BB
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
20 tháng 10 2022
a: \(=\left(a+b+c\right)^2+2\left(a+b+c\right)\left(b-c\right)+\left(b-c\right)^2\)
\(=\left(a+b+c+b-c\right)^2=\left(a+2b\right)^2\)
b: \(=\left[\left(x^2+2\right)^2-4x^2\right]\left(x^4-4\right)\)
\(=\left(x^4+4\right)\left(x^4-4\right)=x^8-16\)
d: \(=x^2+y^2+z^2+2\left(xy+yz+xz\right)+2x^2+2y^2+2z^2-2\left(xy+yz+xz\right)-3\left(x^2+y^2+z^2\right)\)
=0
HH
21 tháng 6 2018
Giải:
1) \(\left(x-6\right)\left(x^2+6x+36\right)-\left(x+4\right)^3=\left(x-2\right)^3+\left(x+5\right)\left(x^2-10x+25\right)-\left(2x^3+6x^2\right)\)
\(\Leftrightarrow x^3-216-\left(x^3+12x^2+48x+64\right)=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow x^3-216-x^3-12x^2-48x-64=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow-280-12x^2-48x=-12x^2+12x+117\)
\(\Leftrightarrow-280-48x-12x-117=0\)
\(\Leftrightarrow-397-60x=0\)
\(\Leftrightarrow-60x=397\)
\(\Leftrightarrow x=-\dfrac{397}{60}\)
Vậy ...
2) \(\left(2x+3\right)^3-\left(2x+5\right)\left(4x^2-10x+25\right)=\left(6x-1\right)^2-\left(x-2\right)\left(x^2+2x+4\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-\left(8x^3+125\right)=36x^2-12x+1-\left(x^3-8\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3-125=36x^2-12x+1-x^3+8+x^3\)
\(\Leftrightarrow54x-98=-12x+9\)
\(\Leftrightarrow54x+12x=9+98\)
\(\Leftrightarrow66x=107\)
\(\Leftrightarrow x=\dfrac{107}{66}\)
Vậy ...
TQ
0
LG
1
A
2
AH
Akai Haruma
Giáo viên
27 tháng 2 2019
10.
\((x^2-2x-3)(x^2+10x+21)=25\)
\(\Leftrightarrow (x-3)(x+1)(x+3)(x+7)=25\)
\(\Leftrightarrow [(x-3)(x+7)][(x+1)(x+3)]=25\)
\(\Leftrightarrow (x^2+4x-21)(x^2+4x+3)=25\)
Đặt \(x^2+4x-21=a\) thì pt trở thành:
\(a(a+24)=25\)
\(\Leftrightarrow a^2+24a-25=0\)
\(\Leftrightarrow (a-1)(a+25)=0\Rightarrow \left[\begin{matrix} a=1\\ a=-25\end{matrix}\right.\)
Nếu \(a=x^2+4x-21=1\Leftrightarrow x^2+4x-22=0\)
\(\Leftrightarrow (x+2)^2=26\Rightarrow x+2=\pm \sqrt{26}\Rightarrow x=-2\pm \sqrt{26}\) (t/m)
Nếu \(a=x^2+4x-21=-25\Leftrightarrow x^2+4x+4=0\Leftrightarrow (x+2)^2=0\Rightarrow x=-2\) (t/m)
Vậy \(x\in \left\{-2\pm \sqrt{26}; -2\right\}\)
AH
Akai Haruma
Giáo viên
27 tháng 2 2019
11.
