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Giúp mình bài này với ạ
Tìm x
a) 3(x-4)+2,6=5,9
b)|1+x|+3=3
c)(\(\frac{-2}{3}\))\(^x\)=\(\frac{-8}{27}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3(x-4)+2,6=5,9
\(3\left(x-4\right)=3,3\)
\(x-4=\frac{11}{10}\)
\(x=\frac{51}{10}\)
b)|1+x|+3=3
\(\Rightarrow|1+x|=0\)
\(1+x=0\)
\(x=-1\)
c) \(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\)
\(x=3\)
chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{x-1}{3}=\frac{x+3}{5}\)
\(\Rightarrow5\left(x-1\right)=3\left(x+3\right)\)
\(\Rightarrow5x-5=3x+9\)
\(\Rightarrow5x-3x=9+5\)
\(\Rightarrow2x=14\)
\(\Rightarrow x=7\)
Vậy x = 7
b) \(\frac{x+1}{1}=\frac{1}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=1\)
\(\Rightarrow x+1=\pm1\)
+) \(x+1=1\Rightarrow x=0\)
+) \(x+1=-1\Rightarrow x=-2\)
Vậy x = 0 hoặc x = -2
- \(\frac{x-1}{3}=\frac{x+3}{5}\)
=> (x - 1).5 = (x + 3).3
=> 5x - 5 = 3x + 9
=> 5x - 3x = 9 + 5
=> 2x = 14
=> x = 14 : 2
=> x = 7
Vậy x = 7
- \(\frac{x+1}{1}=\frac{1}{x+1}\)
=> (x + 1)2 = 1
\(\Rightarrow\left[\begin{array}{nghiempt}x+1=1\\x+1=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=0\\x=-2\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3|x - 7| - 3 = 32
<=> 3|x - 7| = 12
<=> |x - 7| = 4
<=> \(\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}}\)
<=> \(\orbr{\begin{cases}x=11\\x=3\end{cases}}\)
Vậy ...
\(3\left|x-7\right|-3=9\)
=> \(3\left|x-7\right|=9+3\)
=> \(3\left|x-7\right|=12\)
=> \(\left|x-7\right|=4\)
=> \(\orbr{\begin{cases}x+7=4\\x+7=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=-11\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(|x+4|=\frac{7}{3}\) \(\Rightarrow x+4=\pm\left(\frac{7}{3}\right)\)
TH1: \(x+4=\frac{7}{3}\)
\(x=\frac{7}{3}-4=-\frac{5}{3}\)
TH2: \(x+4=-\frac{7}{3}\)
\(x=-\frac{7}{3}-4=-\frac{19}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. Thay \(x=-\frac{2}{3}\) vào \(C=6x^3-3x^2+2\left|x\right|+4\), ta có :
\(C=6\left(-\frac{2}{3}\right)^3-3\left(-\frac{2}{3}\right)^2+2\left|-\frac{2}{3}\right|+4\)
\(\Rightarrow C=6.\frac{-8}{27}-3.\frac{4}{9}+2.\frac{2}{3}+4\)
\(\Rightarrow C=-\frac{16}{9}-\frac{4}{3}+\frac{8}{3}+4\)
\(\Rightarrow C=\frac{32}{9}\)
b. Thay \(x=\frac{1}{2};y=-3\)vào \(D=2\left|x\right|-3\left|y\right|\), ta có :
\(D=2\left|\frac{1}{2}\right|-3\left|-3\right|\)
\(\Rightarrow D=2.\frac{1}{2}-3.3\)
\(\Rightarrow D=2-9\)
\(\Rightarrow D=-7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)
⇔ | 3 - 2x | + 3/4 = 11/4
⇔ | 3 - 2x | = 8/4 = 2
⇔ \(\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}}\text{⇔}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)
b) 2x+2 - 2x = 96
⇔ 2x( 22 - 1 ) = 96
⇔ 2x.3 = 96
⇔ 2x = 32
⇔ 2x = 25
⇔ x = 5
c) ( 2x + 5 )3 = -27
⇔ ( 2x + 5 )3 = (-3)3
⇔ 2x + 5 = -3
⇔ 2x = -8
⇔ x = -4
a. \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)
\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\left|-\frac{11}{4}\right|\)
\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\frac{11}{4}\)
\(\Rightarrow\left|3-2x\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)
b. 2x+2 - 2x = 96
<=> 2x . 22 - 2x = 96
<=> 2x ( 22 - 1 ) = 96
<=> 2x . 3 = 96
<=> 2x = 32 = 25
<=> x = 5
c. ( 2x + 5 )3 = - 27
<=> ( 2x + 5 )3 = ( - 3 )3
<=> 2x + 5 = - 3
<=> 2x = - 8
<=> x = - 4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^{x+1}.3^y=3^x.4^x\)
\(\Rightarrow\hept{\begin{cases}2^{x+1}=4^x\\3^y=3^x\end{cases}}\Rightarrow\hept{\begin{cases}x+1=x+2\\x=y\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\)
vậy x=1,y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
1) a) x=0 hoặc x=4 hoặc x=-4
b) x=-3 hoặc x=1 hoặc x=-1
c) x=1 hoặc x=4
d) x=1 hoặc x=-1/6
2) a) m(x) = 3x
b) x=-2 hoặc x=-1