a)x(x-5)=0\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

b)\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

c)\(\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x=3\)

A=x(y+2)+2(y+2)=(x+2)(y+2)=100.100=10000

B=9\(^2\)(9+1) +11(-9-1)=\(9^2\).10+11.(-10)=810-110=700

\((6x^4-4x^3+x^2+x):(2x^2-2x+1)=3x^2+x\)

mak x>4

\(\Rightarrow x\in\left\{5,6,8,12,20\right\}\)

a)\(A=\dfrac{x\left(\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(2-\sqrt{x-4}\right)^2}\right)}{\sqrt{\left(x-4\right)^2}}\)

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

\(=\dfrac{4x}{x-4}=4+\dfrac{16}{x-4}\)

Để A có giá trị nguyên

\(\Rightarrow x-4\in U\left(16\right)=\left\{1,-1,2,-2,4,-4,8,-8,16,-16\right\}\)

\(\Rightarrow x\in\left\{5,3,6,2,0,8,12,-4,20,-12\right\}\)

Ta có \(f\left(x\right)=x^4+ax^3+bx^2+cx^1+d\)(Gt)

Theo đê ta có

\(f\left(1\right)=3=a+b+c+d+1\)

\(f\left(3\right)=11=27a+9b+3c+d+81\)

\(f\left(5\right)=27=125a+25b+5c+d+625\)

\(f\left(-2\right)=-8a+4b-2c+d+16\)

\(f\left(6\right)=216a+36b+6c+d+1296\)

Khi đó

\(f\left(-2\right)+7f\left(6\right)=-8a+4b-2c+d+16+7\left(216a+36b+6c+d+1296\right)\)

\(=1504a + 256b + 40c+ 8d + 1312\)

\(7\left(1\right)-14f\left(3\right)+15f\left(5\right)=(1504a+256b+40c+8d)+( 7 − 14.81 + 15.625 )\)

\(⇔ 7.3 − 14.11 + 15.27 = ( 1504 a + 256 b + 40 c + 8 d ) + 8248\)

\(⇔ 1504 a + 256 b + 40 c + 8 d = − 7976\)

\(\Rightarrow f ( − 2 ) + 7 f ( 6 ) = − 7976 + 1312 = − 6664\)

Đợi tý mik đưa lại cho tại m gửi nó k đc

đề sai kìa bn

c)\(TacoP>\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}>\dfrac{1}{2}\)

\(\Leftrightarrow2\sqrt{x}-2>\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)

Vậy\(x>9\)Thì \(P>\dfrac{1}{2}\)