a) \(2.\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2.\left(x+5\right)-x.\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Vậy \(S=\left\{-5,2\right\}\)

b) \(x^3-5x^2-4x+20=0\)

\(\Leftrightarrow x^2\left(x-5\right)-4.\left(x-5\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x^2-4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=\pm2\end{cases}}\)

Vậy \(S=\left\{5,\pm2\right\}\)

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

\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\3x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=-\frac{3}{2}\end{cases}}\)

Vậy \(S=\left\{4,-\frac{3}{2}\right\}\)

a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)

b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)

c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)

d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)

e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\frac{4}{9}x^2-\frac{9}{4}y^2\)

a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)

b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)

c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)

d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)

e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\left(\frac{2}{3}x\right)^2-\left(\frac{3}{2}y\right)^2=\frac{4}{9}x^2-\frac{9}{4}y^2\)

Bài làm:

Ta có: \(\left(2x+5\right)\left(x-2\right)-3\left(x+2\right)^2+\left(x+1\right)^2\)

\(=2x^2+x-10-3\left(x^2+4x+4\right)+x^2+2x+1\)

\(=3x^2+3x-9-3x^2-12x-12\)

\(=-9x-21\)

\(\left(2x+5\right)\left(x-2\right)-3\left(x+2\right)^2+\left(x+1\right)^2\)

\(=2x\left(x-2\right)+5\left(x-2\right)-3\left(x^2+4x+4\right)+\left(x^2+2x+1\right)\)

\(=2x^2-4x+5x-10-3x^2-12x-12+x^2+2x+1\)

\(=\left(2x^2-3x^2+x^2\right)+\left(5x-4x-12x+2x\right)-\left(10+12-1\right)\)

\(=0+x-10x-21=-9x-21\)

a) \(A=4x^2+7x+13=4\left(x^2+\frac{7}{4}x+\frac{49}{64}\right)+\frac{159}{16}\)

\(=4\left(x+\frac{7}{8}\right)^2+\frac{159}{16}\ge\frac{159}{16}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(4\left(x+\frac{7}{8}\right)^2=0\Rightarrow x=-\frac{7}{8}\)

Vậy \(A_{Min}=\frac{159}{16}\Leftrightarrow x=-\frac{7}{8}\)

b) \(B=5-8x+x^2=\left(x^2-8x+16\right)-11\)

\(=\left(x-4\right)^2-11\ge-11\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-4\right)^2=0\Rightarrow x=4\)

Vậy \(B_{Min}=-11\Leftrightarrow x=4\)

a. \(2x\left(x-5\right)-\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)\)

\(=2x^2-10x-x^2+4x-4-x^2+9\)

\(=-6x+5\)

b. \(\left(x+1\right)^2+3\left(x-5\right)\left(x+5\right)-\left(2x-1\right)^2\)

\(=x^2+2x+1+3x^2-75-4x^2+4x-1\)

\(=6x-75\)

c. \(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)

\(=2x^2-14x-x^2-x+6-x^2+16\)

\(=-15x+22\)

d. \(\left(x+3\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)

\(=x^2-9-x^2-4x+5-x^2+8x-16\)

\(=-x^2+4x-20\)

Bài làm:

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

\(=2x^2-10x-x^2+4x-4-x^2+9\)

\(=-6x+5\)

b) \(\left(x+1\right)^2+3\left(x-5\right)\left(x+5\right)-\left(2x-1\right)^2\)

\(=x^2+2x+1+3x^2-75-4x^2+4x-1\)

\(=6x-75\)

c) \(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)

\(=2x^2-14x-x^2-x+6-x^2+16\)

\(=-15x+22\)

d) \(\left(x+3\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)

\(=x^2-9-x^2-4x+5-x^2+8x-16\)

\(=-x^2-4x-20\)

lm trên symbolab.com

đặt \(A=x^2+y^2+2x\left(y-1\right)+2y=x^2+y^2+2xy-2x+2y=\left(x+y\right)^2-2\left(x-y\right)\)

do A là số chính phương => \(\left(x+y\right)^2-2\left(x+y\right)\)cũng là số chính phương

\(\Leftrightarrow-2\left(x-y\right)=0\)

\(\Leftrightarrow x=y\)