Tìm x
a, (3x -5). (7- 5x) - (5x+ 2). (2- 3x)= 4
b, (x+ 2). (x2- 2x+ 4)- (x3+ 3). x =14
Chứng minh rằng
a, \(\left(2n-3\right).n-2n.\left(n+2\right)⋮7\forall n\inℤ\)
b, \(n.\left(2n-3\right)-2n.\left(n+1\right)⋮5\forall n\inℤ\)
Chứng minh rằng
a, \(\left(2n-3\right).n-2n.\left(n+2\right)⋮7\forall n\in Z\)
b, \(n.\left(2n-3\right)-2n.\left(n+1\right)⋮5\forall n\in Z\)
Rút gọn
a, (3x-5) . (2x+11) - (2x+3) . (3x+7)
b, (x+2) . (2x2-3x+4) - (x2-1) . (2x+1)
c, 3x2 .(x2+2) + 4x. (x2-1) - (x2+2x+3) . (3x2-2x+1)
\(a,\left(2x-3\right)n-2n\left(n+2\right)\)
\(=n\left(2x-3-2n-4\right)\)
\(=-7n\)
Vì \(-7⋮7\Rightarrow-7n⋮7\) => ĐPCM
\(b,n\left(2n-3\right)-2n\left(n+1\right)\)
\(=n\left(2n-3-2n-2\right)\)
\(=-5n⋮5\) (ĐPCM)
Rút gọn
\(a,\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+33x-10x-55-6x^2-14x-9x-21\)
\(=-76\)
\(b,\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\)
\(=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1\)
\(=9\)
\(c,3x^2\left(x^2+2\right)+4x\left(x^2-1\right)-\left(x^2+2x+3\right)\left(3x^2-2x+1\right)\)
\(=3x^4+6x^2+4x^3-4x-3x^4+2x^3-x^2-6x^3+4x^2-2x-9x^2+6x-3\)
= -3
Bài 1: Làm tính nhân
a. \(\left(2x^{2n}+3x^{2n-1}\right)\left(x^{1-2n}-3x^{2-2n}\right)\)
b. \(\left(25x^2+10xy+4y^2\right)\left(5x-2y\right)\)
c. \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
a: \(=2x^{2n+1-2n}-2\cdot x^{2n}\cdot3\cdot x^{2-2n}+3\cdot x^{2n-1+1-2n}-9\cdot x^{2n-1+2-2n}\)
\(=2x-6x^2+3-9x\)
\(=-6x^2-7x+3\)
b: \(=\left(5x\right)^3-\left(2y\right)^3=125x^3-8y^3\)
Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.
Làm tính nhân: a. \(\left(3x^{2m-1}-\dfrac{3}{7}y^{3n-5}+x^{2m}y^{3m}-3y^2\right)8x^{3-2m}y^{6-3n}\)
b.\(\left(2x^{2n}+3x^{2n-1}\right)\left(x^{1-2n}-3x^{2-2n}\right)\)
a: \(=24x^{2m-1+3-2m}y^{6-3m}-\dfrac{24}{7}y^{3n-7+6-3n}\cdot x^{3-2m}+8x^{3-2m+2m}\cdot y^{6-3n+3m}-24x^{3-2m}y^{6-2n+2}\)
\(=24x^2y^{6-3m}-\dfrac{24}{7}x^{3-2m}\cdot y^{-1}+8x^3y^{-3n+3m+6}-24x^{3-2m}y^{-2n+8}\)
b: \(=2x^{2n+1-2n}-6x^{2n+2-2n}+3x^{2n-1+1-2n}-9x^{2n-1+2-2n}\)
\(=2x-6x^2+3-9x\)
\(=-6x^2-7x+3\)
1) \(\left(\dfrac{-3}{4}\right)^{3x+1}=\dfrac{81}{256}\) 6) \(\left(8x-1\right)^{2n-4}=5^{2n-4}\)
2) \(172.x^2-\dfrac{7^9}{98^3}=\dfrac{1}{2^3}\) 7) \(\left(\dfrac{1}{2x}-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
3) \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
4) \(\left(x+2\right)^2+\left(y-\dfrac{1}{10}\right)^2=0\)
5) \(\left(x-7\right)^{n+1}-\left(x-7\right)^{n+11}=0\)
Giúp mk với!!!!!
