\(\frac{x^2+6x+9}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4x^2+24x+36}\)

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

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

\(=\frac{2x^2-4x-2}{4x^2-8x+4}\)

\(\frac{x^2+6x+9}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4x^2+24x+36}\)

\(=\frac{x^2+2\left(x\right)\left(3\right)+3^2}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4x^2+24x+36}\)

\(=\frac{\left(x+3\right)^2}{\left(x-1\right)^2}.\frac{2x^2+4x-2}{4x^2+24x+36}\)

\(=\frac{\left(x+3\right)^2}{\left(x-1\right)^2}.\frac{2\left(x^2-2x-1\right)}{4x^2+24x+36}\)

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

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

\(=\frac{1.2\left(x^2-2x-1\right)}{\left(x-1\right)^2.4}\)

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

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

\(\left(ac+bd\right)^2=a^2c^2+2abcd+b^2d^2\)

\(\left(ad-bc\right)^2=a^2d^2-2abcd+b^2c^2\)

\(\Rightarrow\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

Mà \(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

Nên \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)(đpcm)

Ta có \(VP=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)(1)

\(VT=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)(2)

từ (1),(2)\(\Rightarrowđpcm\)

Mình nghĩ đề như này

\(\frac{7x+2}{5xy^3}.\frac{x^2y^3}{21x+6}\)

\(=\frac{7x+2}{5}.\frac{x}{3\left(7x+2\right)}\)

\(=\frac{x}{15}\)

Hì...sorry tui bấm thiếu.

Cảm ơn bạn

Trong 3 ngày liên tiếp tui sẽ k cho bạn

\(2=x+y\ge2\sqrt{xy}\)(cô - si)

\(\Rightarrow\sqrt{xy}\le1\Rightarrow xy\le1\)

Ta có \(S=x^2+y^2=\left(x+y\right)^2-2xy\)

\(=4-2xy\ge4-2=2\)

Dấu "=" khi x = y = 1

Ta có: \(\left(x-y\right)^2\ge0\)\(\Leftrightarrow x^2-2xy+y^2\ge0\)

\(\Leftrightarrow x^2+y^2\ge2xy\)\(\Leftrightarrow2\left(x^2+y^2\right)\ge x^2+y^2+2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)\(\Leftrightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

Thay \(x+y=2\)vào bất phương trình ta được:\(x^2+y^2\ge\frac{2^2}{2}=\frac{4}{2}=2\)

Dấu " = " xảy ra \(\Leftrightarrow x-y=0\)\(\Leftrightarrow x=y\)

mà \(x+y=2\)\(\Rightarrow x=y=1\)

Vậy \(minS=2\)\(\Leftrightarrow x=y=1\)

Ta có \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(=\left(3^{64}-1\right)\left(3^{64}+1\right)=\left(3^{128}-1\right)\)

\(\Rightarrow A=\frac{3^{128}-1}{2}\)

Ta có 

\(A=x^2+2\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)

\(=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\)

\(=10x^2+40x+50\)

\(=x^2+10x+25+9x^2+30x+25\)

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

\(ĐKXĐ:x\ne0;x\ne-3;x\ne-6;x\ne-9\)

\(\frac{1}{x^2+3x}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}=\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+9}=\frac{9}{10}\)

\(\Leftrightarrow\frac{9}{x\left(x+9\right)}=\frac{9}{10}\)

\(\Leftrightarrow x^2+9x-10=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-10\end{cases}\left(tm\right)}\)

O.O chắc ko 

\(\frac{1}{x^2+3x}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{3}{10}\)\(ĐKXĐ:x\ne-3;-6\)

\(\frac{1}{x\left(x+3\right)}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{3}{10}\)

\(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{x^2+15x+54}=\frac{3}{10}\)

\(10\left(x+6\right)\left(x+9\right)+10x\left(x+9\right)+10x\left(x+3\right)=3x\left(x+3\right)\left(x+6\right)\left(x+9\right)\)

\(30x^2+270x+540=3x^4+54x^3+297x^2+486x\)

\(30x^2+270x+540-3x^4-54x^3-297x^2-486x=0\)

\(-3\left(89x^2+72x-180+x^4+18x^3\right)=0\)

\(-3\left(x^2+16x+60\right)\left(x-1\right)=0\)

\(-3\left(x+6\right)\left(x+10\right)\left(x+3\right)\left(x-1\right)=0\)

\(\left(x+6\right)\left(x+10\right)\left(x+3\right)\left(x-1\right)=0\)

\(x=-10,1\)

a, CM: AD//AB=AE//AC

Xét tam giác ABC có:

AD//AB vì đề bài cho cạnh BC lấy D ( lấy sao cho AD=AB)

AE//AC vì đề bài cho cạnh AC lấy E  ( lấy sao cho AE=AC)

VÌ ĐỀU CHUNG MỘT TAM GIÁC NÊN 3 CẠNH = NHAU 

\(\Rightarrow\) AD/AB=AE/AC.

b, AB = 2cm vì AD= 2cm( AD//AB \(\Rightarrow=\)nhau và = 2 cm)

đêm hôm khuya khoắt đăng lên lm j :v 

\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left(a+b+c-a\right)\left(3x^2+b^2+c^2+3ab+2bc+3ac\right)-\left(b^3+c^3\right)\)

\(=\left(3a^2+b^2+c^2+3ab+2ac-b^2+bc-c^2\right)\left(b+c\right)\)

\(=\left(3a^2+3ab+3ac+3bc\right)\left(b+c\right)\)

\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

\(\text{ (a+b+c)^3−a^3−b^3−c^3 =a^3+3a^2(b+c)+3a(b+c)^2+(b+c)^2−a^3−b^3−c^3 =3(b+c)(a^2+ab+ac)+b^3+3b^2c+3bc^2+c^3−b^3−c^3 =3(b+c)(a^2+ab+ac+bc) =3(b+c)[a(a+b)+c(a+b)] =3(b+c)(a+b)(a+c)}\)