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

\(\Leftrightarrow\left(16x^2-8x+1\right)-4\left(4x^2+4x+1\right)-x-4=0\)

\(\Leftrightarrow16x^2-8x+1-16x^2-16x-4-x-4=0\)

\(\Leftrightarrow25x-7=0\)

\(\Leftrightarrow25x=7\)

\(\Leftrightarrow x=\dfrac{7}{25}\)

Ta có:

\(\dfrac{1}{2}=0,5\)

\(\dfrac{2}{3}=0,666...\)

\(\dfrac{3}{4}=0,75\)

\(\dfrac{5}{4}=1,25\)

Sắp xếp theo thứ tự từ lớn đến bé là:
\(\dfrac{5}{4};\dfrac{3}{4};\dfrac{2}{3};\dfrac{1}{2}\)

\(C=4x^2+y^2-4x+8y+12\)

\(C=4x^2-4x+1+y^2+8y+16-5\)
\(C=\left(4x^2-4x+1\right)+\left(y^2+8y+16\right)-5\)

\(C=\left(2x-1\right)^2+\left(y+4\right)^2-5\)

Mà: \(\left\{{}\begin{matrix}\left(2x-1\right)^2\ge0\forall x\\\left(y+4\right)^2\ge0\forall x\end{matrix}\right.\)

Nên: \(C=\left(2x-1\right)^2+\left(y+4\right)^2-5\ge-5\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2x-1=0\\y+4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-4\end{matrix}\right.\)

Vậy: \(C_{min}=-5\) khi \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-4\end{matrix}\right.\)

1% x 100000 = 1000 đồng

Chọn D

Ta có:

\(45=3^2\cdot5\)

\(36=2^2\cdot3^2\)

\(\Rightarrow BCNN\left(45,36\right)=3^2\cdot2^2\cdot5=180\)

\(\dfrac{99}{98}-\dfrac{98}{97}+\dfrac{1}{97\cdot98}\)

\(=\dfrac{99\cdot97}{98\cdot97}-\dfrac{98\cdot98}{97\cdot98}+\dfrac{1}{97\cdot98}\)

\(=\dfrac{99\cdot97-98^2+1}{98\cdot97}\)

\(=\dfrac{\left(98+1\right)\left(98-1\right)-98^2+1}{98\cdot97}\)

\(=\dfrac{98^2-1-98^2+1}{98\cdot97}\)

\(=\dfrac{0}{97\cdot98}\)

\(=0\)

x - 15 = ( - 7) - 30

x - 15 = - 37

x = - 37 + 15

x = - 22 

a) \(A=3+3^2+..+3^{60}\)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)

\(A=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+...+3^{59}\cdot\left(1+3\right)\)

\(A=4\cdot\left(3+3^3+...+3^{59}\right)\)

Vậy A chia hết cho 4

b) \(A=3+3^2+3^3+...+3^{60}\)

\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(A=3\cdot\left(1+3+3^2\right)+...+3^{58}\cdot\left(1+3+3^2\right)\)

\(A=13\cdot\left(3+..+3^{58}\right)\)

Vậy A chia hết cho 13

a) Ta có: 

\(10^{10}=10...0\Rightarrow10^{10}-1=10..0-1=9..99\)

Nên \(10^{10}-1\) ⋮ 9

b) Ta có:

\(10^{10}=10...0\Rightarrow10^{10}+2=10..0+2=10..2\)

Mà: \(1+0+0+...+2=3\) ⋮ 3

Nên: \(10^{10}+2\) ⋮ 3