c) \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=18\)

\(\Leftrightarrow2x-4x^2+1-2x+4x^2-4x+1=18\)

\(\Leftrightarrow-4x+2=18\Leftrightarrow-4x=16\Leftrightarrow x=-4\)

Vậy x=-4

d) \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2+3x-3x-9\right)-\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow2x^2+4x+2-x^2-3x+3x+9-x^2+8x-16=0\)

\(\Leftrightarrow12x-5=0\Leftrightarrow12x=5\Leftrightarrow x=\frac{5}{12}\)

e) \(\left(x-5\right)^2-x\left(x-4\right)=9\)

\(\Leftrightarrow x^2-10x+25-x^2+4x=9\)

\(\Leftrightarrow-6x+25=9\Leftrightarrow-6x=-16\Leftrightarrow x=\frac{8}{3}\)

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

\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)

\(\Leftrightarrow-5x+21=0\Leftrightarrow-5x=-21\Leftrightarrow x=\frac{21}{5}\)

Trả lời:

9 người

HT

Trả lời:

Gia đình đó có 9 người

Chúc bn học tốt

K mik nhoa

Trả lời:

1, 9y² - 12xy + 6y

= 3y ( 3y - 4x + 2 )

2, 25x² - y² + 5x - y

= ( 25x² - y² ) + ( 5x - y )

= ( 5x - y ) ( 5x + y ) + ( 5x - y )

= ( 5x - y ) ( 5x + y + 1 )

3, 18xy - 12xy² + 36x³y

= 6xy ( 3 - 2y + 6x² )

4, 16 + 8x + x² - z²  ???

5, 3x ( 2 - y ) - 4 + y²

= 3x ( 2 - y ) - ( 4 - y² )

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

= ( 2 - y ) ( 3x - 2 - y )

6, 9x² - y² + 4yz - 4z²

= 9x² - ( y² - 4yz + 4z² )

= ( 3x )² - ( y - 2z )²

= ( 3x - y + 2z ) ( 3x + y - 2z )

oki bẹn

đợi chút

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

\(\Leftrightarrow A=x^2+6x+9+x^2+3x-3x-9-2\left(x^2-4x+2x-8\right)\)

\(\Leftrightarrow A=x^2+6x+9+x^2+3x-3x-9-2x^2+8x-4x+16\)

\(\Leftrightarrow A=10x+16\)

Thay \(x=-\frac{1}{2}\) vào biểu thức ta có:

\(A=10.\frac{-1}{2}+16=11\)

Vậy...

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

\(\Leftrightarrow B=9x^2+24x+16-\left(x^2+4x-4x-16\right)-10x\)

\(\Leftrightarrow B=9x^2+24x+16-x^2-4x+4x+16-10x\)

\(\Leftrightarrow B=8x^2+14x+32\)

Thay \(x=-\frac{1}{10}\) vào biểu thức ta có:

\(8.\left(\frac{-1}{10}\right)^2+14.\frac{-1}{10}+32=\frac{767}{25}\)

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

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

\(\Leftrightarrow C=x^2+2x+1-4x^2+4x-1+3x^2+6x-6x-12\)

\(\Leftrightarrow C=6x-12\)

Thay x=1 vào biểu thức ta có:

\(6.1-12=-6\)

Vậy....

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

\(\Leftrightarrow D=x^2+3x-3x-9+x^2-4x+4-2x^2+8x\)

\(\Leftrightarrow D=4x-5\)

Thay x=-1 vào biểu thức ta có: 

4.(-1)-5=-9

Vậy....

anh dùng hằng đẳng thức đó là ra 

=A^+B^2+C^2+2AB+2BC+2AC

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

em nghĩ là = abc hì