b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

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

Bài 1:

a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$

$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$

b.

$(x+1)(x+2)(x+3)(x+4)-24$

$=[(x+1)(x+4)][(x+2)(x+3)]-24$

$=(x^2+5x+4)(x^2+5x+6)-24$

$=a(a+2)-24$ (đặt $x^2+5x+4=a$)

$=a^2+2a-24=(a^2-4a)+(6a-24)$

$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$

$=x(x+5)(x^2+5x+10)$

Bài 2:

a. ĐKXĐ: $x\neq 3; 4$

\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)

b. $x^2+20=9x$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Rightarrow x=5$ (do $x\neq 4$)

Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$

Bài 3:

$(2x^2-7x^2:13x:2):(2x-1)=(2x^2-\frac{7}{26}x):(2x-1)$

$=[x(2x-1)+\frac{19}{52}(2x-1)+\frac{19}{52}]:(2x-1)$

$=[(2x-1)(x+\frac{19}{52})+\frac{19}{52}]: (2x-1)$

$\Rightarrow$ thương là $x+\frac{19}{52}$ và thương là $\frac{19}{52}$

A=(x+1)(x+2)(x+3)(x+4)-24

=(x²+5x+4)(x²+5x+6)-24

Đặt t=(x²+5x+4) ta có:

t(t+2)-24=t²+6t-2t-24

=t(t+6)-4(t+6)

=(t-4)(t+6).Thay vào ta đc:

(x²+5x+4-4)(x²+5x+4+6)=(x²+5x)(x²+5x+10) 

=x(x+5)(x²+5x+10)

B=(x²+3x+2)(x²+7x+120-24)

=(x²+3x+2)(x²+7x+96)

=(x²+2x+x+2)(x²+7x+96)

=[x(x+2)+(x+2)](x²+7x+96)

=(x+1)(x+2)(x²+7x+96)

C và D bn cx lm tương tự

\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow E=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)

\(=1+\frac{1}{2}\left(3+4+5+...+201\right)\)

\(=1+\frac{1}{2}\left(1+2+3+...+201-1-2\right)\)

\(=1+\frac{1}{2}\left(\frac{201.202}{2}-3\right)=10150\)

\(\frac{21}{5}\left|x\right|< 2019\Rightarrow\left|x\right|< 2019\div\frac{21}{5}=\frac{3365}{7}\)

\(\Rightarrow-480\le x\le480\)

\(\Rightarrow\sum x=-480+480-479+479+...+-1+1+0=0\)

\(\frac{2^{24}\left(x-3\right)}{\frac{81}{35}.\left(6.2^{24}-2^{26}\right)}=\frac{25}{9}\)

\(\Leftrightarrow\frac{2^{24}\left(x-3\right)}{2^{24}\left(6-2^2\right)}=\frac{25}{9}.\frac{81}{35}\)

\(\Leftrightarrow\frac{x-3}{2}=\frac{45}{7}\)

\(\Leftrightarrow x-3=\frac{90}{7}\)

\(\Rightarrow x=\frac{111}{7}\)

a) \(\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)

Đặt \(x^2+x=y\) ta được:

\(y^2-14y+24\)

\(=x\left(y-12\right)-2\left(y-12\right)\)

\(=\left(y-2\right)\left(y-12\right)\)

Thay ngược trở lại:

\(\left(x^2+x-2\right)\left(x^2+x-12\right)\)

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

d) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+10\right)+1\)

Đặt \(x^2+5x+4=a\) được:

\(a\left(a+6\right)+1\)

\(=a^2+6a+1\)

\(=a^2+2.a.3+3^2-8\)

\(=\left(a+3\right)^2-\left(\sqrt{8}\right)^2\)

\(=\left(a+3-\sqrt{8}\right)\left(a+3+\sqrt{8}\right)\)

Mấy câu kia tương tự.

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

Cauchy-schwarz:

\(\left(1+1+1\right)\left(x^4+x^2+1\right)\ge\left(x^2+x+1\right)^2\)

"="<=>\(x=1\)

b)\(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

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

\(\Rightarrow\left(t-1\right)\left(t+1\right)=24\)

\(\Leftrightarrow t^2-25=0\)

\(\Leftrightarrow t=\pm5\)

t=5\(\Leftrightarrow x^2+x-1=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

t=-5<=> pt vô nghiệm

a)

\((x^2+x)^2+4x^2+4x=(x^2+x)^2+4(x^2+x)\)

\(=(x^2+x)(x^2+x+4)\)

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

b) \(x(x+1)(x+2)(x+3)+1\)

\(=[x(x+3)][(x+1)(x+2)]+1\)

\(=(x^2+3x)(x^2+3x+2)+1\)

\(=(x^2+3x)^2+2(x^2+3x)+1\)

\(=(x^2+3x+1)^2\)

c)

\((x+2)(x+3)(x+4)(x+5)-24\)

\(=[(x+2)(x+5)][(x+3)(x+4)]-24\)

\(=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=a\)

Khi đó biểu thức bằng:

\(a(a+2)-24=a^2+2a-24=a^2-4a+6a-24\)

\(=a(a-4)+6(a-4)=(a-4)(a+6)\)

\(=(x^2+7x+10-4)(x^2+7x+10+6)\)

\(=(x^2+7x+6)(x^2+7x+16)\)

\(=(x^2+x+6x+6)(x^2+7x+16)\)

\(=(x+1)(x+6)(x^2+7x+16)\)

a ) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=24\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=24\)

\(\Leftrightarrow2x=-231\Leftrightarrow x=\dfrac{-231}{2}\)

b ) \(\left(x+3\right)^2-\left(x-4\right)\left(x-8\right)=1\)

\(\Leftrightarrow x^2+6x+9-x^2+12x-32=1\)

\(\Leftrightarrow18x=24\Leftrightarrow x=\dfrac{4}{3}\)

Chúc bạn học tốt !!!!!!!!!!!!