\(a,A=\left(2x-5\right)^2-\left(2x+5\right)^2+40x-1\)

\(=\left(2x-5-2x-5\right)\left(2x-5+2x+5\right)+40x-1\)

\(=-10.4x^2+40x-1\)

\(=-40x^2+40x-1=-1\)

\(b,B=\left(3x-2y\right)^2+\left(3x+2y\right)^2-18x-8y^2+1\)

\(=9x^2-12xy+4y^2+9x^2+12xy+4y^2-18x-8y^2+1\)

\(=18x^2-18x+1\)

\(c,C=\left(2+x\right)^2-\left(2-x\right)^2-8x+3\)

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

\(=2x.4-8x+3=3\)

\(\left(2x-5\right)^2-\left(2x+5\right)^2+40x-1=\left(2x-5-2x-5\right)\left(2x-5+2x+5\right)+40x-1=-40x+40x-1=-1\)

a/ \(=\left(9x^2+30x+25\right)+\left(x^2+10x+25\right)=\)

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

b/ \(=\left(16x^2+8x+1\right)+\left(y^2-4y+4\right)=\left(4x+1\right)^2+\left(y-2\right)^2\)

c/ 

1)6x-8=3x+1

6x-3x=1+8

3x=9

x=3

Vậy x=3

2: 12-10x=25-30x

=>20x=13

=>x=13/20

3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)

=>6x+9-8x+10=10x+21

=>10x+21=-2x+19

=>12x=-2

=>x=-1/6

4: \(\Leftrightarrow25x-15-6x+12=11-5x\)

=>19x-3=11-5x

=>24x=14

=>x=7/12

5: \(\Leftrightarrow8-12x-5+10x=4-6x\)

=>4-6x=-2x+3

=>-4x=-1

=>x=1/4

6: \(\Leftrightarrow32x-24-6+9x=13-40x\)

=>41x-30=13-40x

=>81x=43

=>x=43/81

7: \(\Leftrightarrow10x-5+20x=5x-11\)

=>30x-5=5x-11

=>25x=-6

=>x=-6/25

Bài 2:

a: \(\dfrac{1}{2x^3y}=\dfrac{6yz^3}{12x^3y^2z^3}\)

\(\dfrac{2}{3xy^2z^3}=\dfrac{2\cdot4x^2}{12x^3y^2z^3}=\dfrac{8x^2}{12x^3y^2z^3}\)

a) ĐKXĐ : \(x\ge5\)

Đặt \(\sqrt{x-5}=a;\sqrt[3]{3-x}=b\)(a \(\ge0\))

Khi đó phương trình thành a + b = 2

Lại có \(b^3+a^2=-2\)

=> HPT : \(\hept{\begin{cases}a+b=2\\b^3+a^2=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b^3+\left(2-b\right)^2=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b^3+b^2-4b+6=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=2-b\\\left(b+3\right)\left(b^2-2b+2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}a=5\\b=-3\end{cases}}\)(tm)

a = 5 => x = 30 (tm) 

Vậy x = 30 là nghiệm phương trình 

d) Ta có \(\sqrt{25x^2-20x+4}+\sqrt{25x^2-40x+16}=0\)

<=> \(\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x-4\right)^2}=2\)

<=> |5x - 2| + |5x - 4| = 2

Lại có |5x - 2| + |5x - 4| = |5x - 2| + |4 - 5x| \(\ge\left|5x-2+4-5x\right|=2\)

Dấu "=" xảy ra <=> \(\left(5x-2\right)\left(4-5x\right)\ge0\Leftrightarrow\frac{2}{5}\le x\le\frac{4}{5}\)

Vậy \(\frac{2}{5}\le x\le\frac{4}{5}\)là nghiệm phương trình 

c) ĐKXĐ : \(\frac{3}{2}\le x\le\frac{5}{2}\)

Phương trình tương đương \(\left(\sqrt{2x-3}+\sqrt{7-x}\right)^2=\left(\sqrt{5-2x}+\sqrt{3x-1}\right)^2\)

\(\Leftrightarrow x+4+2\sqrt{\left(2x-3\right)\left(7-x\right)}=x+4+2\sqrt{\left(5-2x\right)\left(3x-1\right)}\)

\(\Leftrightarrow\sqrt{\left(2x-3\right)\left(7-x\right)}=\sqrt{\left(5-2x\right)\left(3x-1\right)}\)

<=> (2x - 3)(7 - x) = (5 - 2x)(3x - 1) 

<=> -2x² + 17x - 21 = -6x² + 17x - 5 

<=> 4x² = 16

<=> x² = 4

<=> \(\orbr{\begin{cases}x=2\\x=-2\left(\text{loại}\right)\end{cases}}\Leftrightarrow x=2\) 

Vậy x = 2 là nghiệm phương trình 

\(A=x^2-20x+101=\left(x-10\right)^2+1\ge1\)

\(minA=1\Leftrightarrow x=10\)

\(B=2x^2+40x-1=2\left(x+10\right)^2-201\ge-201\)

\(minB=-201\Leftrightarrow x=-10\)

\(C=x^2-4xy+5y^2-2y+28=\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+27=\left(x-2y\right)^2+\left(y-1\right)^2+27\ge27\)

\(minC=27\Leftrightarrow\)\(\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

\(D=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x+10\right)=\left(x^2-7x\right)^2-100\ge-100\)

\(minD=100\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

a: Ta có: \(A=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=10

b: ta có: \(B=2x^2+40x-1\)

\(=2\left(x^2+20x-\dfrac{1}{2}\right)\)

\(=2\left(x^2+20x+100-\dfrac{201}{2}\right)\)

\(=2\left(x+10\right)^2-201\ge-201\forall x\)

Dấu '=' xảy ra khi x=-10