\(x^3-3x^2-6x+8=\left(x^3+8\right)-\left(3x^2+6x\right)\)

\(=\left(x+2\right)\left(x^2-2x+4\right)-3x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-2x+4-3x\right)=\left(x+2\right)\left(x^2-5x+4\right)\)

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

Bài chưa hỏi tính gì mà!

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\(A=\frac{5x^2-26x+41}{\left(x-2\right)^2}=\frac{4\left(x^2-4x+4\right)+\left(x^2-10x+25\right)}{\left(x-2\right)^2}=4+\frac{\left(x-5\right)^2}{\left(x-2\right)^2}\ge4\forall x\)

Dấu "=" xảy ra khi \(x-5=0\Rightarrow x=5\)

Vậy GTNN của A là 4 khi x = 5

a) \(\frac{5x^2-20x+20-6x+21}{\left(x-2\right)^2}=\frac{5\left(x^2-4x+4\right)-6\left(x-2\right)+9}{\left(x-2\right)^2}\)

=\(\frac{5\left(x-2\right)^2-6\left(x-2\right)+9}{\left(x-2\right)^2}=5-\frac{6}{\left(x-2\right)}+\frac{9}{\left(x-2\right)^2}=\left(\frac{3}{x-2}-1\right)^2+4\ge4\)

'=' xảy ra \(\Leftrightarrow\frac{3}{x-2}-1=0\Leftrightarrow x=5\)

Vậy ...

=123

=123

k nhé

ĐK: \(x\ne-4\)

\(A=\frac{x}{\left(x+4\right)^2}=\frac{16x}{16\left(x^2+8x+16\right)}=\frac{x^2+8x+16-x^2+8x-16}{16\left(x^2+8x+16\right)}=\frac{1}{16}-\frac{\left(x-4\right)^2}{16\left(x+4\right)^2}\le\frac{1}{16}\forall x\)

Dấu "=" xảy ra khi: \(x-4=0\Rightarrow x=4\) (thỏa mãn ĐKXĐ)

Vậy \(A_{max}=\frac{1}{16}\Leftrightarrow x=4\)

Có bạn ạ

Giải thích: C1, nhân chéo

C2 , Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=> a = bk

    c = dk

Có: ad = bc

<=> bk . d = b . dk

<=> bdk = bdk ( luôn đúng )

Vậy đó !

Có bằng

CM:

Có : a/b=c/d

=>ad/bd=bc/bd

=> ad=bc