Tính:
\(a,\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(b,\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
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19 tháng 11 2018
b) (x+2).(x+3).(x+4).(x+5) - 24
= [(x+2).(x+5)].[(x+3).(x+4)] - 24
= [ x2 + 7x + 10].[x2 + 7x + 12] - 24
= [x2 + 7x + 11 -1].[x2 + 7x + 11+1] - 24
= (x2 +7x+11)2 - 12 - 24
= (x2 +7x+11)2 - 25
= (x2 +7x + 11 - 25).(x2 +7x + 11 + 25)
= (x2 + 7x - 14).(x2 + 7x + 36)
19 tháng 11 2018
a) \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(m=x^2+x+1\)ta có :
\(A=m\left(m+1\right)-12\)
\(A=m^2+m-12\)
\(A=m^2+4m-3m-12\)
\(A=m\left(m+4\right)-3\left(m+4\right)\)
\(A=\left(m+4\right)\left(m-3\right)\)
Lại thay m vào ta có :
\(A=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
b) \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(A=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(A=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(p=x^2+7x+11\)ta có :
\(A=\left(p-1\right)\left(p+1\right)-24\)
\(A=p^2-1^2-24\)
\(A=p^2-25\)
\(A=\left(p-5\right)\left(p+5\right)\)
Lại thay p vào A ta có :
\(A=\left(x^2+7x+5\right)\left(x^2+7x+15\right)\)
25 tháng 7 2021
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
4 tháng 10 2017
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.
28 tháng 8 2021
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
TN
8 tháng 8 2016
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
22 tháng 12 2018
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
a ) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+\dfrac{3}{2}-\dfrac{1}{2}\right)\left(x^2+x+\dfrac{3}{2}+\dfrac{1}{2}\right)-12\)
\(=\left(x^2+x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}-12\)
\(=\left(x^2+x+\dfrac{3}{2}\right)^2-\dfrac{49}{4}\)
\(=\left(x^2+x+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x^2+x+\dfrac{3}{2}+\dfrac{7}{2}\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(=\left[x\left(x+2\right)-\left(x+2\right)\right]\left(x^2+x+5\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)
b ) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-24\)
\(=\left(x^2+7x+11\right)^2-1-24\)
\(=\left(x^2+7x+11\right)^2-25\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left[x\left(x+1\right)+6\left(x+1\right)\right]\left(x^2+7x+16\right)\)
\(=\left(x+6\right)\left(x+1\right)\left(x^2+7x+16\right)\)
a ) Đặt \(t=x^2+x+1\)
b ) \(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)\)
Đặt \(t=x^2+7x+10\)