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5 tháng 4 2020
a) 7(x - 5) + 2 = 51
\(\Leftrightarrow\) 7(x - 5) = 51 - 2
\(\Leftrightarrow\) 7(x - 5) = 49
\(\Leftrightarrow\) x - 5 = 49 : 7
\(\Leftrightarrow\) x - 5 = 7
\(\Leftrightarrow\) x = 7 + 5
\(\Leftrightarrow\) x = 12.
Vậy x = 12.
k) 2412 : (3x + 147) = |-38| + (-26)
\(\Leftrightarrow\) 2412 : (3x + 147)= 38 + (-26)
\(\Leftrightarrow\) 2412 : (3x + 147)= 12
\(\Leftrightarrow\) 3x + 147 = 2412 : 12
\(\Leftrightarrow\) 3x + 147 = 201
\(\Leftrightarrow\) 3x = 201 - 147
\(\Leftrightarrow\) 3x = 54
\(\Leftrightarrow\) x = 54 : 3
\(\Leftrightarrow\) x = 18.
Vậy x = 18.
I) 4824 : (4x + 137) = |-59| + (-35)
\(\Leftrightarrow\) 4824 :(4x + 137) = 59 + (-35)
\(\Leftrightarrow\) 4824 :(4x + 137) = 24
\(\Leftrightarrow\) 4x + 137 = 4824 : 24
\(\Leftrightarrow\) 4x + 137 = 201
\(\Leftrightarrow\) 4x = 201 - 137
\(\Leftrightarrow\) 4x = 64
\(\Leftrightarrow\) x = 64 : 4
\(\Leftrightarrow\) x = 16.
Vậy x = 16.
c) |-123| - 5(x - 3) = (-28) + 66
\(\Leftrightarrow\) 123 - 5(x - 3) = 38
\(\Leftrightarrow\) 5(x - 3) = 123 - 38
\(\Leftrightarrow\) 5(x - 3) = 85
\(\Leftrightarrow\) x - 3 = 85 : 5
\(\Leftrightarrow\) x - 3 = 17
\(\Leftrightarrow\) x = 17 + 3
\(\Leftrightarrow\) x = 20.
Vậy x = 20.
m) 7x - 4 . 6 = 2058
\(\Leftrightarrow\) 7x - 24 = 2058
\(\Leftrightarrow\) 7x = 2058 + 24
\(\Leftrightarrow\) 7x = 2082
\(\Leftrightarrow\) x = 2082 : 7
\(\Leftrightarrow\) x = \(\frac{2058}{7}\).
Vậy x = \(\frac{2058}{7}\).
Phần b) ra phân số nên mình để mai làm và cả những bài còn lại nữa.
Chúc bạn học tốt !!!
TD
5 tháng 4 2020
---------------NHANH NHA MK ĐANG CẦN GẤP--------------
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19 tháng 1 2022
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
f) (x – 5)2 = 9
TG
9 tháng 2 2021
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
9 tháng 2 2021
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
TL
1
11 tháng 3 2018
\(a,3x+17x=340\)
\(x\left(17+3\right)=340\)
\(x20=340\)
\(x=340:20=17\)
\(b,\left|2x+1\right|=3\\ \Rightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3-1=2\\2x=-3-1=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(c,3^x+3^{x+1}+3^{x+2}=1053\\ 3^x\left(1+3+9\right)=1053\\ 3^x.13=1053\\ 3^x=1053:13=81=3^4\\ \Rightarrow x=4\)
SS
25 tháng 6 2019
\(\left(x-2\right)^3+\left(3\text{x}-1\right)\left(3\text{x}+1\right)=\left(x+1\right)^3\)
\(\Leftrightarrow\left(x-2\right)^3+\left(3\text{x}-1\right)\left(3\text{x}+1\right)-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x^3-6\text{x}^2+12\text{x}-8\right)+\left(9\text{x}^2-1\right)-\left(x^3+3\text{x}^2+3\text{x}+1\right)=0\)
\(\Leftrightarrow x^3-6\text{x}^2+12\text{x}-8+9\text{x}^2-1-x^3-3\text{x}^2-3\text{x}-1=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(-6\text{x}^2+9\text{x}^2-3\text{x}^2\right)+\left(12\text{x}-3\text{x}\right)+\left(-8-1-1\right)=0\)
\(\Leftrightarrow9\text{x}-10=0\)
\(\Leftrightarrow9\text{x}=10\Leftrightarrow x=\frac{10}{9}\)
Vậy x = \(\frac{10}{9}\)
ND
2
5 tháng 5 2021
Sửa đề: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415
a) Ta có: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415
⇔3x4(41⋅5+45⋅9+49⋅13+...+481⋅85)=415⇔3x4(41⋅5+45⋅9+49⋅13+...+481⋅85)=415
⇔x⋅34(1−15+15−19+19−113+...+181−185)=415⇔x⋅34(1−15+15−19+19−113+...+181−185)=415
⇔x⋅34(1−185)=415⇔x⋅34(1−185)=415
⇔x⋅6385=415⇔x⋅6385=415
hay x=68189x=68189
Vậy: x=68189
OD
1
5 tháng 5 2021
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
OD
1
5 tháng 5 2021
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
OD
1
5 tháng 5 2021
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
ta có :
3x + 3x + 1 + 3x + 2 = 1053
=> 3x . 1 + 3x . 3 + 3x . 32 = 1053
=> 3x (1 + 3 + 32) = 1053
=> 3x . 13 = 1053
=> 3x = 1053 : 13 = 81
=> x = 4
3x+3x-1+3x-2=1053