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N
2
12 tháng 8 2018
1, x3+ 6x2+11x+6
= x3 + 2x2 + 4x2 + 8x + 3x + 6
= x2(x + 2) + 4x(x + 2) + 3(x + 2)
= (x + 2)(x2 + 4x + 3)
2, x4+3x3-7x2-27x-18
= x4 + 3x3 - 9x2 + 2x2 - 27x -18
= (x4 - 9x2) + (3x3 - 27x) + (2x2 - 18)
= x2(x2 - 9) + 3x(x2 - 9) + 2(x2 - 9)
= (x2 - 9)(x2 + 3x + 2)
= (x + 3)(x - 3)(x2 + 3x + 2)
3, x3-8x2+x+42
= x3 - 7x2 - x2 + 7x - 6x + 42
= (x3 - 7x2) - (x2 - 7x) - (6x - 42)
= x2(x - 7) - x(x - 7) - 6(x - 7)
= (x - 7)(x2 - x - 6)
4, x4+5x3-7x2-41x-30
= x4 + x3 + 4x3 - 4x2 - 11x2 - 11x - 30x - 30
= (x4 + x3) + (4x3 - 4x2) - (11x2 + 11x) - (30x + 30)
= x3(x + 1) + 4x2(x + 1) - 11x(x + 1) - 30(x + 1)
= (x3 + 4x2 - 11x - 30)(x + 1)
5, x5+x-1
= x5 - x4 + x3 + x4 - x3 + x2 - x2+ x -1
= x3(x2 - x + 1)+ x2(x2 - x + 1)- (x2 - x + 1)
= (x2 - x + 1)(x3 + x2 - 1)
6, x5-x4-1
= x5 - x3 - x2 - x4 + x2 + x + x3 - x - 1
= x2(x3 - x - 1) - x(x3 - x - 1) + (x3 - x - 1)
= (x2 - x + 1)(x3 - x - 1)
BD
12 tháng 8 2018
1, x 3+ 6x 2+11x+6
= x 3 + 2x 2 + 4x 2 + 8x + 3x + 6
= x 2 ﴾x + 2﴿ + 4x﴾x + 2﴿ + 3﴾x + 2﴿
= ﴾x + 2﴿﴾x 2 + 4x + 3﴿
2, x 4+3x 3‐7x 2‐27x‐18
= x 4 + 3x 3 ‐ 9x 2 + 2x 2 ‐ 27x ‐18
= ﴾x 4 ‐ 9x 2 ﴿ + ﴾3x 3 ‐ 27x﴿ + ﴾2x 2 ‐ 18﴿
= x 2 ﴾x 2 ‐ 9﴿ + 3x﴾x 2 ‐ 9﴿ + 2﴾x 2 ‐ 9﴿
= ﴾x 2 ‐ 9﴿﴾x 2 + 3x + 2﴿
=﴾x + 3﴿﴾x ‐ 3﴿﴾x 2 + 3x + 2﴿
3, x 3‐8x 2+x+42
= x 3 ‐ 7x 2 ‐ x 2 + 7x ‐ 6x + 42
= ﴾x 3 ‐ 7x 2 ﴿ ‐ ﴾x 2 ‐ 7x﴿ ‐ ﴾6x ‐ 42﴿
= x 2 ﴾x ‐ 7﴿ ‐ x﴾x ‐ 7﴿ ‐ 6﴾x ‐ 7﴿
= ﴾x ‐ 7﴿﴾x 2 ‐ x ‐ 6﴿
4, x 4+5x 3‐7x 2‐41x‐30
= x 4 + x 3 + 4x 3 ‐ 4x 2 ‐ 11x 2 ‐ 11x ‐ 30x ‐ 30
= ﴾x 4 + x 3 ﴿ + ﴾4x 3 ‐ 4x 2 ﴿ ‐ ﴾11x 2 + 11x﴿ ‐ ﴾30x + 30﴿
= x 3 ﴾x + 1﴿ + 4x 2 ﴾x + 1﴿ ‐ 11x﴾x + 1﴿ ‐ 30﴾x + 1﴿
= ﴾x 3 + 4x 2 ‐ 11x ‐ 30﴿﴾x + 1﴿
5, x 5+x‐1
= x 5 ‐ x 4 + x 3 + x 4 ‐ x 3 + x 2 ‐ x 2+ x ‐1
= x 3 ﴾x 2 ‐ x + 1﴿+ x 2 ﴾x 2 ‐ x + 1﴿‐ ﴾x 2 ‐ x + 1﴿
= ﴾x 2 ‐ x + 1﴿﴾x 3 + x 2 ‐ 1﴿ 6, x 5‐x 4‐1
= x 5 ‐ x 3 ‐ x 2 ‐ x 4 + x 2 + x + x 3 ‐ x ‐ 1
= x 2 ﴾x 3 ‐ x ‐ 1﴿ ‐ x﴾x 3 ‐ x ‐ 1﴿ + ﴾x 3 ‐ x ‐ 1﴿
= ﴾x 2 ‐ x + 1﴿﴾x 3 ‐ x ‐ 1﴿
16 tháng 6 2018
a) a3+a2c-abc+b2c+b3 =(a3+b3)+(a2c-abc+b2c)=(a+b)(a2-ab+b2)+c(a2-ab+b2)=(a2-ab+b2)(a+b-c)
b) x3-7x-6 = x3+x2-x2-x-6x-6=x2(x+1)-x(x+1)-6(x+1)=(x+1)(x2-x-6)=(x+1)(x-3)(x+2)
c) x3-x2-14x+24=x3-2x2+x2-2x-12x+24=x2(x-2)+x(x-2)-12(x-2)=(x-2)(x2+x-12)=(x-2)(x+4)(x-3)
27 tháng 7 2018
Ukm
It's very hard
l can't do it
Sorry!
27 tháng 7 2018
a) \(x^4-x^3-7x^2+x+6=0\)
\(\Leftrightarrow x^4+2x^3-3x^3-6x^2-x^2-2x+3x+6=0\)
