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Mấy bài này khá đơn giản .
Bạn chỉ cần áp dụng hằng đẳng thức \(x^2-y^2=\left(x+y\right)\left(x-y\right)\) là được nhs =))
a)
\(\left(xy+4\right)^2-4\left(x+y\right)^2\)
\(=\left(xy+4\right)^2-\left[2\left(x+y\right)\right]^2\)
\(=\left[xy+4-2\left(x+y\right)\right]\left[xy+4+2\left(x+y\right)\right]\)
\(=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)
\(=\left[y\left(x-2\right)-2\left(x-2\right)\right]\left[y\left(x+2\right)+2\left(x+2\right)\right]\)
\(=\left(x-2\right)\left(y-2\right)\left(x+2\right)\left(y+2\right)\)
b)
\(\left(ab-xy\right)^2-\left(bx-ay\right)^2\)
\(=\left(ab-xy-bx+ay\right)\left(ab-xy+bx-ay\right)\)
\(=\left[a\left(b+y\right)-x\left(b+y\right)\right]\left[a\left(b-y\right)+x\left(b-y\right)\right]\)
\(=\left(b+y\right)\left(a-x\right)\left(a+x\right)\left(b-y\right)\)
c)
\(=\left(x^2+8x-34+3x^2-8x-2\right)\left(x^2+8x-34-3x^2+8x+2\right)\)
\(=\left(4x^2-36\right)\left(-2x^2+16x-32\right)\)
\(=\left(2x-6\right)\left(2x+6\right)\left(-2\right)\left(x^2-8x+16\right)\)
\(=\left(2x-6\right)\left(2x+6\right)\left(-2\right)\left(x-4\right)^2\)
Bạn liểm tra lại nhs
Mk lm hay nhấm lắm
=))
ảnh giỏi lắm đừng coi thường à. olm tới 11000 điểm mà chỉ 1 tuần dc 1000 đỉm á
a: \(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
b: \(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)
\(=x^2\cdot\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)
\(=x^2\left(x-1\right)\left(x^3+x-9\right)\)
d: \(=\left(xy+4\right)^2-\left(2x+2y\right)^2\)
\(=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)
\(=\left[x\left(y-2\right)-2\left(y-2\right)\right]\left[x\left(y+2\right)+2\left(y+2\right)\right]\)
\(=\left(y-2\right)\left(x-2\right)\left(y+2\right)\left(x+2\right)\)
e: \(=\left(ab-xy-bx+ay\right)\left(ab-xy+bx-ay\right)\)
\(=\left[a\left(b+y\right)-x\left(b+y\right)\right]\left[b\left(a+x\right)-y\left(a+x\right)\right]\)
\(=\left(b+y\right)\left(a-x\right)\left(a+x\right)\left(b-y\right)\)
a ) \(x^3+3x^2-3x+1\)
\(=x^3-3x+3x^2-1\)
\(=\left(x-1\right)^3\)
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
Lời giải:
a.
$(xy)^2-xy-2=(x^2y^2+xy)-(2xy+2)$
$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$
b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?
bạn ra 2 câu 1 bài chứ nhìu thế ko ai giải đâu
a = x( x3 - 4x2 + 8x - 16)
c, (ab - xy)2 - (bx - ay)2
dựa theo hằng đẳng thức a2 - b2 = ( a + b)(a - b)
=> = [ ( ab - xy) + ( bx - ay)] . [ ( ab - xy) - ( bx - ay) ]
tương tự làm câu d theo câu c