\(x^4-4x^3+10x^2+37x-14=0\)
\(\Leftrightarrow (x^4-4x^3+4x^2)+6x^2+37x-14=0\)
\(\Leftrightarrow x^4+2x^3-(6x^3+12x^2)+(22x^2+44x)-(7x+14)=0\)
\(\Leftrightarrow x^3(x+2)-6x^2(x+2)+22x(x+2)-7(x+2)=0\)
\((x+2)(x^3-6x^2+22x-7)=0\)
\(\Rightarrow \left[\begin{matrix} x+2=0\\ x^3-6x^2+22x-7=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x^3-6x^2+22x-7=0(*)\end{matrix}\right.\)
Đối với pt $(*)$ (ta sử dụng pp Cardano)
\(\Leftrightarrow (x^3-6x^2+12x-8)+10x+1=0\)
\(\Leftrightarrow (x-2)^3+10(x-2)+21=0\)
Đặt \(x-2=a-\frac{10}{3a}\) thì PT trở thành:
\((a-\frac{10}{3a})^3+10(a-\frac{10}{3a})+21=0\)
\(\Leftrightarrow a^3-\frac{1000}{27a^3}+21=0\)
\(\Leftrightarrow 27a^6+576a^3-1000=0\). Đặt \(a^3=t\) thì:
\(27t^2+576t-1000=0\)
\(\Rightarrow 27(t^2+\frac{64}{3}t+\frac{32^2}{3^2})=4072\)
\(\Leftrightarrow 27(t+\frac{32}{3})^2=4072\Rightarrow t=\pm\sqrt{\frac{4072}{27}}-\frac{32}{3}\)
\(\Rightarrow a=\sqrt[3]{\pm \sqrt{\frac{4072}{27}}-\frac{32}{3}}\)
\(x=2+a-\frac{10}{3a}\) với giá trị $a$ như trên.
P/s: Bài này mình thấy có vẻ không phù hợp với lớp 8.
19 tháng 4 2020
a/ 12-3(x-2)=(x+2)(1-3x)+2x
\(\Leftrightarrow18-3x=-3x^2-3x+2\)
\(\Leftrightarrow3x^2=-16\left(vl\right)\)
=> phương trình vô nghiệm
b/\(\left(x+5\right)\left(x+2\right)\) =3(4x-2)+(x-5)
\(\Leftrightarrow x^2+3x+10=13x-11\)
\(\Leftrightarrow x^2-10x+21=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
c/\(\frac{x-5}{x^2-5x}-\frac{x-5}{2x^2-10x}=\frac{x+25}{2x^2-50}\)(x khác 0)
\(\Leftrightarrow\frac{x-5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x-5\right)}=\frac{x^2+25}{2x^2-50}\)
\(\frac{\Leftrightarrow1}{x}-\frac{1}{2x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{1}{2x}=\frac{x+25}{2x^2-50}\Leftrightarrow2x^2-50=2x^2+50x\)
\(\Leftrightarrow50x=-50\Leftrightarrow x=-1\)(tm)
d/4x2-1=(2x+1)(3x-5)
\(\Leftrightarrow4x^2-1=6x^2-7x-5\)
\(\Leftrightarrow2x^2-7x-4=0\Leftrightarrow\left(x-4\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\frac{1}{2}\end{matrix}\right.\)
e/ \(x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
TH
6
24 tháng 6 2018
......................?
mik ko biết
mong bn thông cảm
nha ................
4 tháng 8 2019
a) 10x(x - y)2 - 5(x - y)3 = [10x - 5(x - y)](x - y)2 = (10x - 5x + y)(x - y)2 = (5x + y)(x - y)2
b) -x2 - 10x - 25 = -(x2 + 10x + 52) = -(x + 5)2
c) 64x6y4 - 81x2y2 = (8x3y2)2 - (9xy)2 = (8x3y2 + 9xy)(8x3y2 - 9xy)
d) x6 - y6 = (x3)2 - (y3)2 = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy + y2)(x + y)(x2 - xy + y2)
e)1/8x3 - 3/4x2y + 3/2xy2 - y3 = (1/2x)3 - 3.(1/2x)2y + 3.1/2xy2 - y3 = (1/2x - y)3
f) (3x + 1)2 - (x - 1)2 = (3x + 1 + x - 1)(3x + 1 - x + 1) = 4x(2x + 2) = 8x(x + 1)
HL
13 tháng 7 2017
a) Sai đề nên sửa luôn\(\left(2x-5\right)\left(4x^2+10x+25\right)-2x\left(2x+1\right)^2+8x^2+23x+125\)
=\(8x^3-125-2x\left(4x^2+4x+1\right)+8x^2+23x+125\)
= \(8x^3-125-8x^3-8x^2-2x+8x^2+23x+125\)
= \(21x\)
b) \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)-2^{32}\)
= \(\left(2^{16}-1\right)\left(2^{16}+1\right)-2^{32}\)
= \(2^{32}-1-2^{32}=-1\)
Ta có: \(\left(2x-5\right)\left(4x^2+10x+25\right)\left(2x+5\right)\left(4x^2-10x+25\right)-64x^6\)
\(=\left(8x^3-125\right)\left(8x^3+125\right)-64x^6\)
\(=64x^6-15625-64x^6\)
=-15625