Cho abcd=1. Tính \(S=\frac{a}{abc+ab+a+1}+\frac{b}{bcd+bc+b+1}+\frac{c}{acd+cd+c+1}+\frac{d}{abd+ad+d+1}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
= 2.x2.(x- 1/2) - 4x(x- 1/2) +6.(x- 1/2)
=(x - 1/2)(2x2 - 4x +6)
K nha bn iu
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Thu gọn đa thức:
\(\left(2x-3\right)^2-\left(2x-5\right)\left(2x+5\right)\)
\(=4x^2-12x+9-4x^2+25\)
\(=-12x+34=2\left(17-6x\right)\)
2. Tìm x
\(a.\left(2x+5\right)^2-2x\left(2x-1\right)=6\left(x+1\right)\)
\(\Leftrightarrow4x^2+20x+25-4x^2+2x=6x+6\)
\(\Leftrightarrow22x+25-6x-6=0\)
\(\Leftrightarrow16x+19x=0\Leftrightarrow x=\frac{-19}{16}\)
\(b.9x^2-6x=8\)
\(\Leftrightarrow9x^2-6x-8=0\Leftrightarrow9x^2+6x-12x-8=0\)
\(\Leftrightarrow3x\left(3x+2\right)-4\left(3x+2\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\hept{\orbr{\begin{cases}3x-4=0\Rightarrow x=\frac{4}{3}\\3x+2=0\Rightarrow x=\frac{-2}{3}\end{cases}}}\)
3.CMR
\(\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\)
Ta có:
\(VT=a^2+2ab+b^2+a^2-2ab+b^2\)
\(=2a^2+2b^2=2\left(a^2+b^2\right)=VP\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=4x^2+4x+1=\left(2x+1\right)^2\ge0\)
Vậy MIN \(A=2\)khi \(x=-\frac{1}{2}\)
b) \(B=9x^2+6x+11=\left(3x+1\right)^2+10\ge10\)
Vậy MIN \(B=10\)khi \(x=-\frac{1}{3}\)
c) \(C=2x^2+3x+4=2\left(x+\frac{3}{4}\right)^2+2,875\ge2,875\)
Vậy MIN \(C=2,875\)khi \(x=-\frac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
tớ biết nhưng k nói đâu
bằng 1 bn nha!!!