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a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
Đề : Tìm x???
\(\frac{2}{3}x+\frac{1}{4}x=\frac{-22}{27}\)
\(x\left(\frac{2}{3}+\frac{1}{4}\right)=\frac{-22}{27}\)
\(x\left(\frac{8}{12}+\frac{3}{12}\right)=\frac{-22}{27}\)
\(x.\frac{11}{12}=\frac{-22}{27}\)
\(x=\frac{-22}{27}\div\frac{11}{12}\)
\(x=\frac{-22}{27}.\frac{12}{11}\)
\(x=\frac{-8}{9}\)
Vậy \(x=-\frac{8}{9}\).
\(x-\frac{10}{3}=\frac{7}{15}.\frac{3}{5}\)
\(x-\frac{10}{3}=\frac{7}{25}\)
\(x=\frac{7}{25}+\frac{10}{3}=\frac{271}{75}\)
Vậy x = 271/75
\(x-\frac{10}{3}\)\(=\)\(\frac{7}{15}\)\(.\)\(\frac{3}{5}\)
\(x-\frac{10}{3}\)\(=\) \(\frac{7}{25}\)
\(x\) \(=\)\(\frac{7}{25}\)\(+\)\(\frac{10}{3}\)
\(x\) \(=\) \(\frac{271}{75}\)
Câu 1:
a: =>-2x-x+17=34+x-25
=>-3x+17=x+9
=>-4x=-8
hay x=2
b: =>17x+16x+27=2x+43
=>33x+27=2x+43
=>31x=16
hay x=16/31
c: =>-2x-3x+51=34+2x-50
=>-5x+51=2x-16
=>-7x=-67
hay x=67/7
e: 3x-32>-5x+1
=>8x>33
hay x>33/8
1).( 27,56 x 35 ) + ( 27,56 x 67 ) - ( 27,56 x 2)
= (964 + 1846,52) - 55,12
=2810,52 - 55,12
= 2755,4
2).( 4x 35 ) x ( 25 x 5 ) x 2
= ( 140 x 125 ) x2
= 17500 x 2
=35000
4). 3/10
5). 1188
6). 61/6
Tìm x:
a. 2x(x - 1) - x(4 - x) = 0
\(< =>2x^2\) - 2x - 4x + x2 = 0
<=> 3x2 - 6x = 0
<=> x2 - 2x = 0 <=> x(x-2) = 0
<=> \(\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\) <=> \(\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
a, \(2x\left(x-1\right)-x\left(4-x\right)=0\\ \Leftrightarrow2x^2-2x-4x+x^2=0\\ \Leftrightarrow3x^2-6x=0\\ \Leftrightarrow3x\left(x-2\right)=0\\ \Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\)
\(\nghiempt{\Leftrightarrow\begin{cases}x=0\\x=2\end{array}\right.\)
a, | x+3 | = -27/11 × 22/-9
=> |x + 3| = 6
=> x + 3 = 6 hoặc x + 3 = -6
=> x = 3 hoặc x = -9
vậy_
b, (x-3) × (2x - 7) = 0
=> x - 3 = 0 hoặc 2x - 7 = 0
=> x = 3 hoặc x = 7/2
vậy_
c, (x-1) + (x-2) + (x-3) + ... + (x-100) = 4950
=> x - 1 + x - 2 + x - 3 + ... + x - 100 = 4950
=> 100x - (1 + 2 + 3 + ... + 100) = 4950
=> 100x - (1 + 100).100 : 2 = 4950
=> 100x - 5050 = 4950
=> 100x = 10000
=> x = 10
\(\frac{2}{3}.x=\frac{1}{3}\)1) x-\(\frac{10}{3}\)=\(\frac{7}{15}.\frac{3}{5}\)
x-10/3=7/25
x=7/25+10/3
x=\(\frac{271}{75}\)
2)\(\frac{8}{23}.\frac{46}{24}.x=\frac{1}{3}\)
2/3.x=1/3
x=1/3:2/3
x=1/2
\(\dfrac{2}{3}\) + \(x+\dfrac{1}{4}\)\(x\) = - \(\dfrac{22}{27}\)
\(x+\dfrac{1}{4}x\) = - \(\dfrac{22}{27}\) - \(\dfrac{2}{3}\)
\(\dfrac{5}{4}x\) = - \(\dfrac{40}{27}\)
\(x=-\dfrac{40}{27}:\dfrac{5}{4}\)
\(x=-\dfrac{32}{27}\)
Vậy \(x=-\dfrac{32}{27}\)