Rút gọn biểu thức
\(3\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
Thu gọn biểu thức sau :
a) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot\left(\frac{1}{16}+1\right)\cdot\cdot\cdot\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)-2^{64}\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
Rút gọn biểu thức
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{1+2+3+4+5+.....+2006}\right)\)
Giúp em với ạ
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\left(\dfrac{1}{1+2+3+...+2006}\right)\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left\{\dfrac{1}{\left(2006+1\right)\left[\left(2006-1\right):1+1\right]}\right\}\)
\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{1}{2007\cdot2006}\)
\(=\dfrac{10}{18}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{9}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{36234378}\)
rút gọn biểu thức sau bằng cách nhanh nhất
A = \(\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
B = \(\left(3x^3+3x+1\right)\cdot\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
C = \(\left(2-6x\right)^2+\left(2-5x\right)^2+2\cdot\left(6x-2\right)\cdot\left(2-5x\right)\)
D = \(5\cdot\left(3x-1\right)^2+4\cdot\left(5x+1\right)^2-12\cdot\left(5x-2\right)\left(5x+2\right)\)
E = \(\left(3x-1\right)^2+\left(2x+4\right)\cdot\left(1-3x\right)+\left(x+2\right)^2\)
G = \(\left(x-1\right)^3+4\cdot\left(x+1\right)\cdot\left(1-x\right)+3\cdot\left(x-1\right)\cdot\left(x^2+x+1\right)\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
Rút gọn biểu thức A= \(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2017}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}-\frac{71}{5}\)
rút gọn biểu thức:
\(3x^2\cdot\left(2y-1\right)-2x^2\cdot\left(5y-3\right)-2x\cdot\left(x-1\right)\)
Rút gọn biểu thức sau:
A=\(\left(2x+y\right)^2-\left(y-2x\right)^2\)
B=\(\left(3x+2\right)^2+2\cdot\left(2+3x\right)\cdot\left(1-2y\right)+\left(2y-1\right)^2\)
a: Ta có: \(A=\left(2x+y\right)^2-\left(2x-y\right)^2\)
\(=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\)
\(=4x\cdot2y=8xy\)
b: Ta có: \(B=\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(2y-1\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
Phân tích đa thức thành nhân tử
a)\(x\cdot\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+3\right)+1\)
b)\(\left(x^2-x+2\right)^2+4\cdot x^2-4\cdot x-4\)
c)\(\left(x+2\right)\cdot\left(x+4\right)\cdot\left(x+6\right)\cdot\left(x+8\right)+16\)
a)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(t=x^2+3x\) thì biểu thức có dạng \(t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2=\left(x^2+3x+1\right)^2\)
b)\(\left(x^2-x+2\right)^2+4x^2-4x-4=\left(x^2-x+2\right)^2+4\left(x^2-x-1\right)\)
Đặt \(k=x^2-x+2\) thì biểu thức có dạng
k2+4(k-3)=k2+4k-12=k2-2k+6k-12=k(k-2)+6(k-2)=(k-2)(k+6)=(x2-x)(x2-x+8)=(x-1)x(x2-x+8)
c)làm tương tự câu a
TÍNH GIÁ TRỊ BIỂU THỨC:\(\left(1+\dfrac{2}{3}\right)\cdot\left(1+\dfrac{2}{4}\right)\cdot\left(1+\dfrac{2}{5}\right)\cdot...\cdot\left(1+\dfrac{2}{97}\right)\cdot\left(1+\dfrac{2}{98}\right)\)
Chứng minh giá trị biểu thức không phụ thuộc x :
1, \(\left(2x+1\right)^3-\left(2x-1\right)^3-2\cdot\left(4x+3\right)^2+8\cdot\left(x+3\right)^2\)
2,\(\left(2x+1\right)^2\cdot\left(x-1\right)-2\cdot\left(x-2\right)^3+x\cdot\left(3-2x\right)\cdot\left(3+x\right)-\left(3x-3\right)^2\)
1: \(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(4x+3\right)^2+8\left(x+3\right)^2\)
\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)
\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)
=56
2: \(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)
\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)
\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)
\(=6\)