a) \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)

b) \(\left(x-2\right)^2-1=0\Leftrightarrow\left(x-2\right)^2=1\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) vậy \(x=3;x=1\)

c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=\sqrt[3]{-8}\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)

d) \(\left(x+2\right)^2+1=0\Leftrightarrow\left(x+2\right)^2=-1\) (vô lí)

vậy phương trình vô nghiệm

a) (x-1)² = 0

<=> x-1 = 0

<=> x = 1

b) (x-2)² - 1 = 0

<=> (x-2)² = 1

<=> \(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c) (2x-1)³ = -8

<=> (2x-1)³ = -2³

<=> 2x - 1 = -2

<=> 2x = -1

<=> x = \(-\dfrac{1}{2}\)

d) (x+2)² + 1 = 0

<=> (x+2)² = -1

<=> x+2 = -1

<=> x = -3

a, \(\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2=0^2\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ......

b, \(\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy .....

c, \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-3\)

\(\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy ...

d, \(\left(x+2\right)^2+1=0\)

\(\Leftrightarrow\left(x+2\right)^2=-1\)

\(\Leftrightarrow\) ko tìm dc giá trị của x thỏa mãn (do \(\left(x+2\right)^2\ge0\))

\(a,0,1^{2-x}>0,1^{4+2x}\\ \Leftrightarrow2-x>2x+4\\ \Leftrightarrow3x< -2\\ \Leftrightarrow x< -\dfrac{2}{3}\)

\(b,2\cdot5^{2x+1}\le3\\ \Leftrightarrow5^{2x+1}\le\dfrac{3}{2}\\ \Leftrightarrow2x+1\le log_5\left(\dfrac{3}{2}\right)\\ \Leftrightarrow2x\le log_5\left(\dfrac{3}{2}\right)-1\\ \Leftrightarrow x\le\dfrac{1}{2}log_5\left(\dfrac{3}{2}\right)-\dfrac{1}{2}\\ \Leftrightarrow x\le log_5\left(\dfrac{\sqrt{30}}{10}\right)\)

c, ĐK: \(x>-7\)

\(log_3\left(x+7\right)\ge-1\\ \Leftrightarrow x+7\ge\dfrac{1}{3}\\ \Leftrightarrow x\ge-\dfrac{20}{3}\)

Kết hợp với ĐKXĐ, ta có:\(x\ge-\dfrac{20}{3}\)

d, ĐK: \(x>\dfrac{1}{2}\)

\(log_{0,5}\left(x+7\right)\ge log_{0,5}\left(2x-1\right)\\ \Leftrightarrow x+7\le2x-1\\ \Leftrightarrow x\ge8\)

Kết hợp với ĐKXĐ, ta được: \(x\ge8\)

a ) \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Leftrightarrow3x^2-3x-3x^2+2x=5\)

\(\Leftrightarrow-x=5\)

\(\Leftrightarrow x=-5\)

Vậy phương trình có nghiệm x = - 5 .

a, \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Rightarrow3x^2-3x-\left(3x^2-2x\right)=5\)

\(\Rightarrow3x^2-3x-3x^2+2x=5\)

\(\Rightarrow5x=5\Rightarrow x=1\)

Câu b,c làm tương tự! Cứ tách ra là làm được à!

b ) \(8x\left(2x+1\right)-4x\left(2x-3\right)=-40\)

\(\Leftrightarrow16x^2+8x-8x^2+12x=-40\)

\(\Leftrightarrow20x=-40\)

\(\Leftrightarrow x=-2\)

Vậy phương trình có nghiệm x = - 2

(x - 3)² = (2x - 1)²

=> |x - 3| = |2x - 1|

=> x - 3 = 2x-1 hoặc x - 3 = -(2x - 1)

x - 2x = -1 + 3 hoặc x - 3 = -2x + 1

-x = 2 hoặc x + 2x = 1 + 3

x = -2 hoặc 3x = 4

x = -2 hoặc x = 4 : 3

x = -2 hoặc x = 1,(3)

\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\)

\(Q=\frac{x^2+2x+17}{2\left(x+1\right)}=\frac{\left(x+1\right)^2+16}{2\left(x+1\right)}=\frac{x+1}{2}+\frac{8}{x+1}\ge2\sqrt{\frac{x+1}{2}.\frac{8}{x+1}}=4\)

Dấu "=" tại x = 3