Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\left(x-1\right)^4-16\left(x-1\right)^2=0\)
\(\left(x-1\right)^2\left(\left(x-1\right)^2-16\right)=0\)
\(\left(x-1\right)^2\left(x^2-2x-15\right)=0\)
\(\left(x-1\right)^2\left(x-5\right)\left(x+3\right)=0\)
x=1
x=5
x=-3
a) x - 8 - (12 - 2x) = -20
=> x - 8 - 12 + 2x = -20
=> (x + 2x) + (-8 - 12) = -20
=> 3x - 20 = -20
=> 3x = 0 => x = 0
b) -27 + (x + 8) - ( +11) = 2
=> -27 + x + 8 - 11 = 2
=> -27 + x = 2 + 11 - 8
=> -27 + x = 5
=> x = 5 - (-27) = 32
c) -2x - 16 = -2 - (3x + 9)
=> -2x - 16 = -2 - 3x - 9
=> -2x - 16 + 2 + 3x + 9 = 0
=> (-2x + 3x) + (-16 + 2 + 9) = 0
=> x - 5 = 0
=> x = 5
\(a,x-8-\left(12-2x\right)=-20\)
\(x-8-12+2x=-20\)
\(x+2x-8-12=-20\)
\(3x-20=-20\)
\(3x=-20+20\)
\(3x=0\)
\(x=0\)
\(b,-27+\left(x+8\right)-\left(+11\right)=2\)
\(-27+x+8-11=2\)
\(x-27+8-11=2\)
\(x-30=2\)
\(x=2+30\)
\(x=32\)
\(c,-2x-16=2-\left(3x+9\right)\)
\(-2x-16=2-3x-9\)
\(-2x+3x=2-9+16\)
\(x=9\)
Học tốt
( x - 2 ) ^2 = 1^2 hoặc ( x - 2 )^2 = -1^2
=> x - 2 = 1 hoặc x - 2 = -1
Ta có : x - 2 = 1 => x = 2 + 1 => x = 3
x - 2 = - 1 => x = - 1 + 2 => x = 1
( 2x -1 )^3 = -8
=> (2x-1)^3 = -2^3
=> 2x-1 = -2 => 2x = -2+1 => 2x = -1 => x = -1 :2 => x = -1/2
(x+1/2)^2 = 1/16
=> (x+1/2)^2 = 1/8^2 hoặc (x+1/2)^2 = -1/8^2
=> x+1/2 = 1/8 hoặc x+1/2 = -1/8
Ta có : x+1/2 = 1/8
x= 1/8 - 1/2
x = 2/16 - 8/16
x = -6/16 = -3/8
x + 1/2 = -1/8
x = -1/8 - 1/2
x = -2/16 -8/16
x= -10/16 = -5/8
* ^ là mũ nhé bạn :))
a, \(\frac{x+1}{5}=\frac{3}{7}\Rightarrow7\left(x+1\right)=15\Rightarrow7x+7=15\Rightarrow7x=8\Rightarrow x=\frac{8}{7}\)
b, \(\frac{x-2}{3}=\frac{3}{8}\Rightarrow8\left(x-2\right)=9\Rightarrow8x-16=9\Rightarrow8x=25\Rightarrow x=\frac{25}{8}\)
c, \(\frac{-x-1}{2}=\frac{-3}{5}\Rightarrow5\left(-x-1\right)=-6\Rightarrow-5x-5=-6\Rightarrow-5x=-1\Rightarrow x=\frac{1}{5}\)
d, \(\frac{4}{5-x}=\frac{1}{3}\Rightarrow5-x=12\Rightarrow x=-7\)
e, \(2x\left(x-\frac{1}{7}\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}}\)
\(\left(\frac{-7}{4}:\frac{5}{8}\right)\cdot\frac{11}{16}=\frac{-7}{4}\cdot\frac{8}{5}\cdot\frac{11}{16}=\frac{-7.11}{4.5.2}=\frac{-77}{40}\)
bài đầu
(x-2)2=1
=mà 12 kiểu gì cũng =1
vậy (x-2)=1
x=1+2=3
Các đề bài trên khi chuyển vế đều bị mất đi x nên không có x thỏa mãn
\(\frac{8}{x+1}=\frac{16}{y+2}=\frac{24}{z+3}=\frac{16+48-24}{2x+2+3y+6-z-3}=\frac{40}{25}=1,6\)
\(\Rightarrow\) x + 1 = 5; y + 2 = 10; z + 3 = 15
\(\Rightarrow\) x = 4; y = 8; z = 12
\(\dfrac{x+1}{16}-\dfrac{2x-1}{8}=\dfrac{-x-8}{2}\)
\(\Rightarrow\dfrac{x+1}{16}-\dfrac{4x-2}{16}+\dfrac{8\left(x+8\right)}{16}=0\)
\(\Rightarrow x+1-4x+2+8x+64=0\)
\(\Rightarrow5x+67=0\)
\(\Rightarrow5x=-67\)
\(\Rightarrow x=-\dfrac{67}{5}\)
`(x+1)/16-(2x-1)/8=(-x-8)/2`
`=>x+1-2(2x-1)=8(-x-8)`
`=>x+1-4x+2=-8x-64`
`=>-3x+3=-8x-64`
`=>5x=67`
`=>x=67/5`
Vậy `x=67/5`