3x+2 - 3x.23 = 27
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x=-\frac{1}{2}\)
b) \(x=-\frac{29}{15}\)
c) \(x=\frac{5}{6}\)
**** cho mình nhé
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a) \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)
\(=>\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(=>\left(4x+2\right)\left(2x-4\right)=0\)
\(=>4\left(2x+1\right)\left(x-2\right)=0\)
\(=>\orbr{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=-\frac{1}{2}\\x=2\end{cases}}\)
b)\(x^3-\frac{x}{49}=0=>x\left(x^2-\frac{1}{49}\right)=0=>x\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)
\(=>x=0\)hoặc \(x=\frac{1}{7}\) hoặc \(x=-\frac{1}{7}\)
a)\(\(\left(3x-1\right)^2-\left(x+3\right)^2=0\)\)
\(\(\Leftrightarrow\left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\)\)
\(\(\Leftrightarrow\left(2x-4\right)\left(4x+2\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\4x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}}\)\)
b)\(\(x^3-\frac{x}{49}=0\)\)
\(\(\Leftrightarrow\frac{49x^3-x}{49}=0\)\)
\(\(\Leftrightarrow x\left(49x^2-1\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\49x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(7x-1\right)\left(7x+1\right)=0\end{cases}}}\)\)\
\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7};x=-\frac{1}{7}\end{cases}}\)\)
c)\(\(x^2-7x+12=0\)\)
\(\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)\)
d) \(\(4x^2-3x-1=0\)\)
\(\(\Leftrightarrow4x^2-4x+x-1=0\)\)
\(\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)\)
\(\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\4x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)\)
e) Tham khảo tại : [Toán 8]Giải phương trình | Cộng đồng học sinh Việt Nam - HOCMAI Forum
https://diendan.hocmai.vn/threads/toan-8-giai-phuong-trinh.290061/
_Y nguyệt_
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)
=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}-\frac{2}{5}\)
=> \(\frac{3}{5}.\left(3x-3,7\right)=\frac{-57}{10}\)
=> \(\left(3x-3,7\right)=\frac{-19}{5}\)
=>\(3x=\frac{-19}{5}+\frac{37}{10}\)
=>\(3x=\frac{-1}{10}\)
=>\(x=\frac{-1}{30}\)
Ko biết đúng hay sai!!?
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$(x-15).27=0$
$x-15=0:27=0$
$x=15+0=15$
b.
$23(42-x)=0$
$42-x=0$
$x=42$
c.
$(9x+2).3=60$
$9x+2=60:3=20$
$9x=18$
$x=2$
d.
$71+(26-3x):5=75$
$(26-3x):5=75-71=4$
$26-3x=4.5=20$
$3x=26-20=6$
$x=6:2=3$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 22 + (2x -13) = 83 => 2x -13 = 61 => x = 37.
b) 51 - (-12 + 3x) = 27 => 63 - 3x = 27 => x = 12.
c) - (2x + 2) + 21 = - 23 => 2x + 2 = 44 => x = 21.
d) 25 - (25 - x) = 0 => 25 - 25 + x = 0 => x = 0.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x – 45).27 = 0
=> x - 45 = 0
=> x = 45
b) 23.(42- x) = 23
=> 42- x = 1
=> x = 41
c. 3x – 5=7
=> 3x = 12
=> x = 4
e. 15 – 5x=10
=> 5x = 5
=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
-27 + 5( x-7 ) = 23 - (71-3x)
-27 + 5( x-7 ) = 23 - 71 + 3x
5(x-7) - 3x = 23 - 71 + 27
5x - 35 - 3x = -21
x( 5-3 ) - 35 = -21
x.2 - 35 = -21
x.2 = -21 + 35
x.2 = 14
x = 7
Vậy x = 7
-27+5(x-7)=23-(71-3x)
=>-27+5x-35=23-71+3x
=>-27-35-23+71=3x-5x
=>-14=-2x
=>-2x=-14
=>x=-14:(-2)
=>x=7
Vậy x=7
\(3^{x+2}-3^x\cdot2^3=27\\\Rightarrow 3^x\cdot3^2-3^x\cdot8=27\\\Rightarrow 3^x\cdot(3^2-8)=27\\\Rightarrow 3^x\cdot(9-8)=3^3\\\Rightarrow 3^x=3^3\\\Rightarrow x=3\)