\(P=\frac{8x^3-12x^2+6x-1}{4x^2-4x+1}\)

a) ĐKXĐ: x \(\ne\pm\frac{1}{2}\)

b) Theo đề bài ta có:

\(2x^2+x=0\)

\(\Rightarrow x\left(2x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{2}\left(Loại\right)\end{cases}}}\)

Thay x = 0 (thỏa mãn điều kiện) vào P ta có:

\(P=\frac{0-0+0-1}{0-0+1}=\frac{-1}{1}=-1\)

Vậy khi x = 0 thì P = -1

c) \(P=\frac{8x^3-12x^2+6x-1}{4x^2-4x+1}=\frac{\left(2x-1\right)^3}{\left(2x-1\right)^2}=2x-1\)

Để P \(\inℤ\Leftrightarrow2x-1\inℤ\)

Mà -1\(\inℤ;x\inℤ\Rightarrow-1⋮2x\)

\(\Rightarrow2x\inƯ\left(-1\right)=\left\{1;-1\right\}\)

Ta có bảng giá trị:

Vậy không có x thỏa mãn P \(\inℤ\)

d) Với x \(\ne\pm\frac{1}{2};P=2\)

\(\Leftrightarrow2x-1=2\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\frac{3}{2}\)

Vậy \(x=\frac{3}{2}\)thì \(P=2\)

a,  4x^2 - 4x = -1

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

\(\Leftrightarrow\)(2x-1)²              =0 

\(\Leftrightarrow\)2x - 1          = 0 

\(\Leftrightarrow\)x                = 1/2

b, \(\Leftrightarrow\)( 2x + 1)^3 = 0

\(\Leftrightarrow\)2x + 1 = 0 

\(\Leftrightarrow\)x       = -1/2

đúng thì

a) \(4x^2-4x=-1\)

\(\Leftrightarrow4x^2-4x+1=0\)

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

\(\Leftrightarrow2x-1=0\)

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

b) \(8x^3+12x^2+6x+1=0\)

\(\Leftrightarrow\left(2x+1\right)^3=0\)

\(\Leftrightarrow2x+1=0\)

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

a) (2x)² - 4x = -1

2x - 4x = -1

-2x = -1

x = -1/-2

x = 1/2

a) ĐKXĐ: 4x²-4x+1>0

(2x)²-2.2x+1= (2x-1)²≥ 0

 a)  ĐKXĐ của phương trình : \(4x^2+4x+1\ne0\)\(\Rightarrow x\ne-\frac{1}{2}\)

b)  \(P=\frac{4x^3+8x^2-x-2}{4x^2+4x+1}\)

\(\Rightarrow P=\frac{\left(4x^3-x\right)+\left(8x^2-2\right)}{\left(2x+1\right)^2}\)

 \(\Rightarrow P=\frac{x\left(4x^2-1\right)+2\left(4x^2-1\right)}{\left(2x+1\right)^2}\)

\(\Rightarrow P\left(x\right)=\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}\)

\(\Rightarrow P\left(x\right)=\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)}=\frac{3}{2}\)\(\Rightarrow P\left(x\right)=2\left(x+2\right)\left(2x-1\right)=3\left(2x+1\right)\)

\(\Rightarrow P\left(x\right)=4x^2+6x-6-\left(6x+3\right)=0\)

 \(\Rightarrow P\left(x\right)=4x^2-9=0\)\(\Rightarrow P\left(x\right)=x^2=\frac{9}{4}\)

\(\Rightarrow P\left(x\right)=x^2=\sqrt{\frac{9}{4}}\)\(\Rightarrow P\left(x\right)=\frac{3}{2}\)

câu c)  cx tương tự 

a, x khác -1/2

b, x=\(\frac{\sqrt{7}}{2}\)

thank

\(a,dkxd:x\ge0,x\ne4\)

