tìm x biết (x+1)^2=4/25
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\(3\left(x-2\right)+4\left(x-1\right)=25\)
\(\Leftrightarrow3x-6+4x-4=25\)
\(\Leftrightarrow7x=35\)
\(\Leftrightarrow x=5\)
\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x-3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\end{matrix}\right.\)
1: =>x=8-35=-27
2: =>15-4+x=6
=>x+11=6
hay x=-5
3: =>-30+25-x=-1
=>x+5=1
hay x=-4
4: =>x-(-13)=-8
=>x+13=-8
hay x=-21
5: =>x-29-17+38=-9
=>x-8=-9
hay x=-1
a) (x+1) ^2 = 25
(x+1) ^2 = 5^2
x + 1 = 5 Hoặc x + 1 = -5
x = 5 - 1 x = -5 - 1
x = 4 x = -6
b) 4^5.4^x-1 = 4^7
4^x-1 = 4^7:4^5
4^x-1 = 4^7-5
4^x-1 = 4^2
x- 1 = 2
x = 2+1
x = 3
a, x+4/7=7/4-1/4
x + 4/7 = 6/4
x = 6/4 - 4/7
x = 26/28
b, 9/2+(x-3/4)=25/4
x - 3/4 = 25/4 - 9/2
x - 3/4 = 7/4
x = 7/4 + 3/4
x = 10/4
a, x+4/7=7/4-1/4
x+4/7= 3/2
x = 3/2-4/7= 13/14
b,9/2+(x-3/4)=25/4
x-3/4 = 25/4 - 9/2 = 7/4
x= 7/4 + 3/4 = 5/2
a) Ta có: 6 x 3 = 48 nên x 3 = 8 . Do đó x = 2.
b) Ta có: ( x - 1 ) 2 = 2 2 nên x -1 = 2. Do đó x = 3.
c) Ta có: ( x + 1 ) 2 = 5 2 nên x +1 = 5. Do đó x = 4.
a) 4x = 64
4x = 43
=> x = 3
c) 5x+2 = 25
5x+2 = 52
=> x + 2 = 2
=> x = 0
mình thấy phần b hình như sai đề
Có:
a)\(4^x=64=4^3=>x=3\)
b)\(2^{x+1}=25=>2^x.2=25\Rightarrow2^x=25:2=12,5\)
=> ....
c)\(5^{x+2}=25\Rightarrow5^x.5^2=25=>5^x.25=25=>5^x=25:25=1\)
=>x=0
1.8100
2. 34 x 18 + 18 x 66
= 18 x ( 34 + 66)
= 18 x 100 = 1800
3. X × ( 8 + 12) = 160 + 20 × 12
= X x 20 = 160 + 240
= X x 20 = 400
X = 400 : 20 = 20
4. X x 12 - X x 2 = 2020
(12 - 2) x X = 2020
10 x X = 2020
X = 2020 : 10 = 202
(\(x\) + 1)2 = \(\dfrac{4}{25}\)
(\(x+1\))2 = (\(\dfrac{2}{5}\))2
\(\left[{}\begin{matrix}x+1=-\dfrac{2}{5}\\x+1=\dfrac{2}{5}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{2}{5}-1\\x=-\dfrac{2}{5}-1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)
Vậy \(x\in\){ \(-\dfrac{7}{5}\) ; - \(\dfrac{3}{5}\)}
`@` `\text {Ans}`
`\downarrow`
`(x+1)^2 = 4/25`
`=> (x+1)^2 = (+-2/5)^2`
`=>`\(\left[{}\begin{matrix}x+1=\dfrac{2}{5}\\x+1=-\dfrac{2}{5}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{2}{5}-1\\x=-\dfrac{2}{5}-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)
Vậy, `x \in {-3/5; -7/5}.`