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Ta có

\(5^x.5^{x+1}.5^{x+2}\le10^{18}:2^{18}=>5^{3x+3}\le5^{18}=>3x+3\le18=>x\le5\)

Đề bài:\(5^x\times5^{x+1}\times5^{x+2}\le100...0\left(18\text{ số }10\right):2^{18}\)

Hay có thể viết thế này:\(5^x.5^{x+1}.5^{x+2}\le10^{18}:2^{18}=5^{18}\)

                            \(\Leftrightarrow5^{x+\left(x+1\right)+\left(x+2\right)}\le5^{18}\)

                          \(\Leftrightarrow x+\left(x+1\right)+\left(x+2\right)\le18\)

                       \(\Leftrightarrow3x+3\le18\)

                        \(\Leftrightarrow3x\le15\Leftrightarrow x\le5\)

bạn ở trường nào

B. 1/6 nhé!

B

\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)

\(=6x^2y\)

\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)

\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1