bn phá hằng đẳng thức ra là xong

Như thế này bn thấy rõ k

\(\left(\dfrac{5}{6}-\dfrac{2}{5}x\right):\dfrac{1}{4}=\dfrac{-1}{5}\Leftrightarrow\dfrac{5}{6}-\dfrac{2}{5}x=\dfrac{-1}{20}\Leftrightarrow\dfrac{2}{5}x=\dfrac{53}{60}\Leftrightarrow x=\dfrac{53}{24}\)

a,Ta có \(\left(3^3\right)^n:3^n=9\Leftrightarrow3^{3n}:3^n=3^2\Leftrightarrow3n-n=2\Leftrightarrow n=1\)

b,TA có \(\dfrac{5^2}{5^n}=5^1\Leftrightarrow2-n=1\Leftrightarrow n=1\)

Các câu sau để bn tự làm

Ta có 3x(x-4y)-\(\dfrac{12}{5}\)y(y-5x)=3\(x^2\)-\(12xy-\dfrac{12}{5}y^2+12xy\)

=\(3x^2-\dfrac{12}{5}y^2\) (*)

Thay x=4,y=5 vào (*) ta đc

(*)\(\Leftrightarrow3.4^2-\dfrac{12}{5}.5^2=12.4-12.5=12.\left(-1\right)=-12\)

ĐKXĐ:x\(\ge0\)

B=\(\left(\dfrac{\sqrt{x}-1}{x-1}+\dfrac{\sqrt{x}+1}{x-1}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)

=\(\left(\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)

=\(\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\sqrt{x}}=\dfrac{2}{\sqrt{x}+1}\)

aaa:37.Y=a

=>37.Y=aaa:a

=>37.Y=111

=>Y=111:37

=>Y=3

\(x^2+2xy+2y^2+2y+5=\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)+4=\left(x+y\right)^2+\left(y+1\right)^2+4\\ \\ Vì\left(x+y\right)^2\ge0\left(\forall x,y\right),\left(y+1\right)^2\ge0\left(\forall y\right)\\ \\ \Rightarrow\left(x+y\right)^2+\left(y+1\right)^2+4\ge4\forall x,y\\ \\ \\ \\ \\ Vậy......................\)