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a, (x - 2)2 = 1
(x - 2)2 = -12
=> x - 2 = -1
x = -1 + 2
x = -1
b, (2x - 1)3 = -27
(2x - 1)3 = -33
=> 2x - 1 = -3
2x = -3 + 1
2x = -2
x = -2 : 2
x = -1
a) (x-2)^2 = 1 = 1^2 = (-1)^2
=> x-2 = 1 => x = 3
x - 2 = -1 => x = 1
.KL:..
b) (2x-1)^3 = -27 = (-3)^3
=> 2x-1 = -3 => 2x = -2 => x = -1
c)16/2^n = 1
2^4 : 2^n = 1
24-n = 1 = 20
=> 4-n = 0 => n = 4
c) (x-1/2)^3 = 1/27 = 1/3^3
=>x-1/2 = 1/3
x = 5/6
d) (x+1/2)^2 = 4/25 = (2/5)^2 = (-2/5)^2
...
rùi bn tự lm như phần a nha
e) (x-1)x+2 = (x-1)x+6
=> (x-1)x+2 - (x-1)x+6 = 0
(x-1)x+2.[1-(x-1)4 ] = 0
=> (x-1)x+2 = 0 => x-1 = 0 => x = 1
1-(x-1)4 = 0 => (x-1)^4 = 1 => x -1 = 1 => x = 2
x -1 = -1 => x = 0
KL:...
f) (x-2)2 + (y-3)2 = 0
=> (x-2)^2 = 0 => x - 2=0 => x = 2
(y-3)^2=0 => y-3 = 0 => y =3
g) 5(x-2).(x+3) = 1 = 50
=> (x-2).(x+3) = 0
=> x-2 = 0 => x = 2
x+3 = 0 => x = -3
KL:...
a) 27x : 3x = 9
(27 : 3)x = 9
9x = 91
x = 1
b) 25 : 5x =5
5x = 25 : 5
5x = 51
x = 1
c) 2 : (x + 2)2 = \(\dfrac{1}{18}\)
(x + 2)2 = 2 : \(\dfrac{1}{18}\)
(x + 2)2 = 36
\(\Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
d) (5x - 1)2 = \(\dfrac{36}{49}\)
(5x - 1)2 = \(\left(\dfrac{6}{7}\right)^2\)
Bạn làm tiếp nha, mình có việc bận :v
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
a)2x+3+2x=144
2x*23+2x=144
2x * (23+1)=144
2x * 9 =144
2x=144/9=16
2x=16 =>x=4
b) 7x+7x+1=392
7x + 7x * 7 =392
7x * (1+7)=392
7x * 8= 392
7x= 392/8=49
7x=49 => x=2
d) 3x+3x+3=2268
3x+ 3x * 33=2268
3x *(1+33)=2268
3x*28=2268
3x=2268/28
3x=81 =>x=4
e) 9x+2+9x-92*82=0
9x*92+9x-92 *82=0
9x*(92+1)-92*82=0
9x*82-92*82=0
82*(9x-92)=0
=>9x-92=0
9x=0+92=92
=>x=2
f)8x. 16-2x=45
23x. 24 . -2x=45
23x+ 4 . -2x =45
23x-8x=45
2-5x =210
=>-5x=10 =>x=-2
a) (x - 1) 3 = 27
( x -1 )3 = 33
=> x - 1= 3
=> x = 4
b) x2 +x = 0
x .( x +1 ) = 0
=> \(\hept{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)
c) (2x -3)3 = 25
(2x - 3) 3 = 53
=> 2x - 3 = 5
=> 2x = 8
=> x =4
d) (2x - 3) 2 = 36
=> (2x - 3)2 = 62
=> 2x - 3 = 6
=> 2x = 9
=> x = 4,5
e) 5x+2 = 625
5x . 52 = 625
5x = 625 : 25
5x =25 = 52
=> x= 2
g) (2x - 1) = -8
2x = (-8) + 1
2x = (-7)
x = -3,5
a)(x-1)3-27=0
\(\Leftrightarrow\)(x-1-3)[(x-1)2+3(x-1)+9)]
b) x2+x=0
\(\Leftrightarrow\)x(x+1)=0
c) câu c hình như phải là bình phương chứ không phải lập phương
d)(2x-3)2-36=0
\(\Leftrightarrow\)(2x-3-6)(2x-3+6)
\(\Leftrightarrow\)(2x-9)(2x+3)
e)5x+2=625
\(\Leftrightarrow\)5x=625/25=25=52
\(\Leftrightarrow\)x=2
g)bạn có thiếu dấu lập phương không vậy
a)x=-2
b)x=1
c)x=1/2
f)x=1 hoặc x=-1
h)x=0 hoặc x=6
i)x=2
hok tốt!
_Lan Lan_
Áp dụng hằng đẳng thức:\(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)
\(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)
Áp dụng vào từng bài là được:
\(VD1:x^3+3x^2+3x+1=-1\)
\(\Rightarrow\left(x+1\right)^3=-1\)
\(\Rightarrow x=-2\)
\(VD2:x^3-9x^2+27x-27=-8\)
\(\Rightarrow\left(x-3\right)^3=-8\)
\(\Rightarrow x=1\)
a) \(x^2=\frac{1}{16}\Rightarrow x^2=\left(\pm\frac{1}{4}\right)^2\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)
b) \(x^3=\frac{27}{64}\Rightarrow x^3=\left(\frac{3}{4}\right)^3\Rightarrow x=\frac{3}{4}\)
c) \(\left(x+1\right)^2-1=\frac{1}{2}\)
=> \(\left(x+1\right)^2=\frac{1}{2}+1=\frac{3}{2}\)
=> vô nghiệm
d) (2x - 1)3 - 27 = 0 => (2x - 1)3 = 27 => (2x - 1)3 = 33 => 2x - 1 = 3 => 2x = 4 => x = 2
e) Vì \(x^2\ge0\forall x\)
\(\left(2,5-y\right)^2\ge0\forall y\)
=> x2 + (2,5 - y)2 \(\ge\)0 với mọi x,y
Dấu " = " xảy ra khi x2 = 0 => x = 0 và (2,5 - y)2 = 0 => y = 2,5
Vậy x = 0,y = 2,5
f) 2x = 8 => 2x = 23 => x = 3
g) (-3)x + 1 = -26
=> (-3)x = -26 - 1 = -27
=> (-3)x = (-3)3
=> x = 3
h) 24-x = 32
=> 24-x = 25
=> 4 - x = 5
=> x = -1
j) 2x + 2x+1 = 18
=> 2x + 2x . 2 = 18
=> 2x (1 + 2) = 18
=> 2x . 3 = 18
=> 2x = 6
=> x không thỏa mãn
j) Thiếu dữ liệu