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\(\left(x+2\right)^2+4\left(x+2\right)\left(x-2\right)+\left(x-4\right)^2\\ =x^2+4x+4+4x^2-16+x^2-8x+16\\ =6x^2-4x+4\)
(x + 2)2 + 4(x + 2)(x - 2) + (x - 4)2
<=> x2 + 4x + 4 + 4(x2 - 4) + x2 - 8x + 16
<=> x2 + 4x + 4 + 4x2 - 16 + x2 - 8x + 16
<=> x2 + 4x2 + x2 + 4x - 8x + 4 - 16 + 16
<=> 6x2 - 4x + 4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^3+2x^2+5x=0\)
\(\Leftrightarrow x\left(3x^2+2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x^2+2x+5=0\left(v\text{ô}nghi\text{ệm}\right)\end{cases}}\)
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\(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)+2ab-2ac\)
\(=a\left(a-2b+2c\right)+2ab-2ac\)
\(=a^2-2ab+2ac+2ab-2ac\)
\(=a^2\)
\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)
\(=\left(3x+1-3x-5\right)^2\)
\(=\left(-4\right)^2=16\)
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a,hđt số 3 = \(\left(a^2+2a\right)^2-9\)
b,hđt số 3=\(\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)(đổi dấu làm ngoặc khi trước nó là dấu trừ)=\(x^2-\left(y-6\right)^2\)
a) \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)
\(=\left(a^2+2a\right)^2+3.\left(-3\right)\)
\(=\left(a^2+2a\right)^2-9\)
b) \(\left(x-y+6\right)\left(x+y-6\right)\)
\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)
\(=x^2-\left(y-6\right)^2\)
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(a + b)2 – (a – b)2
= [(a + b) – (a – b)].[(a + b) + (a – b)]
(Áp dụng HĐT (3) với A = a + b; B = a – b)
= 2b.2a
= 4ab
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Bài 2:
a: Ta có: \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)
\(=\left(x+y\right)^3+2\cdot\left(x+y\right)^2\)
\(=7^3+2\cdot7^2=441\)
=a^2+2ab+b^2-(a^2-2ab+b^2)
=4ab
Học tốt
\(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left[a+b-\left(a-b\right)\right]\left[a+b+a-b\right]\)
\(=2b\cdot2a\)
\(=4ab\)