a) 6x : 2 = 16 ; b) 3^x-1 : 8 = 81 . Mn giúp mik câu này nhé . cảm ơn mn
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1-6x+1\right)^2\)
\(=2^2=4\)
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
\(\text{a) }\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(6x+1\right)\left(6x-1\right)\\ =\left[\left(6x+1\right)-\left(6x-1\right)\right]^2\\ =\left[6x+1-6x+1\right]^2\\ =2^2\\ =4\)
\(\text{b) }3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\left(2^{16}-1\right)\left(2^{16}+1\right)\\ =2^{32}-1\)
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left(6x-1\right)-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left[\left(6x-1\right)-2\left(1+6x\right)\right]\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left(6x-1-2-12x\right)\)
\(=36x^2+12x+1+\left(6x-1\right)\left(-6x-3\right)\)
\(=36x^2+12x+1+\left(-36x^2-12x+3\right)\)
\(=36x^2+12x+1-36x^2-12x+3\)
\(=4\)
\(3x^3-6x^3y+3xy=3x\left(x^2-2x^2y+y\right)\)
\(a^2-4ab+4b^2-16=\left(a-2b\right)^2-4^2=\left(a-2b-4\right)\left(a-2b+4\right)\)
\(a,\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+12x+1+36x^2-12x+1-2\cdot\left(36x^2-1\right)\)
\(=72x^2+2-72x^2+2=4\)
\(A=\dfrac{6x}{5x-20}-\dfrac{x}{x^2-8x+16}\)
\(ĐKXĐ:x\ne\pm4\)
\(\Leftrightarrow A=\dfrac{6x}{5\left(x-4\right)}-\dfrac{x}{\left(x-4\right)^2}\)
\(\Leftrightarrow A=\dfrac{6x^2-24x-5x}{5\left(x-4\right)^2}\)
\(\Leftrightarrow\dfrac{6x^2-29x}{5\left(x-4\right)^2}\)
\(\Leftrightarrow\dfrac{x\left(6x-29\right)}{5\left(x-4\right)^2}\)
a) \(2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Vậy GTNN của biểu thức là \(-\dfrac{9}{2}\) khi x = \(\dfrac{3}{2}\)
a ) 6x : 2 = 16
\(6x=16:2\)
\(6x=8\)
\(x=\frac{8}{6}=\frac{4}{3}\)
b) \(3^{x+1}=81\)
\(3^{x+1}=3^4\)
\(x+1=4\)
\(x=3.\)
HT
lớp 4 nói làm gì????