nhanh giùm mình được không

Bài 1: 

a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)

\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)

x bằng bao nhiêu

B = x15 - 8x14 + 8x13 - 8x2 + ... - 8x2 + 8x - 5

B = x^15 - 7x^14 -x^14+7x^13+x^13-7x^12-...-x^2+7x+x-5

B = x^14(x-7) - x^14(x-7) +...+x^2(x-7)-x(x-7)+x-5

B = 7-5=2

`B = x^15 - 7x^14 - x^14 + 7x^13 + x^13 - .... +7x + x - 7 + 2`

`<=> x^14(x-7) - x^13(x-7) + ... + x - 7 + 2`

`<=> (x^14-x^13 + ... + 1)(x-7) + 2`

Thay `x = 7 <=> (x^14 - x^13 + ... + 1) xx 0 + 2 = 2`.

B = \(x^{15}-8x^{14}+8x^{13}+8x^2+...-8x^2+8x-5\)

B = \(x^{15}-7x^{14}+x^{14}+7x^{13}+x^{13}+7x^{12}-...-x^2+7x+x-5\)

B = \(x^{14}\left(x-7\right)-x^{14}\left(x-7\right)+...+x^2\left(x-7\right)-x\left(x-7\right)-5\)

B = \(7-5=2\)

\(\left(x-2\right)^3+\left(x+2\right)^3-x^3-8x^2+10\)

\(=x^3-6x^2+12x-8+x^3+6x^2+12x+8-x^3-8x^2+10\)

\(=x^3-2x^2+24x+10\)

ta có: 8=7+1=x+1

\(B=x^{15}-8x^{14}+8x^{13}-...-8x^2+8x-5\)

\(\Rightarrow B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-...-\left(x+1\right)x^2+\left(x+1\right)x-5\)

\(\Rightarrow B=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-...-x^3-x^2+x^2+x-5\)

\(\Rightarrow B=x-5\)

\(\Rightarrow B=7-5\)

\(\Rightarrow B=2\)

B = x15 - 8x14 + 8x13 - 8x2 + ... - 8x2 + 8x - 5

B = x^15 - 7x^14 -x^14+7x^13+x^13-7x^12-...-x^2+7x+x-5

B = x^14(x-7) - x^14(x-7) +...+x^2(x-7)-x(x-7)+x-5

B = 7-5=2

B = x15 - 8x14 + 8x13 - 8x2 + ... - 8x2 + 8x - 5

B = x^15 - 7x^14 -x^14+7x^13+x^13-7x^12-...-x^2+7x+x-5

B = x^14(x-7) - x^14(x-7) +...+x^2(x-7)-x(x-7)+x-5

B = 7-5=2

Tham khảo cách này nhoá~

Lời giải:

a. PT $\Leftrightarrow (3-2x-3-2x)(3-2x+3+2x)=8$

$\Leftrightarrow -4x.6=8$

$\Leftrightarrow -24x=8\Leftrightarrow x=\frac{-1}{3}$

b.

$9x^5-72x^2=0$

$\Leftrightarrow 9x^2(x^3-8)=0$

$\Leftrightarrow x^2=0$ hoặc $x^3=8$

$\Leftrightarrow x=0$ hoặc $x=2$

c.

$5x^4-8x^2-4=0$

$\Leftrightarrow 5x^4-10x^2+2x^2-4=0$

$\Leftrightarrow 5x^2(x^2-2)+2(x^2-2)=0$

$\Leftrightarrow (5x^2+2)(x^2-2)=0$

$\Leftrightarrow 5x^2+2=0$ (loại) hoặc $x^2-2=0$ (chọn)

$\Leftrightarrow x=\pm \sqrt{2}$

d.

PT $\Leftrightarrow [x^2(x+1)-4(x+1)]:(x-2)=0$

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

$\Leftrightarrow (x-2)(x+2)(x+1):(x-2)=0$
$\Leftrightarrow (x+2)(x+1)=0$

$\Leftrightarrow x+2=0$ hoặc $x+1=0$

$\Leftrightarrow x=-2$ hoặc $x=-1$

a: Ta có: \(\left(3-2x\right)^2-\left(3+2x\right)^2=8\)

\(\Leftrightarrow9-12x+4x^2-9-12x-4x^2=8\)

\(\Leftrightarrow-24x=8\)

hay \(x=-\dfrac{1}{3}\)

b: Ta có: \(9x^5-72x^2=0\)

\(\Leftrightarrow9x^2\left(x^3-8\right)=0\)

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

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

c: Ta có: \(5x^4-8x^2-4=0\)

\(\Leftrightarrow5x^4-10x^2+2x^2-4=0\)

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

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

a) (x - 1) - (2x - 1) = 9 - x

<=> x - 1 - 2x + 1 = 9 - x

<=> x - 2x + x = 9 + 1 - 1

<=> 0x = 9 (vô lý vì 0x = 0 với mọi x)

Vậy PT vô nghiệm

b) 3 - 4x. (25 - 2x) = 8x²

<=> 3 - 100x + 8x² = 8x²

<=> 3 - 100x = 0

<=> -100x = -3

<=> 100x = 3

\(\Leftrightarrow x=\dfrac{3}{100}\)

Vậy: \(S=\left\{\dfrac{3}{100}\right\}\)

a) (x-1) - (2x-1) = 9-x

<=> x - 1 - 2x + 1= 9-x

<=> -x = 9-x

<=> -x + x = 9 

<=> 0 = 9 ( sai )

Vậy tập nghiệm S ={\(\varnothing\)}

b) 3 - 4x(25 - 2x) = 8x²

<=> 3 - 100x + 8x² = 8x²

<=> 3 = 100x

<=> \(\dfrac{3}{100}\)= x = 0,03

Vập tập nghiệm S ={0,03}