\(\dfrac{27^3x\left(-9\right)^5}{3^{16}}\)
Rút gọn biểu thức chứa lũy thừa sau:
\(x\) là nhân ạ
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Những câu hỏi liên quan
25 tháng 2 2023
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
5 tháng 10 2021
\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)
18 tháng 11 2021
\(E=\dfrac{\left|x-3\right|}{\left(x-3\right)\left(x+3\right)}\left(x+3\right)^2=\dfrac{\left|x-3\right|\left(x+3\right)}{x-3}\left(x\ne\pm3\right)\)
Với \(x>3\Leftrightarrow E=x+3\)
Với \(x< 3\Leftrightarrow E=-x-3\)
\(F=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\left(x\ge0;x\ne25\right)\\ F=\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
24 tháng 12 2023
a: \(\left(3x+4y\right)\left(9x^2-12y+16y^2\right)\)
\(=27x^3-36xy+48xy^2+36x^2y-48y^2+64y^3\)
b: \(\left(x+3\right)^3-\left(3x-1\right)^2\)
\(=x^3+9x^2+27x+27-\left(9x^2-6x+1\right)\)
\(=x^3+9x^2+27x+27-9x^2+6x-1\)
\(=x^3+33x+26\)
24 tháng 12 2023
`#3107.101107`
`1.`
`a,`
`(3x + 4y)(9x^2 - 12xy + 16y^2)?`
`= (3x)^3 + (4y)^3`
`= 27x^3 + 64y^3`
`b,`
`(x + 3)^3 - (3x - 1)^2`
`= x^3 + 9x^2 + 27x + 27 - (9x^2 - 6x + 1)`
`= x^3 + 9x^2 + 27x + 27 - 9x^2 + 6x - 1`
`= x^3 + 33x + 26`
_____
Sử dụng HĐT:
`A^3 + B^3 = (A + B)(A^2 + AB + B^2)`
`(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`
`(A - B)^2 = A^2 - 2AB + B^2.`
23 tháng 7 2021
d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5\sqrt{x}-6-2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{x-3}\)
\(=\dfrac{3\sqrt{x}}{x-3}\)
f) Ta có: \(\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)
\(=\dfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}:\dfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)
\(=\dfrac{\sqrt{1-x^2}}{\sqrt{1+x}}=\sqrt{1-x}\)
16 tháng 8 2023
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
2
3 tháng 7 2023
`\sqrt{[27(x-1)^2]/12} +3/2 - (x - 2)\sqrt{[50x^2]/[8(x-2)^2]}` `(1 < x < 2)`
`=\sqrt{[3(x-1)]^2 .3}/\sqrt{2^2 .3} + 3/2 - (x - 2) \sqrt{(5x)^2 . 2}/\sqrt{[2(x - 2)]^2 . 2}`
`=[3\sqrt{3}|x-1|]/[2\sqrt{3}]+3/2-(x-2)[5\sqrt{2}|x|]/[2\sqrt{2}|x-2|]`
`=[3(x-1)]/2+3/2-[5x(x-2)]/[2(2-x)]` (Vì `1 < x < 2`)
`=3/2x - 3/2 + 3/2 + 5/2x`
`=4x`
10 tháng 8 2021
\(x\ge0,x\ne9\)
\(A=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right]:\)
\(\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\left[\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right].\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right).\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)
\(\dfrac{27^3\cdot\left(-9\right)^5}{3^{16}}\\ =-\dfrac{27^3\cdot9^5}{3^{16}}\\ =-\dfrac{\left(3^3\right)^3\cdot\left(3^2\right)^5}{3^{16}}\\-\dfrac{3^9\cdot3^{10}}{3^{16}} \\ =-\dfrac{3^{19}}{3^{16}}\\ -3^3\\ =-27\)
\(\dfrac{27^3\cdot\left(-9\right)^5}{3^{16}}\\ =\dfrac{-\left(27^3\cdot9^5\right)}{3^{16}}\\ =\dfrac{-\left[\left(3^3\right)^3\cdot\left(3^2\right)^5\right]}{3^{16}}\\ =\dfrac{-\left(3^9\cdot3^{10}\right)}{3^{16}}\\ =\dfrac{\left(-3\right)^{19}}{3^{16}}\\ =\left(-3\right)^3\\ =-27\)