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a,
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)
\(\Rightarrow5^{x+3}2=2.5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
b, (Check lai xem de sai o dau khong nhe)
\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)
Dat 5x ra ben ngoai
\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)
\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)
\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)
\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)
\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)
\(\Rightarrow5^x\left(5^{-3}\right).9379\)
=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)
1: x=3/4-1/2=3/4-2/4=1/4
2: x-1/5=2/11
=>x=2/11+1/5=21/55
3: x-5/6=16/42-8/56
=>x-5/6=8/21-4/28=5/21
=>x=5/21+5/6=15/14
4: x/5=5/6-19/30
=>x/5=25/30-19/30=6/30=1/5
=>x=1
5: =>|x|=1/3+1/4=7/12
=>x=7/12 hoặc x=-7/12
6: x=-1/2+3/4
=>x=3/4-1/2=1/4
11: x-(-6/12)=9/48
=>x+1/2=3/16
=>x=3/16-1/2=-5/16
1)x= 1/4
2)x= 2/11+ 1/5
x= 21/55
3)x - 5/6 = 5/21
x = 5/21+5/6
x = 15/14
4)x/5 = 5/6 + -19/30
x:5 = 1/5
x = 1/5.5
x = 1
5) |x| - 1/4 = 6/18
|x| = 6/18 - 1/4
|x| =7/12
⇒x= 7/12 hoặc -7/12
6)x = -1/2 +3/4
x= 1/4
7) x/15 = 3/5 + -2/3
x:15 = -1/15
x = -1/15. 15
x = -1
8)11/8 + 13/6 = 85/x
85/24 = 85/x
⇒ x = 24
9) x - 7/8 = 13/12
x = 13/12 + 7/8
x = 47/24
10)x - -6/15 = 4/27
x = 4/27 + (-6/15)
x = -34/135
11) -(-6/12)+x = 9/48
x= 9/48 - 6/12
x = -5/16
12) x - 4/6 = 5/25 + -7/15
x -4/6 = -4/15
x = -4/15 + 4/6
x = 2/5
b) Ta có: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow2\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow x+3=11\)
hay x=8
c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)
\(\Leftrightarrow18\cdot3^x+12\cdot3^x=10\cdot3^6\)
\(\Leftrightarrow30\cdot3^x=30\cdot3^5\)
Suy ra: x=5
d) Ta có: \(6\cdot8^{x-1}+8^{x+1}=6\cdot8^{19}+8^{21}\)
\(\Leftrightarrow6\cdot\dfrac{8^x}{8}+8^x\cdot8=6\cdot8^{19}+64\cdot8^{19}\)
\(\Leftrightarrow8^x\cdot\dfrac{35}{4}=70\cdot8^{19}\)
\(\Leftrightarrow8^x=8^{20}\)
Suy ra: x=20
`@` `\text {Ans}`
`\downarrow`
`a,`
`2/5 + x = 2/7`
`=> x = 2/7 -2/5`
`=> x= - 4/35`
Vậy, `x=-4/35`
`b,`
`x + 3/5 = -2/5`
`=> x = -2/5 - 3/5`
`=> x=-1`
Vậy, `x=-1`
`c, `
`x - 8/5 = 3/7`
`=> x=3/7 + 8/5`
`=> x=71/35`
Vậy, `x=71/35`
`d,`
`3/5 - x = 7/3`
`=> x=3/5 - 7/3`
`=> x=-26/15`
Vậy, `x=-26/15`
a) \(\dfrac{2}{5}+x=\dfrac{2}{7}\)
\(\Rightarrow x=\dfrac{2}{7}-\dfrac{2}{5}\)
\(\Rightarrow x=-\dfrac{4}{35}\)
b) \(x+\dfrac{3}{5}=-\dfrac{2}{5}\)
\(\Rightarrow x=-\dfrac{2}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=-1\)
c) \(x-\dfrac{8}{5}=\dfrac{3}{7}\)
\(\Rightarrow x=\dfrac{3}{7}+\dfrac{8}{5}\)
\(\Rightarrow x=\dfrac{71}{35}\)
d) \(\dfrac{3}{5}-x=\dfrac{7}{3}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{7}{3}\)
