Tính nhanh:
a) 52 3 ; b) 499 3 ;
c) 120 3 – 60 . 120 2 + 1200.120.-7999; d) 48 3 + 6 . 48 2 + 12.48+9.
a)
52
3...">
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. Tính nhanh: a)
52
3
; b)
499
3
; c)
120
3
–
60
.
120
2
+ 1200.120.-7999; d)
48
3
+
6
.
48
2
+ 12.48+9. \(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\) \(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\) \(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\) \(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\) \(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\) \(=\frac{2x+1}{x-3}\) b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\) thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\) c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\) \(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\) \(\Leftrightarrow4x+2=x^2-3x\) \(\Leftrightarrow x^2-7x-2=0\) \(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\) \(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\) \(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\) bạn tự giải nốt nhé!! d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\) \(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\) bạn tự làm nốt nhé a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\) \(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\) b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\) Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\) Bài 209 : đăng tách ra cho mn cùng làm nhé a,sửa đề : \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\) \(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\) b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\) \(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\) \(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\) c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\) \(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\) \(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\) \(A=x^2-2xy-4z^2+y^2\) \(=\left(x-y\right)^2-\left(2z\right)^2\) \(=\left(x-y+2z\right)\left(x-y-2z\right)\) \(=\left(6+4+45\right)\left(6+4-45\right)\) \(=-1925\) a: \(=3n^4-3n^3-11n^3+11n^2+10n^2-10n\) \(=\left(n-1\right)\left(3n^3-11n^2+10n\right)\) \(=n\left(n-1\right)\left(n-2\right)\left(3n-5\right)\) \(=n\left(n-1\right)\left(n-2\right)\left(3n+3-8\right)\) \(=3n\left(n-1\right)\left(n+1\right)\left(n-2\right)-8n\left(n-2\right)\left(n-1\right)\) Vì n;n-1;n+1;n-2 là 4 số liên tiếp nên n(n-1)(n+1)(n+2) chia hết cho 4!=24 mà -8n(n-2)(n-1) chia hết cho 24 nên A chia hết cho 24 b: \(=n\left(n^4-5n^2+4\right)\) \(=n\left(n-1\right)\left(n-2\right)\left(n+1\right)\left(n+2\right)\) Vì đây là 5 số liên tiếp nên \(n\left(n-1\right)\cdot\left(n-2\right)\left(n+1\right)\left(n+2\right)⋮5!=120\)