câu 1: Rút gọn: \(A=\frac{x\left|x-2\right|}{x^2+8x-20}\)
câu 2: tìm x biết:
a) |2x - 3| > 5
b) |3x - 1| \(\le7\)
c) |3x - 5| + |2x + 3| = 7
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn
\(A=\left(\dfrac{x^2-2x+1}{x^2+x+1}-\dfrac{-2x^2+4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right):\dfrac{2x}{x^3+x}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}=\dfrac{x^2+1}{2}\)
Áp dụng : \(\frac{1}{\sqrt{1}.2}< 2.\left(1-\frac{1}{\sqrt{2}}\right)\)
\(\frac{1}{\sqrt{2}.3}< 2.\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)\)
...................................
\(\frac{1}{\sqrt{2015}.2016}< 2.\left(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)\)
Cộng các BĐT trên với nhau được : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}}< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)=2\left(1-\frac{1}{\sqrt{2016}}\right)< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)
Từ đó suy ra đpcm
Cái ............... là gì vậy bn
Bài 1:
\(D=-3x^2+x+15x-5-3\left(2x^2-5x+2\right)\)
\(=-3x^2+16x-5-6x^2+15x-6\)
\(=-9x^2+31x-11\)
\(=-9\cdot\dfrac{1}{9}+\dfrac{31}{3}-11\)
=-11-1+31/3=-12+31/3=-5/3
b: \(E=x^2+x-56-x^2+7x-10=8x-66\)
\(=-\dfrac{8}{5}-66=-\dfrac{338}{5}\)
c: \(F=-3\left(2x^2+x-16x-8\right)-\left(-3x^2+2x-15x+10\right)-4x^2+24x\)
\(=-6x^2+45x+24+3x^2+13x-10-4x^2+24x\)
\(=-4x^2+82x+14\)
\(=-4\cdot9-82\cdot3+14=-268\)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right):\frac{4}{4x^2-4}\)
\(=\left(\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+2\right)}+\frac{6}{2.\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\frac{4}{4\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}.\left(x-1\right)\left(x+1\right)=\frac{4}{2}=2\)
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
\(a,\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)
\(=\left(5\sqrt{2}+4\sqrt{3}-6\sqrt{2}\right)2\sqrt{3}\)
\(=\left(4\sqrt{3}-\sqrt{2}\right)2\sqrt{3}\)
\(=24-2\sqrt{6}\)
\(A=\frac{x\left|x-2\right|}{x^2+8x-20}=\frac{x\left|x-2\right|}{x^2-2x+10x-20}=\frac{x\left|x-2\right|}{x\left(x-2\right)+10\left(x-2\right)}=\frac{x\left|x-2\right|}{\left(x+10\right)\left(x-2\right)}\)
Xét \(x-2\ge0\Leftrightarrow x\ge2\) ta có :
\(A=\frac{x\left(x-2\right)}{\left(x+10\right)\left(x-2\right)}=\frac{x}{x+10}\)
Xét \(x-2< 0\Leftrightarrow x< 2\) ta có :
\(A=\frac{x\left(2-x\right)}{\left(x+10\right)\left(x-2\right)}=\frac{-x}{x+10}\)
bạn làm hộ mk câu 2 luôn đc ko
mk đang cần gấy câu đấy