1)tìm x biết:
a)(x-2)^2=1
b)(2x-1)^3=-8
c)(x+1/2)^2=1/16
d)(x-2)^3=-27
e(2x-3)^2=25
g)3^x-1=1/243
h)2^x+2^x+3=144
k)81^-2x*27^x=95
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29 tháng 7 2021
a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
29 tháng 7 2021
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
NM
4
2 tháng 9 2021
a) \(3\left(x-2\right)+2\left(x-3\right)=5\)
\(\Rightarrow3x-6+2x-6=5\)
\(\Rightarrow5x=17\Rightarrow x=\dfrac{17}{5}\)
b) \(\left(2x-8\right)^2-16=0\)
\(\Rightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)
\(\Rightarrow\left(2x-12\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=12\\2x=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
c) \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)
\(\Rightarrow4x^2-4x+1-4x^2+12x-x+3=3\)
\(\Rightarrow7x=-1\Rightarrow x=-\dfrac{1}{7}\)
2 tháng 9 2021
a: Ta có: \(3\left(x-2\right)+2\left(x-3\right)=5\)
\(\Leftrightarrow3x-6+2x-6=5\)
\(\Leftrightarrow5x=17\)
hay \(x=\dfrac{17}{5}\)
b: Ta có: \(\left(2x-8\right)^2-16=0\)
\(\Leftrightarrow\left(2x-4\right)\left(2x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
14 tháng 10 2021
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
14 tháng 10 2021
Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
5 tháng 9 2021
a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Rightarrow\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\\ \Rightarrow\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\\ \Rightarrow-3\left(4x+5\right)=9\\ \Rightarrow4x+5=-3\\ \Rightarrow4x=-8\\ \Rightarrow x=-2\)
a. (x - 2)2 = 1
<=> (x - 2)2 = 12 = (-1)2
<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)
Vậy x \(\in\){1; 3}.
b. (2x - 1)3 = -8
<=> (2x - 1)3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -2 + 1
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2.
c. (x + 1/2)2 = 1/16
<=> (x + 1/2)2 = (1/4)2 = (-1/4)2
<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
Vậy x \(\in\){-1/4; -3/4}.
d. (x - 2)3 = -27
<=> (x - 2)3 = (-3)3
<=> x - 2 = -3
<=> x = -3 + 2
<=> x = -1
Vậy x = -1.
a.\(\left(x-2\right)^2\)=1
<=> x-2=1 hoặc x-2=-1
<=> x= 3 hoặc x=1
b.\(\left(2x-1\right)^3\)=-8
\(\left(2x-1\right)^3\)=\(\left(-2\right)^3\)
2x-1=-2
2x=-1
x=-1/2
c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)
x+\(\frac{1}{2}\)=\(\frac{1}{4}\) hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)
x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)
d.\(\left(x-2\right)^3\)=-27
\(\left(x-2\right)^3\)=\(\left(-3\right)^3\)
x-2=-3
x=-1