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A.\(\left(\frac{4}{7}.x-1\right).3=\frac{1}{21}\)
\(\frac{4}{7}.x-1=\frac{1}{21}:3\)
\(\frac{4}{7}.x-1=\frac{1}{21}.3\)
\(\frac{4}{7}.x-1=\frac{1}{7}\)
\(\frac{4}{7}.x=1+\frac{1}{7}\)
\(\frac{4}{7}.x=\frac{8}{7}\)
\(x=\frac{8}{7}:\frac{4}{7}\)
\(x=\frac{8}{7}.\frac{7}{4}\)
\(x=2\)
Vậy x=2
C.\(\frac{2}{3}.x-\frac{1}{3}.x=\frac{5}{9}\)
\(\left(\frac{2}{3}-\frac{1}{3}\right).x=\frac{5}{9}\)
\(\frac{1}{3}.x=\frac{5}{9}\)
\(x=\frac{5}{9}:\frac{1}{3}\)
\(x=\frac{5}{9}.3\)
\(x=\frac{5}{3}\)
Vậy \(x=\frac{5}{3}\)
D. \(\left(x-2\frac{1}{4}\right).\left(\frac{-2}{3}\right).50\%=2\frac{5}{6}\)
\(x.\frac{-2}{3}-\frac{9}{4}.\frac{-2}{3}=\frac{17}{6}:\frac{1}{2}\)
\(x.\frac{-2}{3}-\frac{-3}{2}=\frac{17}{6}.2\)
\(x.\frac{-2}{3}+\frac{3}{2}=\frac{17}{3}\)
\(x.\frac{-2}{3}=\frac{17}{3}-\frac{3}{2}\)
\(x.\frac{-2}{3}=\frac{34}{6}-\frac{9}{6}\)
\(x.\frac{-2}{3}=\frac{25}{6}\)
\(x=\frac{25}{6}:\frac{-2}{3}\)
\(x=\frac{25}{6}.\frac{3}{-2}\)
\(x=\frac{25}{-4}\)
Vậy \(x=\frac{25}{-4}\)
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`@` `\text {Ans}`
`\downarrow`
\(\dfrac{2}{3}\times x-\dfrac{x}{2}-\dfrac{1}{2}=\dfrac{1}{4}?\)
`=>`\(\dfrac{2}{3}x-\dfrac{x}{2}-\dfrac{1}{2}=\dfrac{1}{4}\)
`=>`\(\dfrac{4x}{6}-\dfrac{3x}{6}-\dfrac{1}{2}=\dfrac{1}{4}\)
`=>`\(\dfrac{x}{6}-\dfrac{1}{2}=\dfrac{1}{4}\)
`=>`\(\dfrac{x}{6}=\dfrac{1}{4}+\dfrac{1}{2}\)
`=>`\(\dfrac{x}{6}=\dfrac{3}{4}\)
`=> 4x=3*6`
`=> 4x=18`
`=> x=`\(\dfrac{18}{4}=\dfrac{9}{2}=4,5\)
Vậy, `x=4,5.`
\(\href{https://hoc24.vn/vip/14874167551162}{\text{_KaizulvG_}}\)
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a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)
<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\) ĐKXĐ: x \(\ne\) 0
<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)
<=> 15x + 28 = 3x
<=> 15x - 3x = -28
<=> 12x = -28
<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)
b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)
<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)
<=> -5x . 6 = 12 . 2
<=> -30x = 24
<=> x = \(-\dfrac{4}{5}\)
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a)\( 3x . ( x- 4 ) = 0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}3x=0 \\x-4=0\end{matrix}\right.\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=4\)
b)\(( x- 2 ) . ( 2x - 4) =0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x-2=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\) \(x=2\)
Vậy \(x=2\)
:v
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b/ x + (x + 1) + (x + 2) + .... + (x + 30) = 1240
x + (x + x + ....+ x) + (1 + 2 + ... + 30) = 1240
x + 30x + 465 = 1240
31x = 1240 - 465
31x = 775
=> x = 775 : 31 = 25
c/ Sửa đề (x + 4) + (x + 6) + (x + 8) + .... + (x + 26) = 210
(x + x + x + .... + x) + (4 + 6 + 8 + .... + 26) = 210
12x + 180 = 210
12x = 210 - 180 = 30
=> x = 30 : 12 = 2,5
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f) 11.(x+2)=5.x+40
=> 11x+22=5x+40
=> 6x=18
=> x=18:6=3
g)11.(x-6)=4.x+11
=> 11.x+(-66)=4x+11
=> 7x=77
=> x=77:7=11
h) (3x-4):x+10=12
=> 3x:x-4:x=12-10=2
=> 3-4:x=2
=> 4:x=3-2=1
=> x=4:1=4
k)(5x+7):x+20=26
=> 5x:x+7:x=26-20=6
=> 5+7:x=6
=> => 7:x=6-5=1
=> x=7:1=7
m) x.1999-x=1999.1997+1999
=> x.1998=1999.(1997+1)
=> x.1998=1999.1998
=> x=1999
**** cho mik nha Nguyễn Thị Thảo Ly
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Đổi x thành a, ta có:
f. 11 x ( a + 2) = 5 x a + 40
11 x a + 11 x 2 = 5 x a + 40
11 x a + 22 = 5 x a + 40
6 x a + 5 x a + 22 = 5 x a + 40
6 x a + 22 = 40(Cùng bỏ 5 x a ở hai vế)
6 x a = 40 - 22
6 x a = 18
a = 18 : 6
a = 3
g. 11 x ( a - 6 ) = 4 x a + 11
11 x a - 11 x 6 = 4 x a + 11
4 x a + 7 x a - 66 = 4 x a + 11
7 x a - 66 = 11(Cùng bỏ đi 4 x a ở 2 vế)
7 x a = 11 + 66
7 x a = 77
a = 77 : 7
a = 11
h. ( 3 x a - 4 ) : a + 10 = 12
3 x a - 4 = ( 12 - 10 ) x a
3 x a - 4 = 2 x a
2 x a + a = 2 x a + 4
a = 4(Cùng bỏ đi 2 x a)
k. Tương tự h
m. a x 1999 - a = 1999 x 1997 + 1999
a x 1998 = 1999 x 1998
=> a = 1999
1/2 * x + 2/3 * x = x + 4
x * (1/2+2/3) = x + 4
7/6 * x = x + 4
7/6 * x - x = 4
1/6 * x = 4
x = 24
50%.x+2/3.x=x+4
=>1/2.x+2/3.x=x+4
(1/2+2/3).x=x+4
7/6.x=x+4
x+4:x=7/6
ae cầu cứu tau làm saI RỒI @@