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

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

\(VT\ge VP\)

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\(\left(1\right)\) ) 

... 

a: ta có: \(\dfrac{\left(x+2\right)^2}{2}+\dfrac{\left(2x+1\right)^2}{4}+\dfrac{\left(2x-1\right)^2}{8}-\left(x+1\right)^2=0\)

\(\Leftrightarrow4\left(x^2+4x+4\right)+2\left(4x^2+4x+1\right)+4x^2-4x+1-8\left(x+1\right)^2=0\)

\(\Leftrightarrow4x^2+16x+16+8x^2+8x+2+4x^2-4x+1-8\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow16x^2+20x+19-8x^2-16x-8=0\)

\(\Leftrightarrow8x^2+4x+11=0\)

\(\text{Δ}=4^2-4\cdot8\cdot11=-336< 0\)

Vì Δ<0 nên phương trình vô nghiệm

b.

PT \(\Leftrightarrow \frac{x^2+2x+1}{2}-\frac{4x^2-4x+1}{3}+\frac{4x^2+4x+1}{4}-\frac{x^2-10x+25}{6}=0\)

\(\Leftrightarrow \left(\frac{x^2+2x+1}{2}+\frac{4x^2+4x+1}{4}\right)-\left(\frac{4x^2-4x+1}{3}+\frac{x^2-10x+25}{6}\right)=0\)

\(\Leftrightarrow \frac{6x^2+8x+3}{4}-\frac{9x^2-18x+27}{6}=0\)

\(\Leftrightarrow \frac{3(6x^2+8x+3)-2(9x^2-18x+27)}{12}=0\)

$\Leftrightarrow 5x-\frac{15}{4}=0$

$\Leftrightarrow x=\frac{3}{4}$

c.

PT $\Leftrightarrow (x^3+9x^2+27x+27)-(3x^3+12x^2)+(x^3+6x^2+12x+8)=(-x^3+3x^2-3x+1)-8$

$\Leftrightarrow 42x+42=0$

$\Leftrightarrow x=-1$

x=1 và 0  

ms thỏa mản đề ra

=)))))))))))))))

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x-1=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}}\)

\(49\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)

\(4x^2+8x+4+4x^2-4x+1-8x^2+8-11=0\)

\(4x+2=0\)

\(4x=2\)

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

<=>4(x²+2x+1)+4x²-4x+1-8x²+8-11=0

<=>4x²+8x+4+4x²-4x+1-8x²+8-11=0

<=>4x+2=0

<=>2(2x+1)=0

<=>2x+1=0

<=>x=-1/2

XL mik nhầm

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

~~~~~~~~~~

..........

////////////////

(2x - 1)⁸ = (2x - 1)¹⁰

(2x - 1)¹⁰ - (2x - 1)⁸ = 0

(2x - 1)⁸.[(2x - 1)² - 1] = 0

(2x - 1)⁸ = 0 hoặc (2x - 1)² - 1 = 0

*) (2x - 1)⁸ = 0

2x - 1 = 0

2x = 1

x = 1/2

*) (2x - 1)² - 1 = 0

(2x - 1)² = 1

2x - 1 = 1 hoặc 2x - 1 = -1

**) 2x - 1 = 1

2x = 2

x = 1

**) 2x - 1 = -1

2x = 0

x = 0

Vậy x = 0; x = 1/2; x = 1

(2x - 1)⁸ = (2x - 1)¹⁰

=) (2x - 1)¹⁰ : (2x - 1)⁸ = 1

    (2x - 1)² = 1 =) = 1²

=) 2x - 1 = 1

    2x = 2

      x = 1.

 các bước giải:

${(2x-1)}^8={(2x-1)}^{10}$
$\Rightarrow {(2x-1)}^8-{(2x-1)}^{10}=0$
$\Rightarrow {(2x-1)}^8.1-{(2x-1)}^8.{(2x-1)}^2=0$
$\Rightarrow {(2x-1)}^8.[1-{(2x-1)}^2]=0$
TH$1$
${(2x-1)}^8=0$
$\Rightarrow 2x-1=0$
$\Rightarrow 2x=0+1$
$\Rightarrow 2x=1$
$\Rightarrow x=1:2$
$\Rightarrow x=\frac{1}{2}$
TH$2$
$1-{(2x-1)}^2=0$
$\Rightarrow {(2x-1)}^2=1$
$\Rightarrow {(2x-1)}^2={(\pm 1)}^2$
$\Rightarrow 2x-1=1$ hoặc $2x-1=-1$
$\Rightarrow 2x=2$ hoặc $2x=0$
$\Rightarrow x=1$ hoặc $x=0$
Vậy $x\in \{\frac{1}{2};1;0\}$

giúp mik với

a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)

\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)

\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)

Bài 1:

\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2\cdot50=100\)

\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2\cdot52=104\)

=>A<B

Bài 2:

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

=>4x+13=11

=>4x=-2

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

giống cái kia thôi bn

Mik làm rồi mà

Mà cái bn Nguyễn Duy Đạt gì đó làm thiếu 1 trường hợp

Mà bn vẫn kik hở

Sao zzzzz??????

chắc nhầm ! Xin lỗi ! Xin lỗi !

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

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

''...'' vế sau á :), ko chắc