100<5.24

Phải điền dấu <

Vì 100<120

Đáp số: <

`(3.10)/(5.24)`

`=30/120`

`=1/4`

\(=\dfrac{1}{4}\)

`1)4x^2+1=4x^2+4x+1-4x=(2x+1)^2-4x=(2x-2\sqrt{x}+1)(2x+2\sqrt{x}+1)` (với `x >= 0`)

  `->` Ko có đ/á

(Câu này mình nghĩ là `4x^4+1` chứ nhỉ?)

`2)4x^4+y^4=4x^4+4x^2y^2+y^4-4x^2y^2`

        `=(2x^2+y^2)-(2xy)^2`

        `=(2x^2-2xy+y^2)(2x^2+2xy+y^2)`

   `->bb C`

Bạn ghi lại đề đi bạn, khó nhìn quá

a)2x( 2x-1) -(2x-1)

=(2x-1)(2x-1)

=(2x-1)²

b)2x( 4x + 2x + 1) - ( 4x + 2x +1)

=(2x-1)(4x+2x+1)

=(2x-1)(6x+1)

a) \(2x\left(2x-1\right)-\left(2x-1\right)=\left(2x-1\right)\left(2x-1\right)\)

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

Mik làm cho vui thôi chứ chẳng ai T mik đâu

Ta có; \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)

\(=\frac{-\left(2x-1\right)^2+2x\cdot2x-1}{2x\left(2x-1\right)}\)

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

\(=\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)

Ta có: \(\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\)

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

\(=\frac{x\cdot\sqrt2-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt2-\sqrt{x}-2x+1}{2x-1}\)

\(=\frac{-4x-2\sqrt{x}}{2x-1}=\frac{-2\sqrt{x}\left(2\sqrt{x}+1\right)}{2x-1}\)

Ta có: \(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\)

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

\(=\frac{2x-1+x\sqrt2-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt2-\sqrt{x}}{2x-1}=\frac{-2\sqrt{x}-2}{2x-1}\)

Ta có: \(\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)

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