câu 3. a) chứng minh IK =\(\frac{CD-AB}{2}\)

giải trên symbolab.com í

chú ý: công thức bậc hai \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) thường dùng trong một số nước

nước việt nam ta dùng là: \(x=\frac{-b\pm\sqrt{\Delta}}{2a}\)

2x( x - 1 ) - x( 1 - x )² - ( 1 - x )³

= 2x( x - 1 ) - x( x - 1 )² + ( x - 1 )³

= ( x - 1 )[ 2x - x( x - 1 ) + ( x - 1 )² ]

= ( x - 1 )( 2x - x² + x + x² - 2x + 1 )

= ( x - 1 )( x + 1 )

Ta có: \(2x\left(x-1\right)-x\left(1-x\right)^2-\left(1-x\right)^3\)

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

\(=\left(x-1\right)\left(x+1\right)\)

A B C M N H

\(\Delta ABC\)vuông cân tại A \(\Rightarrow AB=AC=4\left(cm\right)\)

Áp dụng định lý Pytago ta có:

\(AB^2+AC^2=BC^2\)\(\Rightarrow BC^2=2AB^2=2.4^2=32\)

\(\Rightarrow BC=\sqrt{32}=4\sqrt{2}\)( cm )

Ta có: \(S=\frac{1}{2}.AB.AC=\frac{1}{2}.AH.BC\)

\(\Rightarrow AB.AC=AH.BC\)\(\Rightarrow AH=\frac{AB.AC}{BC}=\frac{4.4}{4\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\)( cm )

Vì \(\widehat{AMH}=\widehat{MAN}=\widehat{ANH}=90^o\)

\(\Rightarrow\)Tứ giác AMHN là hình chữ nhật

\(\Rightarrow MN=AH=2\sqrt{2}\)( cm )

\(\frac{x+3}{x^2+x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)+1\)

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)

\(=\frac{x+3}{x^2+x+6}\div\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)+1\)

\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)+1\)

\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)+1\)

\(=\frac{x+3}{x^2+x+6}\div\left(\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\right)+1\)

\(=\frac{x+3}{x^2+x+6}\div\frac{6}{\left(x-2\right)\left(x+2\right)}+1\)

\(=\frac{\left(x+3\right)\left(x-2\right)\left(x+2\right)}{6\left(x^2+x+6\right)}+1\)

\(=\frac{x^3+3x^2-4x-12}{6x^2+6x+36}+\frac{6x^2+6x+36}{6x^2+6x+36}\)

\(=\frac{x^3+3x^2-4x-12+6x^2+6x+36}{6x^2+3x+36}\)

\(=\frac{x^3+9x^2+2x+24}{6x^2+3x+36}\)

ơ lag hai dòng cuối :v

bạn sửa mẫu thức ở hai dòng cuối thành 6x² + 6x + 36 dùm mình nhé -.- dạo này lú quá

(8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0

=> 8(x + 2) - 5x(x + 2) + 4[x(x + 1) - 2(x + 1)] + 2(x² - 4) = 0

=> 8x + 16 - 5x² - 10x + 4(x² + x - 2x - 2) + 2x² - 8 = 0

=> 8x + 16 - 5x² - 10x + 4x² + 4x - 8x - 8 + 2x² - 8 = 0

=> (8x - 10x + 4x - 8x) + (16 - 8 - 8) + (-5x² + 4x² + 2x²)  = 0

=> 0 + x² = 0

=> x² = 0 => x = 0

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(-5x^2-2x+16+4\left(x^2-x-2\right)+2\left(x^2-4\right)=0\)

\(-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)

\(x^2-6x=0\)

\(x\left(x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

3425627829^{44525266272177437674677263632625654354}=\(\infty\)

(x - 1)³ - (x + 1)³ + 6(x + 1)(x - 1)

= x³ - 3x² + 3x - 1 - (x³ + 3x² + 3x + 1) + 6(x²  - 1)

= x³ - 3x² + 3x - 1 - x³ - 3x² - 3x  - 1 +6x² - 6

= (x³ - x³) + (-3x² - 3x² + 6x²) + (3x - 3x) + (-1 - 1 - 6) = -8

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

\(=x^3-3x^2+3x-1-\left(x^3+3x^2+3x+1\right)+6.\left(x^2-1\right)\)

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

\(=-8\)