a: \(x^2+4x-1=0\)

=>\(x^2+4x+4-5=0\)

=>\(\left(x+2\right)^2=5\)

=>\(\left[\begin{array}{l}x+2=\sqrt5\\ x+2=-\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt5-2\\ x=-\sqrt5-2\end{array}\right.\)

b: \(2x^2-4x+1=0\)

=>\(2\left(x^2-2x+\frac12\right)=0\)

=>\(x^2-2x+\frac12=0\)

=>\(x^2-2x+1-\frac12=0\)

=>\(\left(x-1\right)^2=\frac12\)

=>\(\left[\begin{array}{l}x-1=\frac{\sqrt2}{2}\\ x-1=-\frac{\sqrt2}{2}\end{array}\right.\Longrightarrow\left[\begin{array}{l}x=\frac{\sqrt2+2}{2}\\ x=\frac{-\sqrt2+2}{2}\end{array}\right.\)

c: \(\left(2x-1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)=2x-9\)

=>\(2x^2+4x-x-2-\left(x^2-3x+2\right)-2x+9=0\)

=>\(2x^2+x+7-x^2+3x-2=0\)

=>\(x^2+4x+5=0\)

=>\(x^2+4x+4+1=0\)

=>\(\left(x+2\right)^2+1=0\) (vô lý)

=>Phương trình vô nghiệm

d: \(\left(x-1\right)\cdot x\cdot\left(x+1\right)\left(x+2\right)=8\)

=>\(x\left(x+1\right)\left(x+2\right)\left(x-1\right)=8\)

=>\(\left(x^2+x\right)\left(x^2+x-2\right)=8\)

=>\(\left(x^2+x\right)^2-2\left(x^2+x\right)-8=0\)

=>\(\left(x^2+x-4\right)\left(x^2+x+2\right)=0\)

mà \(x^2+x+2=x^2+x+\frac14+\frac74=\left(x+\frac12\right)^2+\frac74\ge\frac74>0\forall x\)

nên \(x^2+x-4=0\)

\(\Delta=1^2-4\cdot1\cdot\left(-4\right)=1+16=17>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{-1-\sqrt{17}}{2\cdot1}=\frac{-1-\sqrt{17}}{2}\\ x=\frac{-1+\sqrt{17}}{2\cdot1}=\frac{-1+\sqrt{17}}{2}\end{array}\right.\)

a) \(\left|2x-3\right|-\dfrac{5}{2}=\dfrac{1}{3}\)

\(\left|2x-3\right|=\dfrac{1}{3}+\dfrac{5}{2}=\dfrac{2}{6}+\dfrac{15}{6}\)

\(\left|2x-3\right|=\dfrac{17}{6}\)

\(+)2x-3=\dfrac{17}{6}\Rightarrow2x=\dfrac{35}{6}\Rightarrow x=\dfrac{35}{12}\)

\(+)2x-3=\dfrac{-17}{6}\Rightarrow2x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{12}\)

vậy...

\(\left|x-1\right|+3x=1\\ \Rightarrow\left|x-1\right|=1-3x\\ \Rightarrow\left\{{}\begin{matrix}x-1=1-3x\\x-1=-1+3x\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x=2\\-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

Dấu ngoặc vuông nhé

thánh bấm nhầm

Câu d:

-1\(\frac23\) - (|2\(x\)| + \(\frac56\)) = - 2

-\(\frac53\) - |2\(x\)| - \(\frac56\) = - 2

|2\(x\)| = - \(\frac53\) - \(\frac56\) + 2

|2\(x\)| = - \(\frac52\) + 2

|2\(x\)| = - \(\frac12\) (vô lí vì trị tuyệt đối của một số luôn là một số không âm)

Không có giá trị nào của x thỏa mãn đề bài.

x ∈ ∅

Câu a:

|\(x\) - 3| = \(x\) + 4

Vì |\(x\) - 3| ≥ 0 ∀ \(x\) nên \(x\) + 4 ≥ 0 ⇒ \(x\) ≥ - 4

Với -4 ≤ \(x\) ≤ 3 ta có:

-\(x\) + 3 = \(x\) + 4

\(x\) + \(x\) = -4 + 3

2\(x\) = -1

\(x=\frac{-1}{2}\)

Với x > 3 ta có:

x - 3 = x + 4

x - x = 3 + 4

0 = 7 (vô lí)

Vậy x = -1/2 là nghiện duy nhất của phương trình.

Vậy \(x\) = -1/2

\(\left(\frac{1}{2}\right)^{2x-1}\) như thế này hả

Tham khảo đây nha: Câu hỏi của le ngoc han       

Ta có : x(x - 2) - x(x - 1) - 15 = 0

<=> x² - 2x - x² + x - 15 = 0

<=> -x - 15 = 0

=> -x = 15

=> x = -15

a) \(\left(2.x-1\right)^6=\left(2.x-1\right)^8\)

\(\Leftrightarrow\left(2.x-1\right)^8-\left(2.x-1\right)^6=0\)

\(\Leftrightarrow\left(2x-1\right)^6.\left[\left(2x-1\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)-1=0\end{matrix}\right.\)

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

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

Vậy : \(x\in\left\{\frac{1}{2},1\right\}\)

b) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)

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

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

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

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

Vậy : \(x\in\left\{0,1,2\right\}\)

Chúc học tốt nhé !!

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

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

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

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

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

Vậy \(x\in\left\{3;1\right\}.\)

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

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

=> \(2x-1=-2\)

=> \(2x=\left(-2\right)+1\)

=> \(2x=-1\)

=> \(x=\left(-1\right):2\)

=> \(x=-\frac{1}{2}\)

Vậy \(x=-\frac{1}{2}.\)

\(\left(2x-3\right)^5=-243\)

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

=> \(2x-3=-3\)

=> \(2x=\left(-3\right)+3\)

=> \(2x=0\)

=> \(x=0:2\)

=> \(x=0\)

Vậy \(x=0.\)

Chúc bạn học tốt!

a) (x - 2)² = 1

=> x - 2 = 1 hoặc x - 2 = -1

x = 3 ; x = 1

Vậy x = 3; x = 1

b) (2x - 1)³ = -8

=> 2x - 1 = -2

2x = -1

x = \(\frac{-1}{2}\)

Vậy x = \(\frac{-1}{2}\)

c) (2x - 3)⁵ = -243

=> (2x - 3)⁵ = (-3)⁵

=> 2x - 3 = -3

2x = 0

x = 0

Vậy x = 0

\(VT=\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=\left|4\right|=4\)

\(VP=\frac{8}{3\left(x+1\right)^2+2}\le\frac{8}{2}=4\)

\(VT\ge VP\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(2x+3\right)\left(1-2x\right)\ge0\left(1\right)\\\left(x+1\right)^2=0\left(2\right)\end{cases}}\)

\(\left(2\right)\)\(\Leftrightarrow\)\(x=-1\) ( thỏa mãn\(\left(1\right)\) ) 

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