\(\Leftrightarrow x^3\left(x+2\right)-3x^2\left(x+2\right)-x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-3x^2-x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-3\right)=0\). Làm nốt
b) \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)
\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-3\right)^2+\left|y+3\right|=0\)
Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)
\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)
c) \(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)^2-2.\left(2x^2+x\right).2+4-1=0\)
\(\Leftrightarrow\left(2x^2+x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}2x^2+x-2=1\\2x^2+x-2=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)
Từ đó tính đc x
d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(x^2+5x+5=a\), khi đó pt có dạng:
\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
LH
2
3 tháng 8 2021
4x^2-81=0
4x^2=81
x^2=81/4
x=\(\mp\frac{9}{2}\)
Vậy............
QA
3 tháng 8 2021
Trả lời:
\(4x^2-81=0\)
\(\Leftrightarrow\left(2x\right)^2-9^2=0\)
\(\Leftrightarrow\left(2x-9\right)\left(2x+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-9=0\\2x+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=-\frac{9}{2}\end{cases}}}\)
Vậy x = 9/2; x = - 9/2 là nghiệm của pt.
4 tháng 8 2017
Ta có : 6x2 - 11x + 3
= 6x2 - 2x - 9x + 3
= (6x2 - 2x) - (9x - 3)
= 2x(3x - 1) - 3(3x - 1)
= (2x - 3)(3x - 1)
19 tháng 6 2021
a, \(\left(x-15\right)\left(x+15\right)-\left(x+2\right)^2-\left(x-5\right)^2\)
\(=x^2-225-x^2-4x-4-x^2+10x-25\)
\(=-x^2+6x-254\)
b, \(\left(2x-1\right)\left(2x+1\right)+\left(x+9\right)^2-\left(x-3\right)^2\)
\(=4x^2-1+x^2+18x+81-x^2+6x-9=4x^2+24x+71\)
c, \(\left(7x-3\right)^2-\left(x-5\right)\left(x+5\right)-\left(2x+4\right)^2\)
\(=49x^2-42x+9-x^2+25-4x^2-16x-16=44x^2-58x+18\)
JS
26 tháng 2 2020
Sửa đề: x2 + 13x + 41 --> x2 + 13x + 42
Giải:
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+41}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{2}\)
(ĐKXĐ: \(x\ne\left\{-1;-2;-3;-4;-5;-6;-7\right\}\))
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+7}=\frac{1}{2}\)
\(\Leftrightarrow\frac{x+7-x-1}{\left(x+1\right)\left(x+7\right)}=\frac{1}{2}\)
\(\Leftrightarrow\left(x+1\right)\left(x+7\right)=12\)
\(\Leftrightarrow x^2+8x+7=12\)
⇔x2-8x=5
⇔ x2-8x+(-4)2=5+(-4)2
⇔ x2-8x+16=21
⇔ (x-4)2=21
⇔ x=±21+4
Vậy...
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