begin{array}{} x=-1\\ x=-\frac{3}{2} \end{array}\right.\)
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
có rút gọn phân số 35/6 về phân số đc ko ? Nếu có thì mn rút gọn cho mik nhé !
Ai xong tr tui k cho
c, \(\left(5+x\right)7=42\Leftrightarrow x+5=6\Leftrightarrow x=1\)
d, \(12:\left(x-4\right)=6\Leftrightarrow x-4=2\Leftrightarrow x=6\)
\(c,\left(5+x\right).7=42\)
\(5+x=42:7\)
\(5+x=6\)
\(x=6-5\)
\(x=1\)
\(d;12:\left(x-4\right)=6\)
\(x-4=12:6\)
\(x-4=2\)
\(x=2+4\)
\(x=6\)