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a) 11x + 11y + x2 + xy
= 11.(x+y) + x.(x+y)
= (x+y).(11+x)
b) 255 + x2 - 4xy + y2
= 255 + 2xy + x2 -2xy + y2
= 255 + 2xy + (x-y)2
...
a, \(11x+11y+x^2+xy=\left(11x+11y\right)+\left(x^2+xy\right)=11\left(x+y\right)+x\left(x+y\right)=\left(x+y\right)\left(x+11\right)\)
b. \(255-4x^2-4xy-y^2=255-\left(4x^2+4xy+y^2\right)=255-\left(2x+y\right)^2=\left(15+2x+y\right)\left(15-2x-y\right)\)
Bài 2:
\(x^2-y^2-4x+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)
\(=\left(72-2\right)\left(102-2\right)=70.100=7000\) ( x+y=102, x-y=72 )
Bài 1:
\(a,=11\left(x+y\right)+x\left(x+y\right)=\left(x+11\right)\left(x+y\right)\\ b,=225-\left(2x+y\right)^2=\left(15-2x-y\right)\left(15+2x+y\right)\)
Bài 2:
\(A=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ A=\left(72-2\right)\left(120-2\right)=70\cdot118=8260\)
Bài 3:
\(a,\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\\ \Leftrightarrow24x+25=49\\ \Leftrightarrow24x=24\Leftrightarrow x=1\)
A/\(4x^2-12+9\)
\(=\left(2x\right)^2-2.2.3+3^2\)
\(=\left(2x+3\right)^2\)
B/\(11x+11y-x^2-xy\)
\(=\left(11x-x^2\right)+\left(11y-xy\right)\)
\(=x\left(11-x\right)+y\left(11-x\right)\)
\(=\left(11-x\right)\left(x+y\right)\)
C/\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.
a) \(x-xy+y-y^2=x\left(1-y\right)+y\left(1-y\right)=\left(x+y\right)\left(1-y\right)\)
b) \(x^2-2x-y^2+1=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
c) \(4x^2-4xy+y^2=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)
d) \(9x^3-9x^2y-4x+4y=9x^2\left(x-y\right)-4\left(x-y\right)=\left(9x^2-4\right)\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\)
e) \(x^3+2+3\left(x^3-2\right)=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)
Câu 1: (2x+y)(y-2x)+4x2=y2-4x2+4x2=y2
Với y=10 giá trị biểu thức trên là 102=100
Câu 2:
a. xy-11x=x.(y-11)
b. x2+4y2+4xy-16=(x2+4xy+4y2)-16
=(x+2y)2-16=(x+2y+4)(x+2y-4)
a) Ta có: \(11x+11y+x^2+xy\)
\(=11\left(x+y\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(11+x\right)\)
b) Ta có: \(225-4x^2-4xy-y^2\)
\(=225-\left(4x^2+4xy+y^2\right)\)
\(=15^2-\left(2x+y\right)^2\)
\(=\left(15-2x-y\right)\left(15+2x+y\right)\)
Câu a không sai đề đâu bạn