Tìm x biết 32/9-x-5/12=1/3+1/3+1/3
Tìm x biết : 32/9 - x - 5/12 = 1/3 + 1/3 + 1/3
giải
32/9-x-5/12= 1/3 + 1/3 +1/3
32/9-x-5/12 =1
32/9-x=1+ 5/12
32/9-x=17/12
x=32/9-17/12
x=77/36
thế đó bạn
Tìm x biết : 32/9 - x - 5/12 = 1/3 + 1/3 + 1/3
Bài 2:Tìm x,biết:
1, 4/5-x=1/3
2, 2/3+5/3x=5/7
3, -12/13x+5=5 1/13
4, x:(-2 1/4)+1,5=-3/4
5,GTTĐ của x -3/4 - 1/4=0
6, 6-GTTĐ của 1/2-x = 2/3
7, (x-1)2=4/9
1) x = 4/5 - 1/3
x = 7/15
2) 5/3.x=1/21
x=1/35
3) -12/13.x = 1/13
x=-1/12
7) th1: x-1=2/3
x = 5/3
Th2: x - 1 = -2/3
x=1/3
1) Tìm x, biết:
a) 3 + |-x| = |-7|
b) -3 + |1 - x| = |-2|
c) 5 + |-1 + x| = -|2|
d) -|3| + |2x| = |-3|
e) 1 + |-x + 2| = -|7|
g) -32 + |x - 16| = |12|
h) |-32| - |x - 16| = |-12|
i) -|32| - |16 - x| = |-32|
2) Tìm x, biết:
a) |x| < 10
b) |x| > 0
c) |x| < -1
3) Tính tổng các số nguyên sau:
a) -9 < x < 9
b) -10 < x \(\le\)10
c) -9 < x < 10
1 a x=4
b x=-4
c x=-7
d x=3
e x=10
g x=60
h x=36
i x=16
2a 1,2,3,4,5,6,7,8,9
b 1,2,3,4,5,6,7,8,9.........
c rỗng
3a 0
b 0
c10
Bạn cho mình các bước giải được không?
tìm số tự nhiên x
a 5/7-3/7x=1 b x+3/4=36/144.-12/9 c 8/23 . 46/24 = 1/3.x d 4/9-[x-1/2]2=1/3 e 3,2x-[4/5+2/3]:32/3=7/20
Tìm x biết a,32-x-5/12=1 b,1+1/12<x<3+6/17 c)4/15/ 4/7 < x < 2/5 . 10/3
Tìm x biết
d) 32%-0,25:x=-17/5
e)(x+1/5)^2+17/25=26/25
f)-32/27-(3x-7/9)^3=-24/27
g)60%x+0,4x+x:3=2
h)|20/9-x|=1/12+1/20+1/30+1/42+1/56+1/72
i)8/5+(2/7+2/17+2/37/5/7+5/17+5/37).x=16/5
Lưu ý: câu i 2/7+2/17+2/37 phần(vế trên) 5/7+5/17+5/37(vế dưới)
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
Bài 1: 1) tìm x biết
A, 9/20 -8/15 . 5/12. B, 2/3÷4/5÷7/12. C, 7/9.1/3+7/9.2/3
2)tìm x biết
A, 2.(x-1)=4026. B, x.3,7+6,3.x=320. C, 0,25.3<3<1,02
Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
Tìm x, biết :
3/2 + 5/4 + 9/8 + 17/16 + 33/32 - 6 + x-1/x+1 = 31/32 - 2/2015
3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015
=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015
=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015
=31/32+5-6+x-1/x+1=31/32-2/2015
=5-6+x-1/x+1=31/32-2/2015-31/32
=-1+x-1/x+1=-2/2015
=x-1/x+1=-2/2015- -1
=x-1/x+1=2013/2015
=>x=2014