\(b,B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\dfrac{1}{\sqrt{x}-2}\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{\sqrt{x^2}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)^2}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(c,x=16\left(tm\right)\Rightarrow B=\dfrac{\sqrt{16}+2}{\sqrt{16}\left(\sqrt{16}-2\right)}=\dfrac{4+2}{4\left(4-2\right)}=\dfrac{6}{8}=\dfrac{3}{4}\)

\(d,B>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Leftrightarrow\sqrt{x}+2>0\Leftrightarrow\sqrt{x}>-2\left(ktm\right)\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)

Kết hợp với \(dk:x\ge0\) ta kết luận \(0\le x< 4\) thì \(B>0\).

a) Điều kiện xác định:

\(\left\{{}\begin{matrix}x-2\sqrt{x}\ne0\\x\ge0\end{matrix}\right.\)\(\Leftrightarrow x>0,x\ne4\)

Vậy...

b) \(B=\dfrac{\sqrt{x}.\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)^2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)^2}\)\(=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

Vậy \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

c) Tại x=16 ( thỏa mãn đk) thay vào B đã rút gọn ta được:

\(B=\dfrac{\sqrt{16}+2}{\sqrt{16}\left(\sqrt{16}-2\right)}=\dfrac{3}{4}\)

d) \(B>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\)

\(\Leftrightarrow\sqrt{x}-2>0\)\(\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)

Vậy x>4 thì B>0

a) (x + 3)² - (x - 2)² = 2x

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

=> 5(2x + 1) = 2x

=> 10x + 5 = 2x

=> 10x - 2x = -5

=> 8x = -5

=> x = -5/8

b) 7x(x - 2) = x - 2

=> 7x(x - 2) - (x - 2) = 0

=> (7x - 1)(x - 2) = 0

=> \(\orbr{\begin{cases}7x-1=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{7}\\x=2\end{cases}}\)

c) 8x³ - 12x² + 6x - 1 = 0

=> (2x - 1)³ = 0

=> 2x - 1 = 0

=> 2x = 1

=> x = 1/2

Bài 1.

a) ( 7x - 3 )² - 5x( 9x + 2 ) - 4x² = 18

<=> 49x² - 42x + 9 - 45x² - 10x - 4x² = 18

<=> -52x + 9 = 18

<=> -52x = 9

<=> x = -9/52 

b) ( x - 7 )² - 9( x + 4 )² = 0

<=> x² - 14x + 49 - 9( x² + 8x + 16 ) = 0

<=> x² - 14x + 49 - 9x² - 72x - 144 = 0

<=> -8x² - 86x - 95 = 0 

<=> -8x² - 10x - 76x - 95 = 0

<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0

<=> ( x + 5/4 )( -8x - 76 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)

c) ( 2x + 1 )² + ( 4x - 1 )( x + 5 ) = 36

<=> 4x² + 4x + 1 + 4x² + 19x - 5 = 36

<=> 8x² + 23x - 4 - 36 = 0

<=> 8x² + 23x - 40 = 0

=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))

Bài 2.

a) x² - 12x + 39 = ( x² - 12x + 36 ) + 3 = ( x - 6 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

b) 17 - 8x + x² = ( x² - 8x + 16 ) + 1 = ( x - 4 )² + 1 ≥ 1 > 0 ∀ x ( đpcm )

c) -x² + 6x - 11 = -( x² - 6x + 9 ) - 2 = -( x - 3 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

d) -x² + 18x - 83 = -( x² - 18x + 81 ) - 2 = -( x - 9 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

a, 7x + 10x  = 5x 

    17x = 5x

17x - 5x = 0

      12x = 0

          x =0

2; 

a, 4x + 7x = 22

    11x = 22

        x = 2

b, 12x - 8x = 25

     4x = 25

       x = \(\dfrac{25}{4}\)

c,  \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\) 

     (\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)

    \(\dfrac{1}{6}\)x     = \(\dfrac{4}{5}\) 

      x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)

     x = \(\dfrac{24}{5}\)