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a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
\(\Leftrightarrow2x=\dfrac{1}{64}\)
hay \(x=\dfrac{1}{128}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\\\Leftrightarrow \left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\\ \Leftrightarrow\left(x-1\right)^{x+2}\left(\left(x-1\right)^{x+2}+1\right)=0\\ \Leftrightarrow\left(x-1\right)^{x+2}=0hoac\left(x-1\right)^{x+2}+1=0\)
Giả tiếp đc x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(2^x=\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{12}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}\)
\(\Leftrightarrow2^x=\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot31}{2\cdot\left(2\cdot3\cdot4\cdot...\cdot31\right)\cdot64}\)
\(\Leftrightarrow2^x=\dfrac{1}{2}\cdot\dfrac{1}{64}=\dfrac{1}{128}\)
\(\Leftrightarrow2^x=\dfrac{1}{2^6}\)
\(\Leftrightarrow2^{x+6}=1\)
\(\Leftrightarrow x+6=0\)
hay x=-6
Vậy: x=-6
`1/4 . 2/6 . 3/8 ... . 30/62 .31/64 =2^x`
`-> (1.2.3....30.31)/(4.6.8....62.64)=2^x`
`-> (1.(2.3...31))/(2.(2.3.4...31).32)=2^x`
`-> 1/(2.32)=2^x`
`-> 1/64=2^x`
`-> 1/(2^6)=2^x`
`-> x=-6`.
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a, 11/13 - ( 5/42 - x ) = - (5/28 - 11/13)
11/13 - (5/42 - x) = - 5/28 + 11/13
- (5/42 - x) + 5/28 = -11/13 + 11/13
- 5/42 + x + 5/28 = 0
- 5/42 + x = 0 - 5/28
- 5/42 + x = - 5/28
x = -5/28 +5/42
x = - 5/84
b, / x + 4/15 \ - / - 3,75 \ = - / - 2,15 \
./ x + 4/15 \ - 3,75 = - 2,15
/ x + 4/15 \ = -2,15 + 3,75
/ x + 4/15 \ = 1,6
x + 4 / 15 = 1,6 hoặc x+ 4/15 = - 1,6
x = 1,6 - 4/15 x = - 1,6 -4/15
x = 4/3 x = -28/15
Vậy x = 4/3 hoặc x = - 28/15
c, ( 0,25 - 30% x ) . 1/3 = 1/4 - 31/6
( 1/4 - 3/10 x ) . 1/3 = - 59/12
( 1/4 - 3/10 x ) = - 59/12 : 1/3
1/4 - 3/10 x = - 59/4
3/10 x = 1/4 + 59/4
3/10 x = 15
x = 15 : 3/10
x = 50
d, ( x - 1/2 ) : 1/3 + 5/7 = 68/7
( x - 1/2 ) : 1/3 = 68/7 - 5/7
( x - 1/2 ) : 1/3 = 63/7
( x - 1/2 ) = 63/7 . 1/3
x -1/2 = 3
x = 3 + 1/2
x = 7/2
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Có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}...\frac{30}{2.31}.\frac{31}{2.32}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}.\frac{1}{32}\)
\(=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^x\)\(\Rightarrow1=2^x.2^{36}=2^{36+x}\)\(\Rightarrow2^{36+x}=2^0\Rightarrow36+x=0\Rightarrow x=-36\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 3|x|=-9
=>|x|=-3(loại)
b: 6(x-2)-(x-3)=31
=>6x-12-x+3=31
=>5x-9=31
=>5x=40
hay x=8
c: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y+z}{7+5+3}=\dfrac{30}{15}=2\)
Do đó: x=14; y=10; z=6
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Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r
a, \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)
\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)
Có : (31 - 1) : 1 + 1 = 31 (thừa số 2)
\(\Rightarrow\frac{1}{2^{31}.32}=2x\)
\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+1=x+4\)
\(\Leftrightarrow0=3\text{ (vô lý) }\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)
=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)
=>x=-36