Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(26^2+52.24+24^2=26^2+2.26.24+24^2\)
= \(\left(26+24\right)^2=50^2=2500\)
b) \(52^2+47^2+94.52\) ( câu này sai đề sửa luôn)
= \(52^2+2.47.52+47^2=\left(52+47\right)^2=99^2\)
= \(9801\)
c) \(50^2-49^2+48^2-47^2+...+2^2-1^2\)
= \(\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
= \(99+95+...+3\)
Dãy số này có : \(\dfrac{99-3}{4}+1=\dfrac{96}{4}+1=25\) số hạng
\(\Rightarrow\) \(99+95+...+3\) = \(\left(99+3\right).25:2=1275\)
d) \(87^2+26.87+13^2=87^2+2.13.87+13^2\)
\(=\left(87+13\right)^2=100^2=10000\)
e) \(3003^2-3^2=\left(3003-3\right)\left(3003+3\right)\)
= \(3000.3006=9018000\)
\(a,26^2+52\cdot24+24^2\\ =26^2+2\cdot26\cdot24+24^2\\ =\left(26+24\right)^2\\ =50^2\\ =2500\)
\(b,53^2+47^2+94\cdot53\\ =53^2+2\cdot47\cdot53+47^2\\ =\left(53+47\right)^2\\ =100^2\\ =10000\)
\(c,50^2-49^2+48^2-47^2+...+2^2-1^2\\ =\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\\ =99\cdot1+97\cdot1+...+3\cdot1\\ =99+97+...+3\\ \)
\(99+97+...+3\) có số số hạng là \(\dfrac{99-3}{2}+1=49\)(số)
\(\Rightarrow99+97+...+3=\dfrac{\left(99+3\right)\cdot49}{2}=2499\)
\(d,87^2+26\cdot87+13^2\\ =87^2+2\cdot13\cdot87+13^2\\ =\left(87+13\right)^2\\ =100^2\\ =10000\)
\(e,3003^2-3^2\\ =\left(3003+3\right)\left(3003-3\right)\\ =3006\cdot3000\\ =9018000\)
\(f,85\cdot12,7+5\cdot3\cdot12,7\\ =85\cdot12,7+15\cdot12,7\\ =12,7\cdot\left(85+15\right)\\ =12,7\cdot100\\ =1270\)
\(\text{Chúc bạn học tốt}\)
Bài : 1 Ta có : (x - 2)3 + 6(x + 1)2 - x3 + 12 = 0
=> x3 - 6x2 + 12x - 8 + 6(x2 + 2x + 1) - x3 + 12 = 0
=> x3 - 6x2 + 12x - 8 + 6x2 + 12x + 6 - x3 + 12 = 0
=> 24x - 10 = 0
=> 24x = 10
=> x = 5/12
Vạy x = 5/12
Bài 4 : Ta có : M = x2 + 6x - 1
=> M = x2 + 6x + 9 - 10
=> M = (x + 3)2 - 10
Vì : \(\left(x+3\right)^2\ge0\forall x\)
Nên : M = (x + 3)2 - 10 \(\ge-10\forall x\)
Vậy Mmin = -10 khi x = -3
1, gọ̣̣i bthứ́c trên là A, ta có:
A=8y3-12y2+6y-1-2y*(4y2-12y+9)-12y2+12y
A=8y3-12y2+6y-1-8y3+24y2-18y-12y2+12y
A=-1
vây bthức A ko phu thuôc vào biến y
u^2v^2(u+v)^2-(u^2v+uv^2)^2 - Step-by-Step Calculator - Symbolab
Tham khảo ở đó nhé!
Bài 1:
Theo bài ra ta có:
\(\left(x-y\right)^2=x^2-2xy+y^2\)
\(=\left(5-y\right)^2-2\times2+\left(5-x\right)^2\)
\(=5^2-2\times5y+y^2-4+5^2-2\times5x+x^2\)
\(=25-10y+y^2+25-10x+x^2-4\)
\(=\left(25+25\right)-\left(10x+10y\right)+x^2+y^2-4\)
\(=50-10\left(x+y\right)+x^2+2xy+y^2-2xy-4\)
\(=50-10\times5+\left(x+y\right)^2-2\times2-4\)
\(=50-50+5^2-4-4\)
\(=25-8=17\)
Vậy giá trị của \(\left(x-y\right)^2\)là 17
Bài 1 :
\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)
\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)
Đặt \(a=x^2+6x-7\)
\(A=a\left(a-9\right)+8\)
\(A=a^2-9a+8\)
\(A=a^2-8a-a+8\)
\(A=a\left(a-8\right)-\left(a-8\right)\)
\(A=\left(a-8\right)\left(a-1\right)\)
Thay a vào là xong bạn :)
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
4a) \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+b^2+2ab\)
=> (a+b)^2=(a-b)^2+4ab
- 2x – x2 + 2 – x – (3x2 + 6x + 5x +10) = – 4x2 + 2
- 2x – x2 + 2 – x – 3x2 – 6x – 5x – 10 = – 4x2 + 2 –10x = 10 x = – 1
- 2x2 – 6x + x – 3 = 0
(x – 3)(2x + 1) = 0
x = 3 hay x = -1/2