\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)

\(\rightarrow\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)

\(\rightarrow\dfrac{3}{5}-\dfrac{2}{3}x=\dfrac{-4}{5}\)

\(\rightarrow x=\dfrac{21}{10}\)

cho mk sửa lại

tacó:

\(\dfrac{-64}{125}=\left(\dfrac{-4}{5}\right)^3\)

suy ra\(\dfrac{2}{3}-\dfrac{3}{5}x=\dfrac{-4}{5}\)

\(\dfrac{3}{5}x=\dfrac{2}{3}-\dfrac{-4}{5}\)

\(\dfrac{3}{5}x=\dfrac{22}{15}\)

\(x=\dfrac{22}{15}:\dfrac{3}{5}\)

\(x=\dfrac{22}{9}\)

ta có:

\(\dfrac{-64}{125}=\left(\dfrac{-16}{5}\right)^3\)

suy ra \(\dfrac{2}{3}-\dfrac{3}{5}x=\dfrac{-16}{5}\)

\(\dfrac{3}{5}x=\dfrac{2}{3}-\dfrac{-16}{5}\)

\(\dfrac{3}{5}x=\dfrac{58}{15}\)

\(x=\dfrac{58}{15}:\dfrac{3}{5}\)

\(x=\dfrac{58}{9}\)

tick cho mk với nha

a. VP: \(\left(x+y\right)^{1999}\cdot\left(x-y\right)^{1999}=\left[\left(x+y\right)\left(x-y\right)\right]^{1999}\)

\(=\left(x^2-xy+xy-y^2\right)^{1999}=\left(x^2-y^2\right)^{1999}=VT\)

--> đpcm

b. VT: \(\dfrac{\left(5^4-5^3\right)^3}{125^4}=\dfrac{500^3}{125^4}=\dfrac{125^3\cdot4^3}{125^4}=\dfrac{4^3}{125}=\dfrac{64}{125}=VP\)

--> đpcm

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

Bài 1 :

a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)

Mà \(x^2+1\ge1>0\)

\(\Rightarrow2x+1\ge0\)

\(\Rightarrow x\ge-\dfrac{1}{2}\)

Vậy ...

b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)

\(=-3+4-\left(-4\right)=-3+4+4=5\)

Bài 2 :

\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)

\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)

\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)

\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)

\(=3\)

a) \(64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)

b) \(125x^6-27x^9=\left(5x^2-3x^3\right)\left(25x^4+15x^5+9x^6\right)\)

c) \(-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left(\dfrac{x^6}{125}+\dfrac{y^3}{64}\right)=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)

\(a,\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=-\dfrac{64}{125}\)

\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)

\(\dfrac{3}{5}-\dfrac{2}{3}x=-\dfrac{4}{5}\)

\(-\dfrac{2}{3}x=-\dfrac{4}{5}-\dfrac{3}{5}\)

\(-\dfrac{2}{3}x=-\dfrac{7}{5}\)

\(x=\dfrac{21}{10}\)

\(b,\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)

\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)

\(x-\dfrac{2}{9}=\dfrac{4}{9}\)

\(x=\dfrac{2}{3}\)

\(c,\left(0,4x-1,3\right)^2=5,29\)

\(\left(0,4x-1,3\right)^2=2,3^2=\left(-2,3\right)^2\)

TH1: \(0,4x-1,3=2,3\)

\(0,4x=3,6\)

\(x=9\)

TH2: \(0,4x-1,3=-2,3\)

\(0,4x=-1\)

\(x=-\dfrac{5}{2}\)

=.= hok tốt!!

a, \(x^3+8=x^3+2x^2-2x^2-4x+4x+8\)

\(=x^2.\left(x+2\right)-2x.\left(x+2\right)+4.\left(x+2\right)\)

\(=\left(x+2\right).\left(x^2-2x+4\right)\)

c) \(-\dfrac{x^6}{125}-\dfrac{y^3}{64}\)

\(=-\left(\dfrac{x^6}{125}+\dfrac{y^3}{64}\right)\)

\(=-\left(\dfrac{x^2}{5}+\dfrac{y^4}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)

\(b)\left(x-3\right)^3=125^2\)

\(\Rightarrow\left(x-3\right)^3=5^{3^2}\)

\(\Rightarrow\left(x-3\right)^3=25^3\)

\(\Rightarrow x-3=25\)

\(\Rightarrow x=28\)