1: =>3x+1=4
=>3x=3
hay x=1
2: \(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^9}{98^3}=\dfrac{1}{2^3}+\dfrac{7^9}{7^6\cdot2^3}\)
\(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^3}{2^3}=\dfrac{344}{2^3}\)
\(\Leftrightarrow x^2=\dfrac{1}{4}\)
=>x=1/2 hoặc x=-1/2
3: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{9}=\dfrac{4}{9}\\x-\dfrac{2}{9}=-\dfrac{4}{9}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{9}\end{matrix}\right.\)
4: =>x+2=0 và y-1/10=0
=>x=-2 và y=1/10
Giúp e vs m.n ơi!!!!
1. tính GTBT:
\(B=\frac{2}{3}x^2y\left(2x^2-\frac{y}{3}\right)-2x^2\left(2x^2-1\right)+\left(2x^2-\frac{y}{3}\right).2x\)
2.tính:
\(P=3x^n\left(4x^{n+1}-1\right)-2x^{n+1}\left(6x^{n-2}-1\right)\)
\(Q=\left(x^{2n}+x^ny^n+y^{2n}\right).x^n.y^n\)
Tính giới hạn :
L = lim \(\dfrac{\left(n^2+2n\right)\left(2n^3+1\right)\left(4n+5\right)}{\left(n^4-3n-1\right)\left(3n^2-7\right)}\)
Dang này thì cứ chọn số hạng có mũ cao nhất trên tử và mẫu là được. Nó là ngắt vô cùng lớn hay bé gì đấy
\(=lim\dfrac{8n^6}{3n^6}=\dfrac{8}{3}\)
1
a,Lim\(\sqrt{1+2n-n^3}\)
b,Lim\(\sqrt{n^2+2n+3}-\sqrt[3]{n^2+n^3}\)
c,Lim\(\dfrac{\left(2\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n+1\right)\left(n+2\right)}\)
d,\(\dfrac{4^{n+1}-3\times2^n}{3^{n+2}+2^n}\)
e,\(\dfrac{7^{n+1}-5^{n+2}+3}{2\times6^{n+1}-3^n+3}\)
f,\(\dfrac{\sqrt{n^4+1}}{n}\) -\(\dfrac{\sqrt{4n^6+1}}{n}\)
\(a=\lim\sqrt{n^3}\sqrt{\dfrac{1}{n^3}+\dfrac{2}{n^2}-1}=\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n^2+2n+3}-n+n-\sqrt[3]{n^2+n^3}\right)\)
\(=\lim\dfrac{2n+3}{\sqrt{n^2+2n+3}+n}+\lim\dfrac{-n^2}{n^2+n\sqrt[3]{n^2+n^3}+\sqrt[3]{\left(n^2+n^3\right)^2}}\)
\(=\lim\dfrac{2+\dfrac{3}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^2}}+1}+\lim\dfrac{-1}{1+\sqrt[3]{\dfrac{1}{n}+1}+\sqrt[3]{\left(\dfrac{1}{n}+1\right)^2}}=\dfrac{2}{2}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(c=\lim\dfrac{\left(\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{0.0}{1.1}=0\)
\(d=\lim\dfrac{4-3\left(\dfrac{2}{4}\right)^n}{9.\left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n}=\dfrac{4}{0}=+\infty\)
\(e=\lim\dfrac{7-25\left(\dfrac{5}{7}\right)^n+3.\left(\dfrac{1}{7}\right)^n}{12.\left(\dfrac{6}{7}\right)^n-\left(\dfrac{3}{7}\right)^n+3\left(\dfrac{1}{7}\right)^n}=\dfrac{7}{0}=+\infty\)
\(f=\lim\dfrac{n^4-4n^6}{n\left(\sqrt{n^4+1}+\sqrt{4n^6+1}\right)}=\lim\dfrac{\dfrac{1}{n^2}-6}{\sqrt{\dfrac{1}{n^6}+\dfrac{1}{n^{10}}}+\sqrt{\dfrac{4}{n^4}+\dfrac{1}{n^{10}}}}=\dfrac{-6}{0}=-\infty\)
Bai 1 : Tính
a) \(\left(3x^n+1-2x^n\right)4x^2\)
b) 2\(\left(x^{2n}+2x^ny^n+y^{2n}\right)-y^n\left(4x^n+2y^n\right)\)
c)\(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\left(x^{3n}+y^{3n}\right)\)
d)\(4^{n+1}-3.4^n\)
[Các bạn giúp mình nha ! Plz ]
a: \(=12x^{n+2}+4x^2-8x^{n+2}\)
\(=4x^{n+2}+4x^2\)
b: \(=2x^{2n}+4x^ny^n+2y^{2n}-4x^ny^n-2y^{2n}\)
\(=2x^{2n}\)
c: \(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)
\(=x^{6n}-y^{6n}\)
d: \(=4^n\cdot4-3\cdot4^n=4^n\)