8x+(-34) = 2x+6
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(8x+\left(-34\right)=-2x+6\)
\(\Leftrightarrow8x+2x=6+34\)
\(\Leftrightarrow10x=40\)
\(\Leftrightarrow x=4\)
\(8x+\left(-34\right)=-2x+6\)
\(\Leftrightarrow8x+2x=6+34\)
\(\Leftrightarrow10x=40\)
\(\Leftrightarrow x=\frac{40}{10}=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
⇔ 8x - 3 - 6x + 4 = 4x - 2 + x + 3
⇔ 2x + 1 = 5x + 1
⇔ 2x - 5x = 1 - 1
⇔ -3x = 0 ⇔ x = 0
Vậy phương trình đã cho có nghiệm là x = 0.
![](https://rs.olm.vn/images/avt/0.png?1311)
4: \(\Leftrightarrow3^{x+4}\cdot\dfrac{1}{3}-4\cdot3^x=3^{16}\left(1-4\cdot3^3\right)\)
=>\(3^x\cdot27-4\cdot3^x=3^{16}\cdot\left(-107\right)\)
=>3^x*23=3^16*(-107)
=>\(x\in\varnothing\)
2: \(\Leftrightarrow2^x\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)=2^{10}\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)\)
=>2^x=2^10
=>x=10
3: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>8^x=8^9
=>x=9
1: \(\Leftrightarrow3^x\cdot\left(4\cdot\dfrac{1}{9}+2\cdot3\right)=3^4\left(4+2\cdot3^3\right)\)
=>3^x=3^4*3^2
=>x=4+2=6
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =x^4-3x^5+4x^8
b: =2x^3+2x^2+4x
c: =4x^2+8x-5
d: =2x+3x^2+7x^4
Ta có: 8x+(-34)=2x+6
\(\Rightarrow\)8x-34-2x-6=0
hay 6x-40=0
\(\Rightarrow\)6x=40
hay x=\(\frac{40}{6}=\frac{20}{3}\)
Vậy: \(x=\frac{20}{3}\)
8x+(-34) = 2x+6
8x -34 = 2x+6
8x -2x=6+34
6x=40
x=40:6
=> x=\(\frac{40}{6}\)=\(\frac{20}{3}\)
Vậy x = \(\frac{20}{3}\)