\(\Rightarrow x=-\dfrac{26}{15}\)
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(x^2=3\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
\(x^2=5\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\end{matrix}\right.\Rightarrow x=-\sqrt{5}\left(vì.x< 0\right)\)
\(x^2=7\Rightarrow\left[{}\begin{matrix}x=-\sqrt{7}\\x=\sqrt{7}\end{matrix}\right.\Rightarrow x=-\sqrt{7}\left(vì.x< 0\right)\)
\(x^2=9\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
\(\left(x-2\right)^2=2\Rightarrow\left[{}\begin{matrix}x-2=-\sqrt{2}\\x-2=\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-\sqrt{2}\\x=2+\sqrt{2}\end{matrix}\right.\)
\(\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x-2=-2\\x-2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\left(x-6\right)^2=6\Rightarrow\left[{}\begin{matrix}x-6=-\sqrt{6}\\x-6=\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{6}\\x=6+\sqrt{6}\end{matrix}\right.\)
\(\left(x-8\right)^2=8\Rightarrow\left[{}\begin{matrix}x-8=-2\sqrt{2}\\x-8=2\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-2\sqrt{2}\\x=2+2\sqrt{2}\end{matrix}\right.\)
\(\left(x-10\right)^2=10\Rightarrow\left[{}\begin{matrix}x-10=-\sqrt{10}\\x-10=\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-\sqrt{10}\\x=10+\sqrt{10}\end{matrix}\right.\)
\(\left(x-\sqrt{3}\right)^2=3\Rightarrow\left[{}\begin{matrix}x-\sqrt{3}=-\sqrt{3}\\x-\sqrt{3}=\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{3}\end{matrix}\right.\)
\(\left(x-\sqrt{5}\right)^2=5\Rightarrow\left[{}\begin{matrix}x-\sqrt{5}=-\sqrt{5}\\x-\sqrt{5}=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{5}\end{matrix}\right.\)
Lời giải:
a.
$\frac{-2}{3}+2x=\frac{4}{3}$
$2x=\frac{4}{3}-\frac{-2}{3}=2$
$x=2:2=1$
b.
$\frac{5}{8}-5:x=\frac{-3}{8}$
$5:x=\frac{5}{8}-\frac{-3}{8}=1$
$x=5:1=5$
c.
$\frac{2}{3}-x=\frac{-1}{2}$
$x=\frac{2}{3}-\frac{-1}{2}=\frac{7}{6}$
d.
$\frac{5}{7}-4x=\frac{-51}{7}$
$4x=\frac{5}{7}-\frac{-51}{7}=8$
$x=8:4=2$
`@` `\text {Ans}`
`\downarrow`
`a,`
`-2/3 + 2x = 4/3`
`=> 2x = 4/3 - (-2/3)`
`=> 2x = 2`
`=> x=2 \div 2`
`=> x=1`
Vậy, `x=1`
`b,`
`5/8 - 5 : x = -3/8`
`=> 5 \div x = 5/8 - (-3/8)`
`=> 5 \div x = 1`
`=> x= 5 \div 1`
`=> x=5`
Vậy, `x=5`
`c,`
`2/3 - x = -1/2`
`=> x=2/3 - (-1/2)`
`=> x=7/6`
Vậy, `x=7/6`
`d,`
`5/7 - 4x = -51/7`
`=> 4x = 5/7 - (-51/7)`
`=> 4x=8`
`=> x=8 \div 4`
`=> x=2`
Vậy, `x=2.`
`@` `\text {Kaizuu lv u}`
\(\dfrac{-2}{3}.\dfrac{2}{5}+\dfrac{-2}{3}.\dfrac{8}{5}\)
\(=\dfrac{-2}{3}.\left(\dfrac{2}{5}+\dfrac{8}{5}\right)\)
\(=\dfrac{-2}{3}.2\)
\(=\dfrac{-4}{3}\)
- \(\dfrac{2}{3}\) x \(\dfrac{2}{5}\) + (- \(\dfrac{2}{3}\)) x \(\dfrac{8}{5}\)
= - \(\dfrac{2}{3}\) x ( \(\dfrac{2}{5}\) + \(\dfrac{8}{5}\))
= - \(\dfrac{2}{3}\) x \(\dfrac{10}{5}\)
= - \(\dfrac{2}{3}\) x 2
= - \(\dfrac{4}{3}\)