câu 1:

x³-1+3x²-3x =(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+x+1+3x)=(x-1)(x^2+4x=1)

Câu 2 :

a) \(\left(x^4-2x^3+2x-1\right):\left(x^2-1\right)\)

\(=\left(x^4-x^2-2x^3+2x+x^2-1\right):\left(x^2-1\right)\)

\(=\left[x^2\left(x^2-1\right)-2x\left(x^2-1\right)+\left(x^2-1\right)\right]:\left(x^2-1\right)\)

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

\(=x^2-2x+1\)

b) \(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)

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

Câu 3 :

Sửa đề :

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

2/ (x² + x + 1) (x²+ x + 2) = 12

đặt x² + x = t

thay vào đc: 

(t + 1) (t + 2) = 12

<=> t² + 3t + 2 = 12

<=> t² + 3t - 10 = 0

<=> t² - 2t + 5t - 10 = 0

<=> t (t - 2) + 5 (t - 2) = 0

<=> (t + 5) (t - 2) = 0

=> \(\hept{\begin{cases}t=-5\\t=2\end{cases}}\)

thay t đc:

*) x² + x = -5  => x loại

*) x² + x = 2 = x² + x - 2 = x² - 1 + x - 1 = (x - 1) (x + 1) + (x - 1) = (x - 1) (x + 2) 

=> x = 1 hoặc x = - 2

S = {-2 ; 1}

3/ (x² - 6x + 4)² - 15(x² - 6x + 10) = 1

đặt x² - 6x + 4 = t

có: t² - 15(t + 6) = 1

<=> t² - 15t - 91 = 0

....

....

số xấu, xem lại đề ~0~

câu 2, a=x2 +x+1 . PHƯƠNG TRÌNH TRỞ THÀNH a x (a +1)=12. giải binh thương 

câu 3, tương tự a= x2 - 6x + 4 .PHƯƠNG TRÌNH TRỞ THÀNH a2 - 15x(a+6)=1. giải bình thương 

nếu cần mình giải chi tiết thì nhắn tin nha

a =  1/(1.2) + 5/(2.3) + ... + 89/(9.10)

a =  [1-1/(1.2)] + [1-1/(2.3)] + ... + [1-1/(9.10)]

\(a=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{9.10}\right)\)

\(a=9-\left[\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right]\)

Ta có:

\(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)

\(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

....

\(\frac{1}{9.10}=\frac{1}{9}-\frac{1}{10}\)

Cộng các vế ở trên lại:

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

Vậy:

a = 9 - 9/10 = 81/10

làm sao mà viết được dưới dạng p/s vậy 

8) \(\left(x+4\right)\left(6x-12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\6x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\6x=12\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\)

Vậy \(x\in\left\{-4;2\right\}\)

11) \(\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{7}{8}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}-0\\3x=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}\\x=-\frac{1}{9}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{16}\\x=-\frac{1}{9}\end{cases}}}\)

Vậy \(x\in\left\{\frac{7}{16};-\frac{1}{9}\right\}\)

12) \(3x-2x^2=0\)

\(\Leftrightarrow x\left(3-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

Vậy \(x\in\left\{0;\frac{3}{2}\right\}\)

13) \(5x+10x^2=0\)

\(\Leftrightarrow5x\left(1+2x\right)=0\)

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

Vậy \(x\in\left\{0;-\frac{1}{2}\right\}\)

8) (x+4)(6x-12)=0

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\6x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\)

Vậy x={-4;2}

11) \(\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{7}{8}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}\\3x=\frac{-1}{3}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{16}\\x=\frac{-1}{9}\end{cases}}}\)

1.

Đặt \(x^2-5x=a\Rightarrow a^2=\left(x^2-5x\right)^2\)

Thay vào pt:

\(\Rightarrow a^2+10a+24=0\)

\(\Leftrightarrow a^2+6a+4a+24=0\)

\(\Leftrightarrow a\left(a+6\right)+4\left(a+6\right)=0\)

\(\Leftrightarrow\left(a+6\right)\left(a+4\right)=0\)

\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)

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

\(\Leftrightarrow\left[x\left(x-3\right)-2\left(x-3\right)\right]\left[x\left(x-4\right)-\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x-4\right)\left(x-1\right)=0\)

\(\Rightarrow x-3=0,x-2=0,x-4=0,x-1=0\)

\(\Rightarrow x=3,x=2,x=4,x=1\)

T I C K mình sẽ giải típ cho cảm ơn

típ nha

Danh ơi ! xem lại đề của bài 2 nha 

\(\dfrac{4x+2}{4x-2}+\dfrac{3-6x}{6x-6}\left(dkxd:x\ne\dfrac{1}{2};x\ne1\right)\)

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

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

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

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

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

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

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

\(=\dfrac{2x-3}{4x^2-6x+2}\)

d: Ta có: \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)

\(=\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11}+1-\sqrt{11}+1}{\sqrt{2}}\)

\(=\sqrt{2}\)

a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)

<=>4x²+4x+1+2(4x²-4x+1)≥12(x²+10x+25)

<=>4x²+4x+1+8x²-8x+2≥12x²+120x+300

<=>4x²+4x+1+8x²-8x+2-12x²-120x-300≥0

<=>-124x-297≥0

<=>124x+297≤0

<=>124x≤-297

<=>x≤\(\frac{-297}{124}\)

b, Tương tự câu a

c,

TH1: 5-3x=2+x

<=> -3x - x = 2 - 5

<=> -4x = -3

<=> x = 3/4

TH2: 5-3x = -2 - x

<=> -3x + x = -2 - 5

<=> -2x = -7

<=> x = 7/2

d: Ta có: \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)

\(=\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11}+1-\sqrt{11}+1}{\sqrt{2}}\)

\(=\sqrt{2}\)