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Trịnh Thị Nguyệt Ánh:
(1 + 2 + 3 + ... + 9)x + 25 = 205
45 x + 25 = 205
45 x = 205 - 25
45 x = 180
x = 180 : 45
x = 4
=> Vậy x = 4
:), Chúc bạn học tốt!!!
`1, -2/9 xx 15/17 + (-2/9) xx 2/17`
`= -2/9 xx (15/17 + 2/17)`
`= -2/9 xx 17/17`
`=-2/9xx1`
`=-2/9`
__
`-5/3 xx 6/5 + (-7/9) xx 3/10`
`= -30/15 + (-21/90)`
`= -2 + (-7/30)`
`=-60/30 +(-7/30)`
`=-67/30`
__
`15/20 xx 7/5 + (-9/7) xx (-6/4)`
`=3/4 xx7/5 + (-9/7) xx(-6/4)`
`= 21/20 + 54/28`
`= 21/20 + 27/14`
`=417/140`
__
`-25/13 xx 5/19 + (-25/13) xx 14/19`
`=-25/13 xx (5/19 +14/19)`
`=-25/13 xx 19/19`
`= -25/13 xx 1`
`=-25/13`
__
`-7/13 xx 13/5 + (-9/7) xx 5/3`
`=-7/5 +(-15/7)`
`=-124/35`
1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)
=>3x-1/5=3/25 hoặc 3x-1/5=-3/25
=>3x=8/25 hoặc 3x=2/25
=>x=8/75 hoặc x=2/75
2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)
=>2x-1/3=2/9 hoặc 2x-1/3=-2/9
=>2x=5/9 hoặc 2x=1/9
=>x=5/18 hoặc x=1/18
1. 11(x-9)=77
x-9=7
x=16
2. 4(x-3)=48
x-3=12
x=15
3.3(x-8)=81
(x-8)=27
x=35
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
1.\(\left(\dfrac{1}{3}-x\right)^2=\dfrac{9}{25}\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}-x=\dfrac{3}{5}\\\dfrac{1}{3}-x=-\dfrac{3}{5}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{15}\\x=\dfrac{14}{15}\end{matrix}\right.\)
2.\(\left(5-x\right)^2=25\Leftrightarrow\left[{}\begin{matrix}5-x=5\\5-x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
1: =-2/9(15/17+2/17)=-2/9
2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)
=-2-7/30=-67/30
3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
1: =-2/9(15/17+2/17)=-2/9
2: =−63+−2190=−63+−2190
=-2-7/30=-67/30
3: =34⋅75+97⋅32=34⋅75+97⋅32
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
a,\(\left(x-15\right):50+22=24\)
\(< =>\frac{\left(x-15\right)}{50}=2< =>x-15=100\)
\(< =>x=100+15=115\)
b,\(42-\left(2x+32\right)+12:2=6\)
\(< =>42-2x-32=0\)
\(< =>10-2x=0< =>x=\frac{10}{2}=5\)
Làm nốt :
c) \(134-2\left\{156-6\cdot\left[54-2\cdot\left(9+6\right)\right]\right\}\cdot x=86\)
=> 134 - 2{156 - 6 . [54 - 2 . 15]} . x = 86
=> 134 - 2{156 - 6 . [54 - 30]} . x = 86
=> 134 - 2{156 - 6. 24} . x = 86
=> 134 - 2{156 - 144} . x = 86
=> 134 - 2.12 . x = 86
=> 134 - 24 . x = 86
=> 24.x = 48
=> x = 2
Bài 2 : a) 120 : [21 - (4x - 4)] = 23.3
=> 120 : [21 - (4x - 4)] = 8.3
=> 120 : [21 - (4x - 4)] = 24
=> 21 - (4x - 4) = 5
=> 4x - 4 = 16
=> 4x = 20
=> x = 5
b) 3.[205 - (x - 9)] - 486 = 0
=> 3.[205 - (x - 9)] = 486
=> 205 - (x - 9) = 162
=> x - 9 = 205 - 162 = 43
=> x = 43 + 9 = 52
c) 204 - 2{200 - 5.[64 - 2.(11 + 6)]} . x = 4
=> 204 - 2{200 - 5.[64 - 2.17]} . x = 4
=> 204 - 2{200 - 5 .[64 - 34]}.x = 4
=> 204 - 2{200 - 5.30} . x = 4
=> 204 - 2{200 - 150}.x = 4
=> 204 - 2.50 . x = 4
=> 2.50.x = 200
=> 100.x = 200
=> x = 2